Simple way of explaining the empty product [duplicate]

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  • Numbers to the Power of Zero

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I understand that $$3^2=9 text because 3times3=9$$



But is it possible to explain in same simple terms how $3^0=1$?










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    It is a mathematical operation: we define it so in order to be consistent with the "usual" properties of raising to power: in "real life" there is no "3 raised to the power of 0".
    – Mauro ALLEGRANZA
    Oct 4 at 14:30






  • 23




    3^0 =3^(1-1)=3/3=1
    – Avinash N
    Oct 4 at 17:46










  • k^x = 1 * k * ...* k (x times) +> when x is 0 only remains 1 ==> k^0 = 1
    – asmgx
    Oct 5 at 2:23






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    Also of: math.stackexchange.com/questions/454670/…
    – Holo
    7 hours ago






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    See also Why is $ large 2^0 = 1 $?
    – Bill Dubuque
    7 hours ago















up vote
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  • Numbers to the Power of Zero

    8 answers



I understand that $$3^2=9 text because 3times3=9$$



But is it possible to explain in same simple terms how $3^0=1$?










share|cite|improve this question















marked as duplicate by Xander Henderson, Holo, Rushabh Mehta, Alex Francisco, Lord Shark the Unknown 3 hours ago


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  • 2




    It is a mathematical operation: we define it so in order to be consistent with the "usual" properties of raising to power: in "real life" there is no "3 raised to the power of 0".
    – Mauro ALLEGRANZA
    Oct 4 at 14:30






  • 23




    3^0 =3^(1-1)=3/3=1
    – Avinash N
    Oct 4 at 17:46










  • k^x = 1 * k * ...* k (x times) +> when x is 0 only remains 1 ==> k^0 = 1
    – asmgx
    Oct 5 at 2:23






  • 1




    Also of: math.stackexchange.com/questions/454670/…
    – Holo
    7 hours ago






  • 1




    See also Why is $ large 2^0 = 1 $?
    – Bill Dubuque
    7 hours ago













up vote
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up vote
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down vote

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4






This question already has an answer here:



  • Numbers to the Power of Zero

    8 answers



I understand that $$3^2=9 text because 3times3=9$$



But is it possible to explain in same simple terms how $3^0=1$?










share|cite|improve this question
















This question already has an answer here:



  • Numbers to the Power of Zero

    8 answers



I understand that $$3^2=9 text because 3times3=9$$



But is it possible to explain in same simple terms how $3^0=1$?





This question already has an answer here:



  • Numbers to the Power of Zero

    8 answers







exponential-function arithmetic






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edited Oct 5 at 6:30









Periodic Sqare well

1427




1427










asked Oct 4 at 13:27









brilliant

28329




28329




marked as duplicate by Xander Henderson, Holo, Rushabh Mehta, Alex Francisco, Lord Shark the Unknown 3 hours ago


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marked as duplicate by Xander Henderson, Holo, Rushabh Mehta, Alex Francisco, Lord Shark the Unknown 3 hours ago


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  • 2




    It is a mathematical operation: we define it so in order to be consistent with the "usual" properties of raising to power: in "real life" there is no "3 raised to the power of 0".
    – Mauro ALLEGRANZA
    Oct 4 at 14:30






  • 23




    3^0 =3^(1-1)=3/3=1
    – Avinash N
    Oct 4 at 17:46










  • k^x = 1 * k * ...* k (x times) +> when x is 0 only remains 1 ==> k^0 = 1
    – asmgx
    Oct 5 at 2:23






  • 1




    Also of: math.stackexchange.com/questions/454670/…
    – Holo
    7 hours ago






  • 1




    See also Why is $ large 2^0 = 1 $?
    – Bill Dubuque
    7 hours ago













  • 2




    It is a mathematical operation: we define it so in order to be consistent with the "usual" properties of raising to power: in "real life" there is no "3 raised to the power of 0".
    – Mauro ALLEGRANZA
    Oct 4 at 14:30






  • 23




    3^0 =3^(1-1)=3/3=1
    – Avinash N
    Oct 4 at 17:46










  • k^x = 1 * k * ...* k (x times) +> when x is 0 only remains 1 ==> k^0 = 1
    – asmgx
    Oct 5 at 2:23






  • 1




    Also of: math.stackexchange.com/questions/454670/…
    – Holo
    7 hours ago






  • 1




    See also Why is $ large 2^0 = 1 $?
    – Bill Dubuque
    7 hours ago








2




2




It is a mathematical operation: we define it so in order to be consistent with the "usual" properties of raising to power: in "real life" there is no "3 raised to the power of 0".
– Mauro ALLEGRANZA
Oct 4 at 14:30




It is a mathematical operation: we define it so in order to be consistent with the "usual" properties of raising to power: in "real life" there is no "3 raised to the power of 0".
– Mauro ALLEGRANZA
Oct 4 at 14:30




23




23




3^0 =3^(1-1)=3/3=1
– Avinash N
Oct 4 at 17:46




3^0 =3^(1-1)=3/3=1
– Avinash N
Oct 4 at 17:46












k^x = 1 * k * ...* k (x times) +> when x is 0 only remains 1 ==> k^0 = 1
– asmgx
Oct 5 at 2:23




k^x = 1 * k * ...* k (x times) +> when x is 0 only remains 1 ==> k^0 = 1
– asmgx
Oct 5 at 2:23




1




1




Also of: math.stackexchange.com/questions/454670/…
– Holo
7 hours ago




Also of: math.stackexchange.com/questions/454670/…
– Holo
7 hours ago




1




1




See also Why is $ large 2^0 = 1 $?
– Bill Dubuque
7 hours ago





See also Why is $ large 2^0 = 1 $?
– Bill Dubuque
7 hours ago











15 Answers
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$x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.



We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.



Using this definition, we can notice the following interesting property:




For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$




Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.






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  • 5




    If we redefine x<sup>k</sup> to be the multiplicative identity (1) multiplied by x, k times, then when k=0, we don't multiply at all. Many people use sloppy language and say x<sup>k</sup> is "x multiplied by itself k times", but that would make 3<sup>2</sup>=3x3x3 (3 multiplied by 3 twice) rather than 3x3.
    – Monty Harder
    Oct 4 at 15:38






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    Important: $x$ must not be zero
    – Barranka
    Oct 5 at 1:04






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    @HRSE I agree with $0^0=1(=exp(0))$ but not on base of what you mention. The middle expression is not defined because $ln(0)$ is not defined. IMV $0^0$ is an empty product (just like $3^0$) hence is a neutral element in multiplication.
    – Vera
    2 days ago






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    For a slightly simpler version, I'd go with "for any pair of positive integers $k_1$ and $k_2$, $x^k_1 cdot x^k_2 = x^k_1+k_2$, so if we extend it to $0$, $x^k = x^k+0 = x^k cdot x^0$ forces $1 = x^0$".
    – Pablo H
    2 days ago







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    This is not the place to "discuss" $0^0$. math.stackexchange.com/questions/11150/…
    – user202729
    2 days ago

















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$$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.






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  • 1




    “implying” is a bit strong here, else what about nonsense like “$-1 = (-1)^tfrac12times(-1)^tfrac12 =$ product of two square roots, square roots are always positive $Longrightarrow$ $-1$ is positive”?
    – leftaroundabout
    Oct 4 at 23:42







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    It's nonsense because $x^1/2$ is not the same thing as $sqrtx$ in the first place, and then because the square root of -1 is $i$, which doesn't follow the ordering of negative/positive at all.
    – Nij
    Oct 4 at 23:56






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    How is $x^1/2$ not the same as $sqrtx$?
    – GraphicsMuncher
    2 days ago










  • @leftaroundabout That's nonsense, but not because of Vera's property. It's nonsense because the middle claim, namely "square roots are always positive", is simply false; in the reals because square roots don't always exist and in the complex plane because non-reals aren't positive and are frequently square roots.
    – Daniel Wagner
    yesterday






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    @DanielWagner my example wasn't particularly well thought through, but the point was that you can't, in general, just toss around calculation rules outside of the domain they were originally defined in, and then use the “results” to prove that such and such previously undefined expression must have a certain value. You can use that as the motivation for extending the definition, but the proof that it's actually consistent needs to be done separately.
    – leftaroundabout
    yesterday

















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5xum answer is the correct one, but I'll try to give you a more intuitive way to get $x^0 = 1$



We have
$$x^k = underbracextimes x cdots x_ktext times=1 times underbrace xtimes x cdots x_ktext times$$



Therefore
$$x^0 = 1 times underbrace xtimes x cdots x_0text times = 1 $$






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  • 2




    This is actually a superior way to define exponentation. Start with the multiplicative identity (1) and multiply by x, k times. If k is zero, there are no multiplications by x.
    – Monty Harder
    Oct 4 at 15:34






  • 6




    I would not call this more intuitive. In fact, it is not intuitive at all that you want to multiply by 1 in the definition. You only realize this is useful after you've gone through the reasoning summarized in 5xum's post.
    – Paul Sinclair
    Oct 4 at 16:30










  • In my intermediate algebra classes I define $x^0 = 1$, and for natural $n$, $x^n = 1 cdot x cdot x...$ [$n$ times], and $x^-n = 1 div x div x...$ [$n$ times]. So the symmetry of each one starting with $1$ seems pretty intuitive. Also there are actually $n$ operations (see elsewhere comment by @MontyHarder).
    – Daniel R. Collins
    Oct 4 at 18:08







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    Couldn't you just as easily say that $x^k = x times x cdots x = 0 + x times x cdots x$, so $x^0 = 0$?
    – Arcanist Lupus
    Oct 5 at 1:00







  • 2




    @ArcanistLupus Well, never thought about it this way. I think that it doesn't really make sense to use the additive identity (0+a=a) for exponentation, since the usual formula is all about multiplicating. But you're right that it's a flaw in my reasoning, thanks for pointing it out!
    – F.Carette
    2 days ago

















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The value of the empty product really is a consequence of the associative property of multiplication: To see this, consider the following: For any nonempty, finite set $A$ of numbers, let's define $$displaystyle prod_a in A a$$ as the product of all numbers in $A$. Note that $A$ really needs to be nonempty (at least for now), otherwise this definition doesn't make any sense. Then, if we have another disjoint set $B$, the product of all numbers in $Acup B$ will be $$displaystyle prod_x in A cup B x= left( displaystyle prod_a in A a right) left( displaystyle prod_b in B b right).$$ For example let's take the two sets $M:=x,y$ and $N:=z$. Then $Mcup N = x,y,z$, and by the associative law $$displaystyle prod_p in M cup Np = x*y*z=(x*y)*(z)= left( displaystyle prod_m in M m right) left( displaystyle prod_n in N n right).$$ Now, let's try to extend this notation to the empty set $$. What should $displaystyle prod_x in x$ be? If we want to keep the rule $$displaystyle prod_x in A cup B x= left( displaystyle prod_a in A a right) left( displaystyle prod_b in B b right),$$ then there is only one natural choice: Since for any set $M$ we have $M = M cup $, it follows that $$displaystyle prod_m in M m = displaystyle prod_x in M cup x = left( displaystyle prod_m in M m right) left( displaystyle prod_x in x right).$$ For this to hold the empty product $displaystyle prod_x in x$ has to be equal to the multiplicative identity, $1$. By the same argument, the empty sum is equal to the additive identity, $0$.






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  • 4




    This product notation probably works nicer if you use multisets instead of sets. Then there are no worries about disjointness.
    – Joonas Ilmavirta
    2 days ago






  • 1




    ...or disjoint union, $sqcup$.
    – AccidentalFourierTransform
    2 days ago

















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To increase power by one, multiply by 3, for example $3^2times3=3^3$. To decrease power by one, divide by 3, for example $3^2div3=3^1$. If you repeat this, you get $3^1div3=3^0$.



You can see it this way:



$$3^3=3times3times3$$



$$3^2=3times3$$



$$3^1=3$$



$$3^0=3div3$$



$$3^-1=3div3div3$$






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    +1; I've successfully explained this to ten year olds by making such a diagram. They didn't have any trouble with it. Good simple answer.
    – Wildcard
    2 days ago

















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Another way to look at the quantity $a^b$, with $a$ and $b$ non-negative integers, is as the number of mappings (functions) that exist from a set with $b$ elements (henceforth $B$) to a set with $a$ elements (henceforth $A$).



Recall further that a mapping from a set $X$ to a set $Y$ is a set $M$ of pairs $(x,y)$ where



  • for all $(x,y) in M$, $x in X$ and $y in y$; and

  • for all $x in X$, there is exactly one $y in Y$ such that $(x,y) in M$.

Now, if $B$ has zero elements (i.e., if it is the empty set $emptyset$), it is clear that there exists exactly one $M$ that satisfies the two conditions above, namely $M = emptyset$.






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    Not the simplest answer, but a way to think about this and related questions about combining a bunch of things.



    If you have a list of numbers and you want to write a computer program to sum them you do something like



    sum = 0
    while there are unsummed numbers in the list
    sum = sum + next number


    Then the sum variable contains the answer. If the list is empty the while loop never does anything and the sum is $0$.



    For multiplication you want



    product = 1
    while there are unmultiplied numbers in the list
    product = product * next number


    That clearly gives the correct answer when the list is not empty and tells you what the answer should be for an empty list.
    This same strategy (called "accumulation") works for combining values whenever you have an associative combiner.



    answer = identity element for operation
    while there are unused objects in the list
    answer= answer (operator) next object


    For example, you can use this pattern to find the union of a set of sets, starting with the empty set, or the intersection of a set of sets, starting with the universe of discourse.






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    • This way the definition for $x^k$ is also simpler. Compare $x^k = 1cdot underbracexcdotldotscdot x_textk times$ with effectively only one base case ($k$ = 0) to the alternative: $x^0 = 1$, $x^k = underbracexcdotldotscdot x_textk times$. Here we have effectively two base cases ($k = 0, 1$).
      – ComFreek
      2 days ago


















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    A definition of powers of natural numbers that is based on maps between finite sets can be formulated in real world terms of fruit and people.



    Say you have an amount $b$ of fruit, each of a different kind.



    • For example an orange and a pear, which means $b = 2$.

    Now say there are $a$ persons between whom you can distribute the fruit.



    • E.g. Alice, Bob and Carol, for $a = 3$. (No, you can't take fruit for yourself.)

    The power $a^b$ is the number of possibilities that you have for distributing the fruit between them.
    You could write up all the different possibilities, and find that there are $9$ of them:



    1. Alice: orange, pear. Bob: Nothing. Carol: Nothing.

    2. Alice: orange. Bob: pear. Carol: Nothing.

    3. Alice: orange. Bob: Nothing. Carol: pear.

    4. Alice: pear. Bob: orange. Carol: Nothing.

    5. Alice: Nothing. Bob: orange, pear. Carol: Nothing.

    6. Alice: Nothing. Bob: orange. Carol: pear.

    7. Alice: pear. Bob: Nothing. Carol: orange.

    8. Alice: Nothing. Bob: pear. Carol: orange.

    9. Alice: Nothing. Bob: Nothing. Carol: orange, pear.

    Of course, you could have found that result by observing that you have $3$ choices for whom to give the orange, and then for each of those choices $3$ more for the $pear$, giving a total of $3 cdot 3 = 9$ possibilities. Either way, we conclude $3^2 = 9$.




    Now for the actual questions:
    You don't have any fruit, and there are three people that are ready to be given fruit.



    Everybody doesn't get a fruit:



    1. Alice: Nothing. Bob: Nothing. Carol: Nothing.

    And this is the only one possibility that you have. Thus $3^0 = 1$.



    While you only have only one option in this case, it is still a valid assignment of $0$ fruits to $3$ people.






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      depends in who you need to explain this to,
      if beginners for abstract algebra then this may work



      $x^k = 1 × k × ... × k $ (x times)



      Therefore
      When the power = 0 the k will not be repeated at all



      $ x^0 = 1 $






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        How is this different from math.stackexchange.com/a/2942188/78700?
        – chepner
        2 days ago

















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      I suppose one explanation is that $1$ is the multiplicative identity, similar to how $0$ is the additive identity. $3 times 0 = 0$. That is, starting with the additive identity, if you add the number $3$ to it zero times, you will get $0$.



      $3^0 = 1$. That is, starting with the multiplicative identity, if you multiply it by $3$ zero times, you will get $1$.



      Could this explanation be understood by someone who doesn't understand additive identities and multiplicative identities? If not, then you should start by explaining those concepts.



      $0$ is the additive identity because if you add $0$ to anything, you get that number back. $1$ is the multiplicative identity because if you multiply anything by it, you get that number back.






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        One intuitive way to define an empty product is as a product of 0 terms. Technically, by that definition, there is no such thing as the empty product because there are 0 ways to bracket a sequence of 0 terms, 1 way to bracket a sequence of 1 term, 1 way to bracket a sequence of 2 terms, and 2 ways to bracket a sequence of 3 terms. We could solve that problem by defining another meaning for the term "empty product." Since multiplication is associative, we can write a string of numbers with the multiplication symbol without brackets to denote multiplication of all of them. Since 1 is the multiplicative identity, sticking a 1 $times$ at the beginning of the expression gives the same result. We could define this to be another notation for the product of all the numbers in the original expression not counting the one and say that $timestext[2][3]$ is another way of saying 1 $times$ 2 $times$ 3. This notation can also be extended to o terms so we can call that the empty product and calculate that $timestext[2][3]$ = 1 $times$ 2 $times$ 3 = 6 and the empty product is $timestext$ = 1 = 1.






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          Since $a^n=1iff nln a=ln 1=0$ so either $n=0$ or $ln a = 0$. In this case, $ln3>0$ so the result follows.






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            The idea is to extend exponents allowing also $0$ and $1$, but maintaining the property
            $$
            x^m+n=x^mx^n
            $$

            Note that also $x^1$ is problematic, because it is not a product to begin with.



            Well, for $x^1$, we want $x^2=x^1x^1$, but also $x^3=x^1x^2$ and we see that defining $x^1=x$ is what we need.



            Next we have $0+2=2$, so we need
            $$
            x^2=x^0x^2
            $$

            and the only way to make it work is defining $x^0=1$.



            At least when $xne0$, but then, why make differences?



            Next we can check that the property $x^m+n=x^mx^n$ is indeed preserved with this definition.



            Actually, this paves the way for a rigorous definition of powers with nonnegative integer exponents, namely
            $$
            x^0=1, qquad x^n+1=x^nxquad (nge0)
            $$

            using recursion. This allows to prove the property $x^m+n=x^mx^n$ without “handwaving”, but with formal induction.






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              If you multiply a number by 3 twice, then you get a number that is 9 times the original number. If you multiply a number by 3 no times, you get a number that is 1 times the original number. This also works for showing 3^1 = 3.



              This might be more clear with prime factorization. For instance, compare the facorizations of 12 and 15. The factorization of 12 has two 2's, one 3, and no 5's. 15 has no 2's, one 3, and one 5.



              You can also show it with blocks: a cube has 3^3 blocks, a square has 3^2 blocks, a line has 3 blocks, and a single block is 3^0=1.






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                Below I explain in simple terms the (abstract) algebraic motivation behind the uniform extension of the power laws from positive powers to zero and negative powers. Most of the post is elementary, so if you encounter unfamiliar terms you can safely skip past them.



                $a^m+n = a^m a^n, $ i.e. $p(m!+!n) = p(m)p(n),$ for $,p(k) = a^kin Bbb Rbackslash 0,$ shows that powers $,a^Bbb N_+$ under multiplication have the same algebraic (semigroup) structure as the positive naturals $Bbb N_+$ under addition. If we enlarge $Bbb N_+$ to a monoid $Bbb N$ or group $Bbb Z$ by adjoining a neutral element $0$ along with additive inverses (negative integers) then there is a unique way of extending this structure preserving (hom) power map, namely:



                $$p(n) = p(0+n) = p(0)p(n) ,Rightarrow, p(0) = 1, rm i.e. a^0= 1$$



                $$1 = p(0) = p(-n+n) = p(-n)p(n),Rightarrow, p(-n) = p(n)^-1, rm i.e. a^-n = (a^n)^-1$$



                The fact that it proves very handy to use $0$ and negative integers when deducing facts about positive integers transfers to the isomorphic structure of powers $a^Bbb Z.,$ Because the power map on $,Bbb Z,$ is an extension of that on positive powers we are guaranteed that proofs about positive powers remain true even if the proof uses negative or zero powers, just as for proofs about positive integers that use negative integers and zero.



                For example using negative integers allows us to concisely state Bezout's lemma for the gcd, i.e. that $,gcd(m,n) = j m + k n,$ for some integers $j,k$ (which may be negative). In particular if $S$ is a group of integers (i.e. closed under subtraction) and it contains two coprime positives then it contains their gcd $= 1$. When translated to the isomorphic power form this says that if a set of powers is closed under division and it contains powers $a^m, a^n$ for coprime $m,n$ then it contains $a^1 = a,,$ e.g. see here where $a^m, a^n$ are integer matrices with determinant $= 1$. However, if we are restricted to positive powers and cancellation (vs. division) then analogous proofs may become much more cumbersome, and the key algebraic structure may become highly obfuscated by the manipulations needed to keep all powers positive, e.g. see here on proving $,a^m = b^m, a^n = b^n,Rightarrow, a = b,$ for integers $a,b,,$ Such enlargments to richer structures with $0$ and inverses allows us to work with objects in simpler forms that better highlight fundamental algebraic structure (here cyclic groups or principal ideals)



                This structure preservation principle is a key property that is employed when enlarging algebraic structures such as groups and rings. If the extended structure preserves the laws (axioms) of the base structure then everything we deduce about the base structure using the extended structure remains valid in the base structure. For example, to solve for integer or rational roots of quadratic and cubics we can employ well-known formulas. Even though these formula may employ complex numbers to solve for integer, rational or real roots, those result are valid in these base number systems because the proofs only employed (ring) axioms that remain valid in the base structures, e.g. associative, commutative, distributive laws. These single uniform formulas greatly simplify ancient methods where the quadratic and cubic formulas bifurcated into motley special cases to avoid untrusted "imaginary" or "negative" numbers, as well as (ab)surds (nowadays, using e.g. set-theoretic foundations, we know rigorous methods to construct such extended numbers systems in a way that proves they remain as consistent as the base number system). Further, as above, instead of appealing to ancient heuristics like the Hankel or Peacock Permanence Principle, we can use the axiomatic method to specify precisely what algebraic structure is preserved in extensions (e.g. the (semi)group structure underlying the power laws).






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                  15 Answers
                  15






                  active

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                  15 Answers
                  15






                  active

                  oldest

                  votes









                  active

                  oldest

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                  active

                  oldest

                  votes








                  up vote
                  66
                  down vote



                  accepted










                  $x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.



                  We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.



                  Using this definition, we can notice the following interesting property:




                  For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$




                  Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.






                  share|cite|improve this answer
















                  • 5




                    If we redefine x<sup>k</sup> to be the multiplicative identity (1) multiplied by x, k times, then when k=0, we don't multiply at all. Many people use sloppy language and say x<sup>k</sup> is "x multiplied by itself k times", but that would make 3<sup>2</sup>=3x3x3 (3 multiplied by 3 twice) rather than 3x3.
                    – Monty Harder
                    Oct 4 at 15:38






                  • 1




                    Important: $x$ must not be zero
                    – Barranka
                    Oct 5 at 1:04






                  • 3




                    @HRSE I agree with $0^0=1(=exp(0))$ but not on base of what you mention. The middle expression is not defined because $ln(0)$ is not defined. IMV $0^0$ is an empty product (just like $3^0$) hence is a neutral element in multiplication.
                    – Vera
                    2 days ago






                  • 1




                    For a slightly simpler version, I'd go with "for any pair of positive integers $k_1$ and $k_2$, $x^k_1 cdot x^k_2 = x^k_1+k_2$, so if we extend it to $0$, $x^k = x^k+0 = x^k cdot x^0$ forces $1 = x^0$".
                    – Pablo H
                    2 days ago







                  • 2




                    This is not the place to "discuss" $0^0$. math.stackexchange.com/questions/11150/…
                    – user202729
                    2 days ago














                  up vote
                  66
                  down vote



                  accepted










                  $x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.



                  We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.



                  Using this definition, we can notice the following interesting property:




                  For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$




                  Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.






                  share|cite|improve this answer
















                  • 5




                    If we redefine x<sup>k</sup> to be the multiplicative identity (1) multiplied by x, k times, then when k=0, we don't multiply at all. Many people use sloppy language and say x<sup>k</sup> is "x multiplied by itself k times", but that would make 3<sup>2</sup>=3x3x3 (3 multiplied by 3 twice) rather than 3x3.
                    – Monty Harder
                    Oct 4 at 15:38






                  • 1




                    Important: $x$ must not be zero
                    – Barranka
                    Oct 5 at 1:04






                  • 3




                    @HRSE I agree with $0^0=1(=exp(0))$ but not on base of what you mention. The middle expression is not defined because $ln(0)$ is not defined. IMV $0^0$ is an empty product (just like $3^0$) hence is a neutral element in multiplication.
                    – Vera
                    2 days ago






                  • 1




                    For a slightly simpler version, I'd go with "for any pair of positive integers $k_1$ and $k_2$, $x^k_1 cdot x^k_2 = x^k_1+k_2$, so if we extend it to $0$, $x^k = x^k+0 = x^k cdot x^0$ forces $1 = x^0$".
                    – Pablo H
                    2 days ago







                  • 2




                    This is not the place to "discuss" $0^0$. math.stackexchange.com/questions/11150/…
                    – user202729
                    2 days ago












                  up vote
                  66
                  down vote



                  accepted







                  up vote
                  66
                  down vote



                  accepted






                  $x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.



                  We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.



                  Using this definition, we can notice the following interesting property:




                  For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$




                  Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.






                  share|cite|improve this answer












                  $x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.



                  We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.



                  Using this definition, we can notice the following interesting property:




                  For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$




                  Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 4 at 13:32









                  5xum

                  84.7k388153




                  84.7k388153







                  • 5




                    If we redefine x<sup>k</sup> to be the multiplicative identity (1) multiplied by x, k times, then when k=0, we don't multiply at all. Many people use sloppy language and say x<sup>k</sup> is "x multiplied by itself k times", but that would make 3<sup>2</sup>=3x3x3 (3 multiplied by 3 twice) rather than 3x3.
                    – Monty Harder
                    Oct 4 at 15:38






                  • 1




                    Important: $x$ must not be zero
                    – Barranka
                    Oct 5 at 1:04






                  • 3




                    @HRSE I agree with $0^0=1(=exp(0))$ but not on base of what you mention. The middle expression is not defined because $ln(0)$ is not defined. IMV $0^0$ is an empty product (just like $3^0$) hence is a neutral element in multiplication.
                    – Vera
                    2 days ago






                  • 1




                    For a slightly simpler version, I'd go with "for any pair of positive integers $k_1$ and $k_2$, $x^k_1 cdot x^k_2 = x^k_1+k_2$, so if we extend it to $0$, $x^k = x^k+0 = x^k cdot x^0$ forces $1 = x^0$".
                    – Pablo H
                    2 days ago







                  • 2




                    This is not the place to "discuss" $0^0$. math.stackexchange.com/questions/11150/…
                    – user202729
                    2 days ago












                  • 5




                    If we redefine x<sup>k</sup> to be the multiplicative identity (1) multiplied by x, k times, then when k=0, we don't multiply at all. Many people use sloppy language and say x<sup>k</sup> is "x multiplied by itself k times", but that would make 3<sup>2</sup>=3x3x3 (3 multiplied by 3 twice) rather than 3x3.
                    – Monty Harder
                    Oct 4 at 15:38






                  • 1




                    Important: $x$ must not be zero
                    – Barranka
                    Oct 5 at 1:04






                  • 3




                    @HRSE I agree with $0^0=1(=exp(0))$ but not on base of what you mention. The middle expression is not defined because $ln(0)$ is not defined. IMV $0^0$ is an empty product (just like $3^0$) hence is a neutral element in multiplication.
                    – Vera
                    2 days ago






                  • 1




                    For a slightly simpler version, I'd go with "for any pair of positive integers $k_1$ and $k_2$, $x^k_1 cdot x^k_2 = x^k_1+k_2$, so if we extend it to $0$, $x^k = x^k+0 = x^k cdot x^0$ forces $1 = x^0$".
                    – Pablo H
                    2 days ago







                  • 2




                    This is not the place to "discuss" $0^0$. math.stackexchange.com/questions/11150/…
                    – user202729
                    2 days ago







                  5




                  5




                  If we redefine x<sup>k</sup> to be the multiplicative identity (1) multiplied by x, k times, then when k=0, we don't multiply at all. Many people use sloppy language and say x<sup>k</sup> is "x multiplied by itself k times", but that would make 3<sup>2</sup>=3x3x3 (3 multiplied by 3 twice) rather than 3x3.
                  – Monty Harder
                  Oct 4 at 15:38




                  If we redefine x<sup>k</sup> to be the multiplicative identity (1) multiplied by x, k times, then when k=0, we don't multiply at all. Many people use sloppy language and say x<sup>k</sup> is "x multiplied by itself k times", but that would make 3<sup>2</sup>=3x3x3 (3 multiplied by 3 twice) rather than 3x3.
                  – Monty Harder
                  Oct 4 at 15:38




                  1




                  1




                  Important: $x$ must not be zero
                  – Barranka
                  Oct 5 at 1:04




                  Important: $x$ must not be zero
                  – Barranka
                  Oct 5 at 1:04




                  3




                  3




                  @HRSE I agree with $0^0=1(=exp(0))$ but not on base of what you mention. The middle expression is not defined because $ln(0)$ is not defined. IMV $0^0$ is an empty product (just like $3^0$) hence is a neutral element in multiplication.
                  – Vera
                  2 days ago




                  @HRSE I agree with $0^0=1(=exp(0))$ but not on base of what you mention. The middle expression is not defined because $ln(0)$ is not defined. IMV $0^0$ is an empty product (just like $3^0$) hence is a neutral element in multiplication.
                  – Vera
                  2 days ago




                  1




                  1




                  For a slightly simpler version, I'd go with "for any pair of positive integers $k_1$ and $k_2$, $x^k_1 cdot x^k_2 = x^k_1+k_2$, so if we extend it to $0$, $x^k = x^k+0 = x^k cdot x^0$ forces $1 = x^0$".
                  – Pablo H
                  2 days ago





                  For a slightly simpler version, I'd go with "for any pair of positive integers $k_1$ and $k_2$, $x^k_1 cdot x^k_2 = x^k_1+k_2$, so if we extend it to $0$, $x^k = x^k+0 = x^k cdot x^0$ forces $1 = x^0$".
                  – Pablo H
                  2 days ago





                  2




                  2




                  This is not the place to "discuss" $0^0$. math.stackexchange.com/questions/11150/…
                  – user202729
                  2 days ago




                  This is not the place to "discuss" $0^0$. math.stackexchange.com/questions/11150/…
                  – user202729
                  2 days ago










                  up vote
                  33
                  down vote













                  $$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.






                  share|cite|improve this answer
















                  • 1




                    “implying” is a bit strong here, else what about nonsense like “$-1 = (-1)^tfrac12times(-1)^tfrac12 =$ product of two square roots, square roots are always positive $Longrightarrow$ $-1$ is positive”?
                    – leftaroundabout
                    Oct 4 at 23:42







                  • 3




                    It's nonsense because $x^1/2$ is not the same thing as $sqrtx$ in the first place, and then because the square root of -1 is $i$, which doesn't follow the ordering of negative/positive at all.
                    – Nij
                    Oct 4 at 23:56






                  • 2




                    How is $x^1/2$ not the same as $sqrtx$?
                    – GraphicsMuncher
                    2 days ago










                  • @leftaroundabout That's nonsense, but not because of Vera's property. It's nonsense because the middle claim, namely "square roots are always positive", is simply false; in the reals because square roots don't always exist and in the complex plane because non-reals aren't positive and are frequently square roots.
                    – Daniel Wagner
                    yesterday






                  • 1




                    @DanielWagner my example wasn't particularly well thought through, but the point was that you can't, in general, just toss around calculation rules outside of the domain they were originally defined in, and then use the “results” to prove that such and such previously undefined expression must have a certain value. You can use that as the motivation for extending the definition, but the proof that it's actually consistent needs to be done separately.
                    – leftaroundabout
                    yesterday














                  up vote
                  33
                  down vote













                  $$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.






                  share|cite|improve this answer
















                  • 1




                    “implying” is a bit strong here, else what about nonsense like “$-1 = (-1)^tfrac12times(-1)^tfrac12 =$ product of two square roots, square roots are always positive $Longrightarrow$ $-1$ is positive”?
                    – leftaroundabout
                    Oct 4 at 23:42







                  • 3




                    It's nonsense because $x^1/2$ is not the same thing as $sqrtx$ in the first place, and then because the square root of -1 is $i$, which doesn't follow the ordering of negative/positive at all.
                    – Nij
                    Oct 4 at 23:56






                  • 2




                    How is $x^1/2$ not the same as $sqrtx$?
                    – GraphicsMuncher
                    2 days ago










                  • @leftaroundabout That's nonsense, but not because of Vera's property. It's nonsense because the middle claim, namely "square roots are always positive", is simply false; in the reals because square roots don't always exist and in the complex plane because non-reals aren't positive and are frequently square roots.
                    – Daniel Wagner
                    yesterday






                  • 1




                    @DanielWagner my example wasn't particularly well thought through, but the point was that you can't, in general, just toss around calculation rules outside of the domain they were originally defined in, and then use the “results” to prove that such and such previously undefined expression must have a certain value. You can use that as the motivation for extending the definition, but the proof that it's actually consistent needs to be done separately.
                    – leftaroundabout
                    yesterday












                  up vote
                  33
                  down vote










                  up vote
                  33
                  down vote









                  $$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.






                  share|cite|improve this answer












                  $$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 4 at 13:31









                  Vera

                  2,312517




                  2,312517







                  • 1




                    “implying” is a bit strong here, else what about nonsense like “$-1 = (-1)^tfrac12times(-1)^tfrac12 =$ product of two square roots, square roots are always positive $Longrightarrow$ $-1$ is positive”?
                    – leftaroundabout
                    Oct 4 at 23:42







                  • 3




                    It's nonsense because $x^1/2$ is not the same thing as $sqrtx$ in the first place, and then because the square root of -1 is $i$, which doesn't follow the ordering of negative/positive at all.
                    – Nij
                    Oct 4 at 23:56






                  • 2




                    How is $x^1/2$ not the same as $sqrtx$?
                    – GraphicsMuncher
                    2 days ago










                  • @leftaroundabout That's nonsense, but not because of Vera's property. It's nonsense because the middle claim, namely "square roots are always positive", is simply false; in the reals because square roots don't always exist and in the complex plane because non-reals aren't positive and are frequently square roots.
                    – Daniel Wagner
                    yesterday






                  • 1




                    @DanielWagner my example wasn't particularly well thought through, but the point was that you can't, in general, just toss around calculation rules outside of the domain they were originally defined in, and then use the “results” to prove that such and such previously undefined expression must have a certain value. You can use that as the motivation for extending the definition, but the proof that it's actually consistent needs to be done separately.
                    – leftaroundabout
                    yesterday












                  • 1




                    “implying” is a bit strong here, else what about nonsense like “$-1 = (-1)^tfrac12times(-1)^tfrac12 =$ product of two square roots, square roots are always positive $Longrightarrow$ $-1$ is positive”?
                    – leftaroundabout
                    Oct 4 at 23:42







                  • 3




                    It's nonsense because $x^1/2$ is not the same thing as $sqrtx$ in the first place, and then because the square root of -1 is $i$, which doesn't follow the ordering of negative/positive at all.
                    – Nij
                    Oct 4 at 23:56






                  • 2




                    How is $x^1/2$ not the same as $sqrtx$?
                    – GraphicsMuncher
                    2 days ago










                  • @leftaroundabout That's nonsense, but not because of Vera's property. It's nonsense because the middle claim, namely "square roots are always positive", is simply false; in the reals because square roots don't always exist and in the complex plane because non-reals aren't positive and are frequently square roots.
                    – Daniel Wagner
                    yesterday






                  • 1




                    @DanielWagner my example wasn't particularly well thought through, but the point was that you can't, in general, just toss around calculation rules outside of the domain they were originally defined in, and then use the “results” to prove that such and such previously undefined expression must have a certain value. You can use that as the motivation for extending the definition, but the proof that it's actually consistent needs to be done separately.
                    – leftaroundabout
                    yesterday







                  1




                  1




                  “implying” is a bit strong here, else what about nonsense like “$-1 = (-1)^tfrac12times(-1)^tfrac12 =$ product of two square roots, square roots are always positive $Longrightarrow$ $-1$ is positive”?
                  – leftaroundabout
                  Oct 4 at 23:42





                  “implying” is a bit strong here, else what about nonsense like “$-1 = (-1)^tfrac12times(-1)^tfrac12 =$ product of two square roots, square roots are always positive $Longrightarrow$ $-1$ is positive”?
                  – leftaroundabout
                  Oct 4 at 23:42





                  3




                  3




                  It's nonsense because $x^1/2$ is not the same thing as $sqrtx$ in the first place, and then because the square root of -1 is $i$, which doesn't follow the ordering of negative/positive at all.
                  – Nij
                  Oct 4 at 23:56




                  It's nonsense because $x^1/2$ is not the same thing as $sqrtx$ in the first place, and then because the square root of -1 is $i$, which doesn't follow the ordering of negative/positive at all.
                  – Nij
                  Oct 4 at 23:56




                  2




                  2




                  How is $x^1/2$ not the same as $sqrtx$?
                  – GraphicsMuncher
                  2 days ago




                  How is $x^1/2$ not the same as $sqrtx$?
                  – GraphicsMuncher
                  2 days ago












                  @leftaroundabout That's nonsense, but not because of Vera's property. It's nonsense because the middle claim, namely "square roots are always positive", is simply false; in the reals because square roots don't always exist and in the complex plane because non-reals aren't positive and are frequently square roots.
                  – Daniel Wagner
                  yesterday




                  @leftaroundabout That's nonsense, but not because of Vera's property. It's nonsense because the middle claim, namely "square roots are always positive", is simply false; in the reals because square roots don't always exist and in the complex plane because non-reals aren't positive and are frequently square roots.
                  – Daniel Wagner
                  yesterday




                  1




                  1




                  @DanielWagner my example wasn't particularly well thought through, but the point was that you can't, in general, just toss around calculation rules outside of the domain they were originally defined in, and then use the “results” to prove that such and such previously undefined expression must have a certain value. You can use that as the motivation for extending the definition, but the proof that it's actually consistent needs to be done separately.
                  – leftaroundabout
                  yesterday




                  @DanielWagner my example wasn't particularly well thought through, but the point was that you can't, in general, just toss around calculation rules outside of the domain they were originally defined in, and then use the “results” to prove that such and such previously undefined expression must have a certain value. You can use that as the motivation for extending the definition, but the proof that it's actually consistent needs to be done separately.
                  – leftaroundabout
                  yesterday










                  up vote
                  23
                  down vote













                  5xum answer is the correct one, but I'll try to give you a more intuitive way to get $x^0 = 1$



                  We have
                  $$x^k = underbracextimes x cdots x_ktext times=1 times underbrace xtimes x cdots x_ktext times$$



                  Therefore
                  $$x^0 = 1 times underbrace xtimes x cdots x_0text times = 1 $$






                  share|cite|improve this answer
















                  • 2




                    This is actually a superior way to define exponentation. Start with the multiplicative identity (1) and multiply by x, k times. If k is zero, there are no multiplications by x.
                    – Monty Harder
                    Oct 4 at 15:34






                  • 6




                    I would not call this more intuitive. In fact, it is not intuitive at all that you want to multiply by 1 in the definition. You only realize this is useful after you've gone through the reasoning summarized in 5xum's post.
                    – Paul Sinclair
                    Oct 4 at 16:30










                  • In my intermediate algebra classes I define $x^0 = 1$, and for natural $n$, $x^n = 1 cdot x cdot x...$ [$n$ times], and $x^-n = 1 div x div x...$ [$n$ times]. So the symmetry of each one starting with $1$ seems pretty intuitive. Also there are actually $n$ operations (see elsewhere comment by @MontyHarder).
                    – Daniel R. Collins
                    Oct 4 at 18:08







                  • 3




                    Couldn't you just as easily say that $x^k = x times x cdots x = 0 + x times x cdots x$, so $x^0 = 0$?
                    – Arcanist Lupus
                    Oct 5 at 1:00







                  • 2




                    @ArcanistLupus Well, never thought about it this way. I think that it doesn't really make sense to use the additive identity (0+a=a) for exponentation, since the usual formula is all about multiplicating. But you're right that it's a flaw in my reasoning, thanks for pointing it out!
                    – F.Carette
                    2 days ago














                  up vote
                  23
                  down vote













                  5xum answer is the correct one, but I'll try to give you a more intuitive way to get $x^0 = 1$



                  We have
                  $$x^k = underbracextimes x cdots x_ktext times=1 times underbrace xtimes x cdots x_ktext times$$



                  Therefore
                  $$x^0 = 1 times underbrace xtimes x cdots x_0text times = 1 $$






                  share|cite|improve this answer
















                  • 2




                    This is actually a superior way to define exponentation. Start with the multiplicative identity (1) and multiply by x, k times. If k is zero, there are no multiplications by x.
                    – Monty Harder
                    Oct 4 at 15:34






                  • 6




                    I would not call this more intuitive. In fact, it is not intuitive at all that you want to multiply by 1 in the definition. You only realize this is useful after you've gone through the reasoning summarized in 5xum's post.
                    – Paul Sinclair
                    Oct 4 at 16:30










                  • In my intermediate algebra classes I define $x^0 = 1$, and for natural $n$, $x^n = 1 cdot x cdot x...$ [$n$ times], and $x^-n = 1 div x div x...$ [$n$ times]. So the symmetry of each one starting with $1$ seems pretty intuitive. Also there are actually $n$ operations (see elsewhere comment by @MontyHarder).
                    – Daniel R. Collins
                    Oct 4 at 18:08







                  • 3




                    Couldn't you just as easily say that $x^k = x times x cdots x = 0 + x times x cdots x$, so $x^0 = 0$?
                    – Arcanist Lupus
                    Oct 5 at 1:00







                  • 2




                    @ArcanistLupus Well, never thought about it this way. I think that it doesn't really make sense to use the additive identity (0+a=a) for exponentation, since the usual formula is all about multiplicating. But you're right that it's a flaw in my reasoning, thanks for pointing it out!
                    – F.Carette
                    2 days ago












                  up vote
                  23
                  down vote










                  up vote
                  23
                  down vote









                  5xum answer is the correct one, but I'll try to give you a more intuitive way to get $x^0 = 1$



                  We have
                  $$x^k = underbracextimes x cdots x_ktext times=1 times underbrace xtimes x cdots x_ktext times$$



                  Therefore
                  $$x^0 = 1 times underbrace xtimes x cdots x_0text times = 1 $$






                  share|cite|improve this answer












                  5xum answer is the correct one, but I'll try to give you a more intuitive way to get $x^0 = 1$



                  We have
                  $$x^k = underbracextimes x cdots x_ktext times=1 times underbrace xtimes x cdots x_ktext times$$



                  Therefore
                  $$x^0 = 1 times underbrace xtimes x cdots x_0text times = 1 $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Oct 4 at 14:14









                  F.Carette

                  9199




                  9199







                  • 2




                    This is actually a superior way to define exponentation. Start with the multiplicative identity (1) and multiply by x, k times. If k is zero, there are no multiplications by x.
                    – Monty Harder
                    Oct 4 at 15:34






                  • 6




                    I would not call this more intuitive. In fact, it is not intuitive at all that you want to multiply by 1 in the definition. You only realize this is useful after you've gone through the reasoning summarized in 5xum's post.
                    – Paul Sinclair
                    Oct 4 at 16:30










                  • In my intermediate algebra classes I define $x^0 = 1$, and for natural $n$, $x^n = 1 cdot x cdot x...$ [$n$ times], and $x^-n = 1 div x div x...$ [$n$ times]. So the symmetry of each one starting with $1$ seems pretty intuitive. Also there are actually $n$ operations (see elsewhere comment by @MontyHarder).
                    – Daniel R. Collins
                    Oct 4 at 18:08







                  • 3




                    Couldn't you just as easily say that $x^k = x times x cdots x = 0 + x times x cdots x$, so $x^0 = 0$?
                    – Arcanist Lupus
                    Oct 5 at 1:00







                  • 2




                    @ArcanistLupus Well, never thought about it this way. I think that it doesn't really make sense to use the additive identity (0+a=a) for exponentation, since the usual formula is all about multiplicating. But you're right that it's a flaw in my reasoning, thanks for pointing it out!
                    – F.Carette
                    2 days ago












                  • 2




                    This is actually a superior way to define exponentation. Start with the multiplicative identity (1) and multiply by x, k times. If k is zero, there are no multiplications by x.
                    – Monty Harder
                    Oct 4 at 15:34






                  • 6




                    I would not call this more intuitive. In fact, it is not intuitive at all that you want to multiply by 1 in the definition. You only realize this is useful after you've gone through the reasoning summarized in 5xum's post.
                    – Paul Sinclair
                    Oct 4 at 16:30










                  • In my intermediate algebra classes I define $x^0 = 1$, and for natural $n$, $x^n = 1 cdot x cdot x...$ [$n$ times], and $x^-n = 1 div x div x...$ [$n$ times]. So the symmetry of each one starting with $1$ seems pretty intuitive. Also there are actually $n$ operations (see elsewhere comment by @MontyHarder).
                    – Daniel R. Collins
                    Oct 4 at 18:08







                  • 3




                    Couldn't you just as easily say that $x^k = x times x cdots x = 0 + x times x cdots x$, so $x^0 = 0$?
                    – Arcanist Lupus
                    Oct 5 at 1:00







                  • 2




                    @ArcanistLupus Well, never thought about it this way. I think that it doesn't really make sense to use the additive identity (0+a=a) for exponentation, since the usual formula is all about multiplicating. But you're right that it's a flaw in my reasoning, thanks for pointing it out!
                    – F.Carette
                    2 days ago







                  2




                  2




                  This is actually a superior way to define exponentation. Start with the multiplicative identity (1) and multiply by x, k times. If k is zero, there are no multiplications by x.
                  – Monty Harder
                  Oct 4 at 15:34




                  This is actually a superior way to define exponentation. Start with the multiplicative identity (1) and multiply by x, k times. If k is zero, there are no multiplications by x.
                  – Monty Harder
                  Oct 4 at 15:34




                  6




                  6




                  I would not call this more intuitive. In fact, it is not intuitive at all that you want to multiply by 1 in the definition. You only realize this is useful after you've gone through the reasoning summarized in 5xum's post.
                  – Paul Sinclair
                  Oct 4 at 16:30




                  I would not call this more intuitive. In fact, it is not intuitive at all that you want to multiply by 1 in the definition. You only realize this is useful after you've gone through the reasoning summarized in 5xum's post.
                  – Paul Sinclair
                  Oct 4 at 16:30












                  In my intermediate algebra classes I define $x^0 = 1$, and for natural $n$, $x^n = 1 cdot x cdot x...$ [$n$ times], and $x^-n = 1 div x div x...$ [$n$ times]. So the symmetry of each one starting with $1$ seems pretty intuitive. Also there are actually $n$ operations (see elsewhere comment by @MontyHarder).
                  – Daniel R. Collins
                  Oct 4 at 18:08





                  In my intermediate algebra classes I define $x^0 = 1$, and for natural $n$, $x^n = 1 cdot x cdot x...$ [$n$ times], and $x^-n = 1 div x div x...$ [$n$ times]. So the symmetry of each one starting with $1$ seems pretty intuitive. Also there are actually $n$ operations (see elsewhere comment by @MontyHarder).
                  – Daniel R. Collins
                  Oct 4 at 18:08





                  3




                  3




                  Couldn't you just as easily say that $x^k = x times x cdots x = 0 + x times x cdots x$, so $x^0 = 0$?
                  – Arcanist Lupus
                  Oct 5 at 1:00





                  Couldn't you just as easily say that $x^k = x times x cdots x = 0 + x times x cdots x$, so $x^0 = 0$?
                  – Arcanist Lupus
                  Oct 5 at 1:00





                  2




                  2




                  @ArcanistLupus Well, never thought about it this way. I think that it doesn't really make sense to use the additive identity (0+a=a) for exponentation, since the usual formula is all about multiplicating. But you're right that it's a flaw in my reasoning, thanks for pointing it out!
                  – F.Carette
                  2 days ago




                  @ArcanistLupus Well, never thought about it this way. I think that it doesn't really make sense to use the additive identity (0+a=a) for exponentation, since the usual formula is all about multiplicating. But you're right that it's a flaw in my reasoning, thanks for pointing it out!
                  – F.Carette
                  2 days ago










                  up vote
                  12
                  down vote













                  The value of the empty product really is a consequence of the associative property of multiplication: To see this, consider the following: For any nonempty, finite set $A$ of numbers, let's define $$displaystyle prod_a in A a$$ as the product of all numbers in $A$. Note that $A$ really needs to be nonempty (at least for now), otherwise this definition doesn't make any sense. Then, if we have another disjoint set $B$, the product of all numbers in $Acup B$ will be $$displaystyle prod_x in A cup B x= left( displaystyle prod_a in A a right) left( displaystyle prod_b in B b right).$$ For example let's take the two sets $M:=x,y$ and $N:=z$. Then $Mcup N = x,y,z$, and by the associative law $$displaystyle prod_p in M cup Np = x*y*z=(x*y)*(z)= left( displaystyle prod_m in M m right) left( displaystyle prod_n in N n right).$$ Now, let's try to extend this notation to the empty set $$. What should $displaystyle prod_x in x$ be? If we want to keep the rule $$displaystyle prod_x in A cup B x= left( displaystyle prod_a in A a right) left( displaystyle prod_b in B b right),$$ then there is only one natural choice: Since for any set $M$ we have $M = M cup $, it follows that $$displaystyle prod_m in M m = displaystyle prod_x in M cup x = left( displaystyle prod_m in M m right) left( displaystyle prod_x in x right).$$ For this to hold the empty product $displaystyle prod_x in x$ has to be equal to the multiplicative identity, $1$. By the same argument, the empty sum is equal to the additive identity, $0$.






                  share|cite|improve this answer


















                  • 4




                    This product notation probably works nicer if you use multisets instead of sets. Then there are no worries about disjointness.
                    – Joonas Ilmavirta
                    2 days ago






                  • 1




                    ...or disjoint union, $sqcup$.
                    – AccidentalFourierTransform
                    2 days ago














                  up vote
                  12
                  down vote













                  The value of the empty product really is a consequence of the associative property of multiplication: To see this, consider the following: For any nonempty, finite set $A$ of numbers, let's define $$displaystyle prod_a in A a$$ as the product of all numbers in $A$. Note that $A$ really needs to be nonempty (at least for now), otherwise this definition doesn't make any sense. Then, if we have another disjoint set $B$, the product of all numbers in $Acup B$ will be $$displaystyle prod_x in A cup B x= left( displaystyle prod_a in A a right) left( displaystyle prod_b in B b right).$$ For example let's take the two sets $M:=x,y$ and $N:=z$. Then $Mcup N = x,y,z$, and by the associative law $$displaystyle prod_p in M cup Np = x*y*z=(x*y)*(z)= left( displaystyle prod_m in M m right) left( displaystyle prod_n in N n right).$$ Now, let's try to extend this notation to the empty set $$. What should $displaystyle prod_x in x$ be? If we want to keep the rule $$displaystyle prod_x in A cup B x= left( displaystyle prod_a in A a right) left( displaystyle prod_b in B b right),$$ then there is only one natural choice: Since for any set $M$ we have $M = M cup $, it follows that $$displaystyle prod_m in M m = displaystyle prod_x in M cup x = left( displaystyle prod_m in M m right) left( displaystyle prod_x in x right).$$ For this to hold the empty product $displaystyle prod_x in x$ has to be equal to the multiplicative identity, $1$. By the same argument, the empty sum is equal to the additive identity, $0$.






                  share|cite|improve this answer


















                  • 4




                    This product notation probably works nicer if you use multisets instead of sets. Then there are no worries about disjointness.
                    – Joonas Ilmavirta
                    2 days ago






                  • 1




                    ...or disjoint union, $sqcup$.
                    – AccidentalFourierTransform
                    2 days ago












                  up vote
                  12
                  down vote










                  up vote
                  12
                  down vote









                  The value of the empty product really is a consequence of the associative property of multiplication: To see this, consider the following: For any nonempty, finite set $A$ of numbers, let's define $$displaystyle prod_a in A a$$ as the product of all numbers in $A$. Note that $A$ really needs to be nonempty (at least for now), otherwise this definition doesn't make any sense. Then, if we have another disjoint set $B$, the product of all numbers in $Acup B$ will be $$displaystyle prod_x in A cup B x= left( displaystyle prod_a in A a right) left( displaystyle prod_b in B b right).$$ For example let's take the two sets $M:=x,y$ and $N:=z$. Then $Mcup N = x,y,z$, and by the associative law $$displaystyle prod_p in M cup Np = x*y*z=(x*y)*(z)= left( displaystyle prod_m in M m right) left( displaystyle prod_n in N n right).$$ Now, let's try to extend this notation to the empty set $$. What should $displaystyle prod_x in x$ be? If we want to keep the rule $$displaystyle prod_x in A cup B x= left( displaystyle prod_a in A a right) left( displaystyle prod_b in B b right),$$ then there is only one natural choice: Since for any set $M$ we have $M = M cup $, it follows that $$displaystyle prod_m in M m = displaystyle prod_x in M cup x = left( displaystyle prod_m in M m right) left( displaystyle prod_x in x right).$$ For this to hold the empty product $displaystyle prod_x in x$ has to be equal to the multiplicative identity, $1$. By the same argument, the empty sum is equal to the additive identity, $0$.






                  share|cite|improve this answer














                  The value of the empty product really is a consequence of the associative property of multiplication: To see this, consider the following: For any nonempty, finite set $A$ of numbers, let's define $$displaystyle prod_a in A a$$ as the product of all numbers in $A$. Note that $A$ really needs to be nonempty (at least for now), otherwise this definition doesn't make any sense. Then, if we have another disjoint set $B$, the product of all numbers in $Acup B$ will be $$displaystyle prod_x in A cup B x= left( displaystyle prod_a in A a right) left( displaystyle prod_b in B b right).$$ For example let's take the two sets $M:=x,y$ and $N:=z$. Then $Mcup N = x,y,z$, and by the associative law $$displaystyle prod_p in M cup Np = x*y*z=(x*y)*(z)= left( displaystyle prod_m in M m right) left( displaystyle prod_n in N n right).$$ Now, let's try to extend this notation to the empty set $$. What should $displaystyle prod_x in x$ be? If we want to keep the rule $$displaystyle prod_x in A cup B x= left( displaystyle prod_a in A a right) left( displaystyle prod_b in B b right),$$ then there is only one natural choice: Since for any set $M$ we have $M = M cup $, it follows that $$displaystyle prod_m in M m = displaystyle prod_x in M cup x = left( displaystyle prod_m in M m right) left( displaystyle prod_x in x right).$$ For this to hold the empty product $displaystyle prod_x in x$ has to be equal to the multiplicative identity, $1$. By the same argument, the empty sum is equal to the additive identity, $0$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 2 days ago

























                  answered Oct 4 at 17:41









                  Jannik Pitt

                  368314




                  368314







                  • 4




                    This product notation probably works nicer if you use multisets instead of sets. Then there are no worries about disjointness.
                    – Joonas Ilmavirta
                    2 days ago






                  • 1




                    ...or disjoint union, $sqcup$.
                    – AccidentalFourierTransform
                    2 days ago












                  • 4




                    This product notation probably works nicer if you use multisets instead of sets. Then there are no worries about disjointness.
                    – Joonas Ilmavirta
                    2 days ago






                  • 1




                    ...or disjoint union, $sqcup$.
                    – AccidentalFourierTransform
                    2 days ago







                  4




                  4




                  This product notation probably works nicer if you use multisets instead of sets. Then there are no worries about disjointness.
                  – Joonas Ilmavirta
                  2 days ago




                  This product notation probably works nicer if you use multisets instead of sets. Then there are no worries about disjointness.
                  – Joonas Ilmavirta
                  2 days ago




                  1




                  1




                  ...or disjoint union, $sqcup$.
                  – AccidentalFourierTransform
                  2 days ago




                  ...or disjoint union, $sqcup$.
                  – AccidentalFourierTransform
                  2 days ago










                  up vote
                  12
                  down vote













                  To increase power by one, multiply by 3, for example $3^2times3=3^3$. To decrease power by one, divide by 3, for example $3^2div3=3^1$. If you repeat this, you get $3^1div3=3^0$.



                  You can see it this way:



                  $$3^3=3times3times3$$



                  $$3^2=3times3$$



                  $$3^1=3$$



                  $$3^0=3div3$$



                  $$3^-1=3div3div3$$






                  share|cite|improve this answer








                  New contributor




                  fhucho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.













                  • 1




                    +1; I've successfully explained this to ten year olds by making such a diagram. They didn't have any trouble with it. Good simple answer.
                    – Wildcard
                    2 days ago














                  up vote
                  12
                  down vote













                  To increase power by one, multiply by 3, for example $3^2times3=3^3$. To decrease power by one, divide by 3, for example $3^2div3=3^1$. If you repeat this, you get $3^1div3=3^0$.



                  You can see it this way:



                  $$3^3=3times3times3$$



                  $$3^2=3times3$$



                  $$3^1=3$$



                  $$3^0=3div3$$



                  $$3^-1=3div3div3$$






                  share|cite|improve this answer








                  New contributor




                  fhucho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.













                  • 1




                    +1; I've successfully explained this to ten year olds by making such a diagram. They didn't have any trouble with it. Good simple answer.
                    – Wildcard
                    2 days ago












                  up vote
                  12
                  down vote










                  up vote
                  12
                  down vote









                  To increase power by one, multiply by 3, for example $3^2times3=3^3$. To decrease power by one, divide by 3, for example $3^2div3=3^1$. If you repeat this, you get $3^1div3=3^0$.



                  You can see it this way:



                  $$3^3=3times3times3$$



                  $$3^2=3times3$$



                  $$3^1=3$$



                  $$3^0=3div3$$



                  $$3^-1=3div3div3$$






                  share|cite|improve this answer








                  New contributor




                  fhucho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  To increase power by one, multiply by 3, for example $3^2times3=3^3$. To decrease power by one, divide by 3, for example $3^2div3=3^1$. If you repeat this, you get $3^1div3=3^0$.



                  You can see it this way:



                  $$3^3=3times3times3$$



                  $$3^2=3times3$$



                  $$3^1=3$$



                  $$3^0=3div3$$



                  $$3^-1=3div3div3$$







                  share|cite|improve this answer








                  New contributor




                  fhucho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|cite|improve this answer



                  share|cite|improve this answer






                  New contributor




                  fhucho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  answered 2 days ago









                  fhucho

                  22113




                  22113




                  New contributor




                  fhucho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





                  New contributor





                  fhucho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  fhucho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.







                  • 1




                    +1; I've successfully explained this to ten year olds by making such a diagram. They didn't have any trouble with it. Good simple answer.
                    – Wildcard
                    2 days ago












                  • 1




                    +1; I've successfully explained this to ten year olds by making such a diagram. They didn't have any trouble with it. Good simple answer.
                    – Wildcard
                    2 days ago







                  1




                  1




                  +1; I've successfully explained this to ten year olds by making such a diagram. They didn't have any trouble with it. Good simple answer.
                  – Wildcard
                  2 days ago




                  +1; I've successfully explained this to ten year olds by making such a diagram. They didn't have any trouble with it. Good simple answer.
                  – Wildcard
                  2 days ago










                  up vote
                  8
                  down vote













                  Another way to look at the quantity $a^b$, with $a$ and $b$ non-negative integers, is as the number of mappings (functions) that exist from a set with $b$ elements (henceforth $B$) to a set with $a$ elements (henceforth $A$).



                  Recall further that a mapping from a set $X$ to a set $Y$ is a set $M$ of pairs $(x,y)$ where



                  • for all $(x,y) in M$, $x in X$ and $y in y$; and

                  • for all $x in X$, there is exactly one $y in Y$ such that $(x,y) in M$.

                  Now, if $B$ has zero elements (i.e., if it is the empty set $emptyset$), it is clear that there exists exactly one $M$ that satisfies the two conditions above, namely $M = emptyset$.






                  share|cite|improve this answer
























                    up vote
                    8
                    down vote













                    Another way to look at the quantity $a^b$, with $a$ and $b$ non-negative integers, is as the number of mappings (functions) that exist from a set with $b$ elements (henceforth $B$) to a set with $a$ elements (henceforth $A$).



                    Recall further that a mapping from a set $X$ to a set $Y$ is a set $M$ of pairs $(x,y)$ where



                    • for all $(x,y) in M$, $x in X$ and $y in y$; and

                    • for all $x in X$, there is exactly one $y in Y$ such that $(x,y) in M$.

                    Now, if $B$ has zero elements (i.e., if it is the empty set $emptyset$), it is clear that there exists exactly one $M$ that satisfies the two conditions above, namely $M = emptyset$.






                    share|cite|improve this answer






















                      up vote
                      8
                      down vote










                      up vote
                      8
                      down vote









                      Another way to look at the quantity $a^b$, with $a$ and $b$ non-negative integers, is as the number of mappings (functions) that exist from a set with $b$ elements (henceforth $B$) to a set with $a$ elements (henceforth $A$).



                      Recall further that a mapping from a set $X$ to a set $Y$ is a set $M$ of pairs $(x,y)$ where



                      • for all $(x,y) in M$, $x in X$ and $y in y$; and

                      • for all $x in X$, there is exactly one $y in Y$ such that $(x,y) in M$.

                      Now, if $B$ has zero elements (i.e., if it is the empty set $emptyset$), it is clear that there exists exactly one $M$ that satisfies the two conditions above, namely $M = emptyset$.






                      share|cite|improve this answer












                      Another way to look at the quantity $a^b$, with $a$ and $b$ non-negative integers, is as the number of mappings (functions) that exist from a set with $b$ elements (henceforth $B$) to a set with $a$ elements (henceforth $A$).



                      Recall further that a mapping from a set $X$ to a set $Y$ is a set $M$ of pairs $(x,y)$ where



                      • for all $(x,y) in M$, $x in X$ and $y in y$; and

                      • for all $x in X$, there is exactly one $y in Y$ such that $(x,y) in M$.

                      Now, if $B$ has zero elements (i.e., if it is the empty set $emptyset$), it is clear that there exists exactly one $M$ that satisfies the two conditions above, namely $M = emptyset$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 days ago









                      fkraiem

                      2,8161510




                      2,8161510




















                          up vote
                          7
                          down vote













                          Not the simplest answer, but a way to think about this and related questions about combining a bunch of things.



                          If you have a list of numbers and you want to write a computer program to sum them you do something like



                          sum = 0
                          while there are unsummed numbers in the list
                          sum = sum + next number


                          Then the sum variable contains the answer. If the list is empty the while loop never does anything and the sum is $0$.



                          For multiplication you want



                          product = 1
                          while there are unmultiplied numbers in the list
                          product = product * next number


                          That clearly gives the correct answer when the list is not empty and tells you what the answer should be for an empty list.
                          This same strategy (called "accumulation") works for combining values whenever you have an associative combiner.



                          answer = identity element for operation
                          while there are unused objects in the list
                          answer= answer (operator) next object


                          For example, you can use this pattern to find the union of a set of sets, starting with the empty set, or the intersection of a set of sets, starting with the universe of discourse.






                          share|cite|improve this answer




















                          • This way the definition for $x^k$ is also simpler. Compare $x^k = 1cdot underbracexcdotldotscdot x_textk times$ with effectively only one base case ($k$ = 0) to the alternative: $x^0 = 1$, $x^k = underbracexcdotldotscdot x_textk times$. Here we have effectively two base cases ($k = 0, 1$).
                            – ComFreek
                            2 days ago















                          up vote
                          7
                          down vote













                          Not the simplest answer, but a way to think about this and related questions about combining a bunch of things.



                          If you have a list of numbers and you want to write a computer program to sum them you do something like



                          sum = 0
                          while there are unsummed numbers in the list
                          sum = sum + next number


                          Then the sum variable contains the answer. If the list is empty the while loop never does anything and the sum is $0$.



                          For multiplication you want



                          product = 1
                          while there are unmultiplied numbers in the list
                          product = product * next number


                          That clearly gives the correct answer when the list is not empty and tells you what the answer should be for an empty list.
                          This same strategy (called "accumulation") works for combining values whenever you have an associative combiner.



                          answer = identity element for operation
                          while there are unused objects in the list
                          answer= answer (operator) next object


                          For example, you can use this pattern to find the union of a set of sets, starting with the empty set, or the intersection of a set of sets, starting with the universe of discourse.






                          share|cite|improve this answer




















                          • This way the definition for $x^k$ is also simpler. Compare $x^k = 1cdot underbracexcdotldotscdot x_textk times$ with effectively only one base case ($k$ = 0) to the alternative: $x^0 = 1$, $x^k = underbracexcdotldotscdot x_textk times$. Here we have effectively two base cases ($k = 0, 1$).
                            – ComFreek
                            2 days ago













                          up vote
                          7
                          down vote










                          up vote
                          7
                          down vote









                          Not the simplest answer, but a way to think about this and related questions about combining a bunch of things.



                          If you have a list of numbers and you want to write a computer program to sum them you do something like



                          sum = 0
                          while there are unsummed numbers in the list
                          sum = sum + next number


                          Then the sum variable contains the answer. If the list is empty the while loop never does anything and the sum is $0$.



                          For multiplication you want



                          product = 1
                          while there are unmultiplied numbers in the list
                          product = product * next number


                          That clearly gives the correct answer when the list is not empty and tells you what the answer should be for an empty list.
                          This same strategy (called "accumulation") works for combining values whenever you have an associative combiner.



                          answer = identity element for operation
                          while there are unused objects in the list
                          answer= answer (operator) next object


                          For example, you can use this pattern to find the union of a set of sets, starting with the empty set, or the intersection of a set of sets, starting with the universe of discourse.






                          share|cite|improve this answer












                          Not the simplest answer, but a way to think about this and related questions about combining a bunch of things.



                          If you have a list of numbers and you want to write a computer program to sum them you do something like



                          sum = 0
                          while there are unsummed numbers in the list
                          sum = sum + next number


                          Then the sum variable contains the answer. If the list is empty the while loop never does anything and the sum is $0$.



                          For multiplication you want



                          product = 1
                          while there are unmultiplied numbers in the list
                          product = product * next number


                          That clearly gives the correct answer when the list is not empty and tells you what the answer should be for an empty list.
                          This same strategy (called "accumulation") works for combining values whenever you have an associative combiner.



                          answer = identity element for operation
                          while there are unused objects in the list
                          answer= answer (operator) next object


                          For example, you can use this pattern to find the union of a set of sets, starting with the empty set, or the intersection of a set of sets, starting with the universe of discourse.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Oct 5 at 0:46









                          Ethan Bolker

                          37.1k54299




                          37.1k54299











                          • This way the definition for $x^k$ is also simpler. Compare $x^k = 1cdot underbracexcdotldotscdot x_textk times$ with effectively only one base case ($k$ = 0) to the alternative: $x^0 = 1$, $x^k = underbracexcdotldotscdot x_textk times$. Here we have effectively two base cases ($k = 0, 1$).
                            – ComFreek
                            2 days ago

















                          • This way the definition for $x^k$ is also simpler. Compare $x^k = 1cdot underbracexcdotldotscdot x_textk times$ with effectively only one base case ($k$ = 0) to the alternative: $x^0 = 1$, $x^k = underbracexcdotldotscdot x_textk times$. Here we have effectively two base cases ($k = 0, 1$).
                            – ComFreek
                            2 days ago
















                          This way the definition for $x^k$ is also simpler. Compare $x^k = 1cdot underbracexcdotldotscdot x_textk times$ with effectively only one base case ($k$ = 0) to the alternative: $x^0 = 1$, $x^k = underbracexcdotldotscdot x_textk times$. Here we have effectively two base cases ($k = 0, 1$).
                          – ComFreek
                          2 days ago





                          This way the definition for $x^k$ is also simpler. Compare $x^k = 1cdot underbracexcdotldotscdot x_textk times$ with effectively only one base case ($k$ = 0) to the alternative: $x^0 = 1$, $x^k = underbracexcdotldotscdot x_textk times$. Here we have effectively two base cases ($k = 0, 1$).
                          – ComFreek
                          2 days ago











                          up vote
                          6
                          down vote













                          A definition of powers of natural numbers that is based on maps between finite sets can be formulated in real world terms of fruit and people.



                          Say you have an amount $b$ of fruit, each of a different kind.



                          • For example an orange and a pear, which means $b = 2$.

                          Now say there are $a$ persons between whom you can distribute the fruit.



                          • E.g. Alice, Bob and Carol, for $a = 3$. (No, you can't take fruit for yourself.)

                          The power $a^b$ is the number of possibilities that you have for distributing the fruit between them.
                          You could write up all the different possibilities, and find that there are $9$ of them:



                          1. Alice: orange, pear. Bob: Nothing. Carol: Nothing.

                          2. Alice: orange. Bob: pear. Carol: Nothing.

                          3. Alice: orange. Bob: Nothing. Carol: pear.

                          4. Alice: pear. Bob: orange. Carol: Nothing.

                          5. Alice: Nothing. Bob: orange, pear. Carol: Nothing.

                          6. Alice: Nothing. Bob: orange. Carol: pear.

                          7. Alice: pear. Bob: Nothing. Carol: orange.

                          8. Alice: Nothing. Bob: pear. Carol: orange.

                          9. Alice: Nothing. Bob: Nothing. Carol: orange, pear.

                          Of course, you could have found that result by observing that you have $3$ choices for whom to give the orange, and then for each of those choices $3$ more for the $pear$, giving a total of $3 cdot 3 = 9$ possibilities. Either way, we conclude $3^2 = 9$.




                          Now for the actual questions:
                          You don't have any fruit, and there are three people that are ready to be given fruit.



                          Everybody doesn't get a fruit:



                          1. Alice: Nothing. Bob: Nothing. Carol: Nothing.

                          And this is the only one possibility that you have. Thus $3^0 = 1$.



                          While you only have only one option in this case, it is still a valid assignment of $0$ fruits to $3$ people.






                          share|cite|improve this answer
























                            up vote
                            6
                            down vote













                            A definition of powers of natural numbers that is based on maps between finite sets can be formulated in real world terms of fruit and people.



                            Say you have an amount $b$ of fruit, each of a different kind.



                            • For example an orange and a pear, which means $b = 2$.

                            Now say there are $a$ persons between whom you can distribute the fruit.



                            • E.g. Alice, Bob and Carol, for $a = 3$. (No, you can't take fruit for yourself.)

                            The power $a^b$ is the number of possibilities that you have for distributing the fruit between them.
                            You could write up all the different possibilities, and find that there are $9$ of them:



                            1. Alice: orange, pear. Bob: Nothing. Carol: Nothing.

                            2. Alice: orange. Bob: pear. Carol: Nothing.

                            3. Alice: orange. Bob: Nothing. Carol: pear.

                            4. Alice: pear. Bob: orange. Carol: Nothing.

                            5. Alice: Nothing. Bob: orange, pear. Carol: Nothing.

                            6. Alice: Nothing. Bob: orange. Carol: pear.

                            7. Alice: pear. Bob: Nothing. Carol: orange.

                            8. Alice: Nothing. Bob: pear. Carol: orange.

                            9. Alice: Nothing. Bob: Nothing. Carol: orange, pear.

                            Of course, you could have found that result by observing that you have $3$ choices for whom to give the orange, and then for each of those choices $3$ more for the $pear$, giving a total of $3 cdot 3 = 9$ possibilities. Either way, we conclude $3^2 = 9$.




                            Now for the actual questions:
                            You don't have any fruit, and there are three people that are ready to be given fruit.



                            Everybody doesn't get a fruit:



                            1. Alice: Nothing. Bob: Nothing. Carol: Nothing.

                            And this is the only one possibility that you have. Thus $3^0 = 1$.



                            While you only have only one option in this case, it is still a valid assignment of $0$ fruits to $3$ people.






                            share|cite|improve this answer






















                              up vote
                              6
                              down vote










                              up vote
                              6
                              down vote









                              A definition of powers of natural numbers that is based on maps between finite sets can be formulated in real world terms of fruit and people.



                              Say you have an amount $b$ of fruit, each of a different kind.



                              • For example an orange and a pear, which means $b = 2$.

                              Now say there are $a$ persons between whom you can distribute the fruit.



                              • E.g. Alice, Bob and Carol, for $a = 3$. (No, you can't take fruit for yourself.)

                              The power $a^b$ is the number of possibilities that you have for distributing the fruit between them.
                              You could write up all the different possibilities, and find that there are $9$ of them:



                              1. Alice: orange, pear. Bob: Nothing. Carol: Nothing.

                              2. Alice: orange. Bob: pear. Carol: Nothing.

                              3. Alice: orange. Bob: Nothing. Carol: pear.

                              4. Alice: pear. Bob: orange. Carol: Nothing.

                              5. Alice: Nothing. Bob: orange, pear. Carol: Nothing.

                              6. Alice: Nothing. Bob: orange. Carol: pear.

                              7. Alice: pear. Bob: Nothing. Carol: orange.

                              8. Alice: Nothing. Bob: pear. Carol: orange.

                              9. Alice: Nothing. Bob: Nothing. Carol: orange, pear.

                              Of course, you could have found that result by observing that you have $3$ choices for whom to give the orange, and then for each of those choices $3$ more for the $pear$, giving a total of $3 cdot 3 = 9$ possibilities. Either way, we conclude $3^2 = 9$.




                              Now for the actual questions:
                              You don't have any fruit, and there are three people that are ready to be given fruit.



                              Everybody doesn't get a fruit:



                              1. Alice: Nothing. Bob: Nothing. Carol: Nothing.

                              And this is the only one possibility that you have. Thus $3^0 = 1$.



                              While you only have only one option in this case, it is still a valid assignment of $0$ fruits to $3$ people.






                              share|cite|improve this answer












                              A definition of powers of natural numbers that is based on maps between finite sets can be formulated in real world terms of fruit and people.



                              Say you have an amount $b$ of fruit, each of a different kind.



                              • For example an orange and a pear, which means $b = 2$.

                              Now say there are $a$ persons between whom you can distribute the fruit.



                              • E.g. Alice, Bob and Carol, for $a = 3$. (No, you can't take fruit for yourself.)

                              The power $a^b$ is the number of possibilities that you have for distributing the fruit between them.
                              You could write up all the different possibilities, and find that there are $9$ of them:



                              1. Alice: orange, pear. Bob: Nothing. Carol: Nothing.

                              2. Alice: orange. Bob: pear. Carol: Nothing.

                              3. Alice: orange. Bob: Nothing. Carol: pear.

                              4. Alice: pear. Bob: orange. Carol: Nothing.

                              5. Alice: Nothing. Bob: orange, pear. Carol: Nothing.

                              6. Alice: Nothing. Bob: orange. Carol: pear.

                              7. Alice: pear. Bob: Nothing. Carol: orange.

                              8. Alice: Nothing. Bob: pear. Carol: orange.

                              9. Alice: Nothing. Bob: Nothing. Carol: orange, pear.

                              Of course, you could have found that result by observing that you have $3$ choices for whom to give the orange, and then for each of those choices $3$ more for the $pear$, giving a total of $3 cdot 3 = 9$ possibilities. Either way, we conclude $3^2 = 9$.




                              Now for the actual questions:
                              You don't have any fruit, and there are three people that are ready to be given fruit.



                              Everybody doesn't get a fruit:



                              1. Alice: Nothing. Bob: Nothing. Carol: Nothing.

                              And this is the only one possibility that you have. Thus $3^0 = 1$.



                              While you only have only one option in this case, it is still a valid assignment of $0$ fruits to $3$ people.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered 2 days ago









                              Julian Bitterwolf

                              635




                              635




















                                  up vote
                                  3
                                  down vote













                                  depends in who you need to explain this to,
                                  if beginners for abstract algebra then this may work



                                  $x^k = 1 × k × ... × k $ (x times)



                                  Therefore
                                  When the power = 0 the k will not be repeated at all



                                  $ x^0 = 1 $






                                  share|cite|improve this answer








                                  New contributor




                                  asmgx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.













                                  • 1




                                    How is this different from math.stackexchange.com/a/2942188/78700?
                                    – chepner
                                    2 days ago














                                  up vote
                                  3
                                  down vote













                                  depends in who you need to explain this to,
                                  if beginners for abstract algebra then this may work



                                  $x^k = 1 × k × ... × k $ (x times)



                                  Therefore
                                  When the power = 0 the k will not be repeated at all



                                  $ x^0 = 1 $






                                  share|cite|improve this answer








                                  New contributor




                                  asmgx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.













                                  • 1




                                    How is this different from math.stackexchange.com/a/2942188/78700?
                                    – chepner
                                    2 days ago












                                  up vote
                                  3
                                  down vote










                                  up vote
                                  3
                                  down vote









                                  depends in who you need to explain this to,
                                  if beginners for abstract algebra then this may work



                                  $x^k = 1 × k × ... × k $ (x times)



                                  Therefore
                                  When the power = 0 the k will not be repeated at all



                                  $ x^0 = 1 $






                                  share|cite|improve this answer








                                  New contributor




                                  asmgx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.









                                  depends in who you need to explain this to,
                                  if beginners for abstract algebra then this may work



                                  $x^k = 1 × k × ... × k $ (x times)



                                  Therefore
                                  When the power = 0 the k will not be repeated at all



                                  $ x^0 = 1 $







                                  share|cite|improve this answer








                                  New contributor




                                  asmgx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.









                                  share|cite|improve this answer



                                  share|cite|improve this answer






                                  New contributor




                                  asmgx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.









                                  answered Oct 5 at 2:28









                                  asmgx

                                  1414




                                  1414




                                  New contributor




                                  asmgx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.





                                  New contributor





                                  asmgx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.






                                  asmgx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                  Check out our Code of Conduct.







                                  • 1




                                    How is this different from math.stackexchange.com/a/2942188/78700?
                                    – chepner
                                    2 days ago












                                  • 1




                                    How is this different from math.stackexchange.com/a/2942188/78700?
                                    – chepner
                                    2 days ago







                                  1




                                  1




                                  How is this different from math.stackexchange.com/a/2942188/78700?
                                  – chepner
                                  2 days ago




                                  How is this different from math.stackexchange.com/a/2942188/78700?
                                  – chepner
                                  2 days ago










                                  up vote
                                  2
                                  down vote













                                  I suppose one explanation is that $1$ is the multiplicative identity, similar to how $0$ is the additive identity. $3 times 0 = 0$. That is, starting with the additive identity, if you add the number $3$ to it zero times, you will get $0$.



                                  $3^0 = 1$. That is, starting with the multiplicative identity, if you multiply it by $3$ zero times, you will get $1$.



                                  Could this explanation be understood by someone who doesn't understand additive identities and multiplicative identities? If not, then you should start by explaining those concepts.



                                  $0$ is the additive identity because if you add $0$ to anything, you get that number back. $1$ is the multiplicative identity because if you multiply anything by it, you get that number back.






                                  share|cite|improve this answer


























                                    up vote
                                    2
                                    down vote













                                    I suppose one explanation is that $1$ is the multiplicative identity, similar to how $0$ is the additive identity. $3 times 0 = 0$. That is, starting with the additive identity, if you add the number $3$ to it zero times, you will get $0$.



                                    $3^0 = 1$. That is, starting with the multiplicative identity, if you multiply it by $3$ zero times, you will get $1$.



                                    Could this explanation be understood by someone who doesn't understand additive identities and multiplicative identities? If not, then you should start by explaining those concepts.



                                    $0$ is the additive identity because if you add $0$ to anything, you get that number back. $1$ is the multiplicative identity because if you multiply anything by it, you get that number back.






                                    share|cite|improve this answer
























                                      up vote
                                      2
                                      down vote










                                      up vote
                                      2
                                      down vote









                                      I suppose one explanation is that $1$ is the multiplicative identity, similar to how $0$ is the additive identity. $3 times 0 = 0$. That is, starting with the additive identity, if you add the number $3$ to it zero times, you will get $0$.



                                      $3^0 = 1$. That is, starting with the multiplicative identity, if you multiply it by $3$ zero times, you will get $1$.



                                      Could this explanation be understood by someone who doesn't understand additive identities and multiplicative identities? If not, then you should start by explaining those concepts.



                                      $0$ is the additive identity because if you add $0$ to anything, you get that number back. $1$ is the multiplicative identity because if you multiply anything by it, you get that number back.






                                      share|cite|improve this answer














                                      I suppose one explanation is that $1$ is the multiplicative identity, similar to how $0$ is the additive identity. $3 times 0 = 0$. That is, starting with the additive identity, if you add the number $3$ to it zero times, you will get $0$.



                                      $3^0 = 1$. That is, starting with the multiplicative identity, if you multiply it by $3$ zero times, you will get $1$.



                                      Could this explanation be understood by someone who doesn't understand additive identities and multiplicative identities? If not, then you should start by explaining those concepts.



                                      $0$ is the additive identity because if you add $0$ to anything, you get that number back. $1$ is the multiplicative identity because if you multiply anything by it, you get that number back.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited yesterday









                                      David R.

                                      3201728




                                      3201728










                                      answered 2 days ago









                                      John

                                      2012




                                      2012




















                                          up vote
                                          2
                                          down vote













                                          One intuitive way to define an empty product is as a product of 0 terms. Technically, by that definition, there is no such thing as the empty product because there are 0 ways to bracket a sequence of 0 terms, 1 way to bracket a sequence of 1 term, 1 way to bracket a sequence of 2 terms, and 2 ways to bracket a sequence of 3 terms. We could solve that problem by defining another meaning for the term "empty product." Since multiplication is associative, we can write a string of numbers with the multiplication symbol without brackets to denote multiplication of all of them. Since 1 is the multiplicative identity, sticking a 1 $times$ at the beginning of the expression gives the same result. We could define this to be another notation for the product of all the numbers in the original expression not counting the one and say that $timestext[2][3]$ is another way of saying 1 $times$ 2 $times$ 3. This notation can also be extended to o terms so we can call that the empty product and calculate that $timestext[2][3]$ = 1 $times$ 2 $times$ 3 = 6 and the empty product is $timestext$ = 1 = 1.






                                          share|cite|improve this answer


























                                            up vote
                                            2
                                            down vote













                                            One intuitive way to define an empty product is as a product of 0 terms. Technically, by that definition, there is no such thing as the empty product because there are 0 ways to bracket a sequence of 0 terms, 1 way to bracket a sequence of 1 term, 1 way to bracket a sequence of 2 terms, and 2 ways to bracket a sequence of 3 terms. We could solve that problem by defining another meaning for the term "empty product." Since multiplication is associative, we can write a string of numbers with the multiplication symbol without brackets to denote multiplication of all of them. Since 1 is the multiplicative identity, sticking a 1 $times$ at the beginning of the expression gives the same result. We could define this to be another notation for the product of all the numbers in the original expression not counting the one and say that $timestext[2][3]$ is another way of saying 1 $times$ 2 $times$ 3. This notation can also be extended to o terms so we can call that the empty product and calculate that $timestext[2][3]$ = 1 $times$ 2 $times$ 3 = 6 and the empty product is $timestext$ = 1 = 1.






                                            share|cite|improve this answer
























                                              up vote
                                              2
                                              down vote










                                              up vote
                                              2
                                              down vote









                                              One intuitive way to define an empty product is as a product of 0 terms. Technically, by that definition, there is no such thing as the empty product because there are 0 ways to bracket a sequence of 0 terms, 1 way to bracket a sequence of 1 term, 1 way to bracket a sequence of 2 terms, and 2 ways to bracket a sequence of 3 terms. We could solve that problem by defining another meaning for the term "empty product." Since multiplication is associative, we can write a string of numbers with the multiplication symbol without brackets to denote multiplication of all of them. Since 1 is the multiplicative identity, sticking a 1 $times$ at the beginning of the expression gives the same result. We could define this to be another notation for the product of all the numbers in the original expression not counting the one and say that $timestext[2][3]$ is another way of saying 1 $times$ 2 $times$ 3. This notation can also be extended to o terms so we can call that the empty product and calculate that $timestext[2][3]$ = 1 $times$ 2 $times$ 3 = 6 and the empty product is $timestext$ = 1 = 1.






                                              share|cite|improve this answer














                                              One intuitive way to define an empty product is as a product of 0 terms. Technically, by that definition, there is no such thing as the empty product because there are 0 ways to bracket a sequence of 0 terms, 1 way to bracket a sequence of 1 term, 1 way to bracket a sequence of 2 terms, and 2 ways to bracket a sequence of 3 terms. We could solve that problem by defining another meaning for the term "empty product." Since multiplication is associative, we can write a string of numbers with the multiplication symbol without brackets to denote multiplication of all of them. Since 1 is the multiplicative identity, sticking a 1 $times$ at the beginning of the expression gives the same result. We could define this to be another notation for the product of all the numbers in the original expression not counting the one and say that $timestext[2][3]$ is another way of saying 1 $times$ 2 $times$ 3. This notation can also be extended to o terms so we can call that the empty product and calculate that $timestext[2][3]$ = 1 $times$ 2 $times$ 3 = 6 and the empty product is $timestext$ = 1 = 1.







                                              share|cite|improve this answer














                                              share|cite|improve this answer



                                              share|cite|improve this answer








                                              edited yesterday

























                                              answered 2 days ago









                                              Timothy

                                              280211




                                              280211




















                                                  up vote
                                                  1
                                                  down vote













                                                  Since $a^n=1iff nln a=ln 1=0$ so either $n=0$ or $ln a = 0$. In this case, $ln3>0$ so the result follows.






                                                  share|cite|improve this answer
























                                                    up vote
                                                    1
                                                    down vote













                                                    Since $a^n=1iff nln a=ln 1=0$ so either $n=0$ or $ln a = 0$. In this case, $ln3>0$ so the result follows.






                                                    share|cite|improve this answer






















                                                      up vote
                                                      1
                                                      down vote










                                                      up vote
                                                      1
                                                      down vote









                                                      Since $a^n=1iff nln a=ln 1=0$ so either $n=0$ or $ln a = 0$. In this case, $ln3>0$ so the result follows.






                                                      share|cite|improve this answer












                                                      Since $a^n=1iff nln a=ln 1=0$ so either $n=0$ or $ln a = 0$. In this case, $ln3>0$ so the result follows.







                                                      share|cite|improve this answer












                                                      share|cite|improve this answer



                                                      share|cite|improve this answer










                                                      answered yesterday









                                                      TheSimpliFire

                                                      11.3k62256




                                                      11.3k62256




















                                                          up vote
                                                          1
                                                          down vote













                                                          The idea is to extend exponents allowing also $0$ and $1$, but maintaining the property
                                                          $$
                                                          x^m+n=x^mx^n
                                                          $$

                                                          Note that also $x^1$ is problematic, because it is not a product to begin with.



                                                          Well, for $x^1$, we want $x^2=x^1x^1$, but also $x^3=x^1x^2$ and we see that defining $x^1=x$ is what we need.



                                                          Next we have $0+2=2$, so we need
                                                          $$
                                                          x^2=x^0x^2
                                                          $$

                                                          and the only way to make it work is defining $x^0=1$.



                                                          At least when $xne0$, but then, why make differences?



                                                          Next we can check that the property $x^m+n=x^mx^n$ is indeed preserved with this definition.



                                                          Actually, this paves the way for a rigorous definition of powers with nonnegative integer exponents, namely
                                                          $$
                                                          x^0=1, qquad x^n+1=x^nxquad (nge0)
                                                          $$

                                                          using recursion. This allows to prove the property $x^m+n=x^mx^n$ without “handwaving”, but with formal induction.






                                                          share|cite|improve this answer
























                                                            up vote
                                                            1
                                                            down vote













                                                            The idea is to extend exponents allowing also $0$ and $1$, but maintaining the property
                                                            $$
                                                            x^m+n=x^mx^n
                                                            $$

                                                            Note that also $x^1$ is problematic, because it is not a product to begin with.



                                                            Well, for $x^1$, we want $x^2=x^1x^1$, but also $x^3=x^1x^2$ and we see that defining $x^1=x$ is what we need.



                                                            Next we have $0+2=2$, so we need
                                                            $$
                                                            x^2=x^0x^2
                                                            $$

                                                            and the only way to make it work is defining $x^0=1$.



                                                            At least when $xne0$, but then, why make differences?



                                                            Next we can check that the property $x^m+n=x^mx^n$ is indeed preserved with this definition.



                                                            Actually, this paves the way for a rigorous definition of powers with nonnegative integer exponents, namely
                                                            $$
                                                            x^0=1, qquad x^n+1=x^nxquad (nge0)
                                                            $$

                                                            using recursion. This allows to prove the property $x^m+n=x^mx^n$ without “handwaving”, but with formal induction.






                                                            share|cite|improve this answer






















                                                              up vote
                                                              1
                                                              down vote










                                                              up vote
                                                              1
                                                              down vote









                                                              The idea is to extend exponents allowing also $0$ and $1$, but maintaining the property
                                                              $$
                                                              x^m+n=x^mx^n
                                                              $$

                                                              Note that also $x^1$ is problematic, because it is not a product to begin with.



                                                              Well, for $x^1$, we want $x^2=x^1x^1$, but also $x^3=x^1x^2$ and we see that defining $x^1=x$ is what we need.



                                                              Next we have $0+2=2$, so we need
                                                              $$
                                                              x^2=x^0x^2
                                                              $$

                                                              and the only way to make it work is defining $x^0=1$.



                                                              At least when $xne0$, but then, why make differences?



                                                              Next we can check that the property $x^m+n=x^mx^n$ is indeed preserved with this definition.



                                                              Actually, this paves the way for a rigorous definition of powers with nonnegative integer exponents, namely
                                                              $$
                                                              x^0=1, qquad x^n+1=x^nxquad (nge0)
                                                              $$

                                                              using recursion. This allows to prove the property $x^m+n=x^mx^n$ without “handwaving”, but with formal induction.






                                                              share|cite|improve this answer












                                                              The idea is to extend exponents allowing also $0$ and $1$, but maintaining the property
                                                              $$
                                                              x^m+n=x^mx^n
                                                              $$

                                                              Note that also $x^1$ is problematic, because it is not a product to begin with.



                                                              Well, for $x^1$, we want $x^2=x^1x^1$, but also $x^3=x^1x^2$ and we see that defining $x^1=x$ is what we need.



                                                              Next we have $0+2=2$, so we need
                                                              $$
                                                              x^2=x^0x^2
                                                              $$

                                                              and the only way to make it work is defining $x^0=1$.



                                                              At least when $xne0$, but then, why make differences?



                                                              Next we can check that the property $x^m+n=x^mx^n$ is indeed preserved with this definition.



                                                              Actually, this paves the way for a rigorous definition of powers with nonnegative integer exponents, namely
                                                              $$
                                                              x^0=1, qquad x^n+1=x^nxquad (nge0)
                                                              $$

                                                              using recursion. This allows to prove the property $x^m+n=x^mx^n$ without “handwaving”, but with formal induction.







                                                              share|cite|improve this answer












                                                              share|cite|improve this answer



                                                              share|cite|improve this answer










                                                              answered 16 hours ago









                                                              egreg

                                                              169k1283191




                                                              169k1283191




















                                                                  up vote
                                                                  1
                                                                  down vote













                                                                  If you multiply a number by 3 twice, then you get a number that is 9 times the original number. If you multiply a number by 3 no times, you get a number that is 1 times the original number. This also works for showing 3^1 = 3.



                                                                  This might be more clear with prime factorization. For instance, compare the facorizations of 12 and 15. The factorization of 12 has two 2's, one 3, and no 5's. 15 has no 2's, one 3, and one 5.



                                                                  You can also show it with blocks: a cube has 3^3 blocks, a square has 3^2 blocks, a line has 3 blocks, and a single block is 3^0=1.






                                                                  share|cite|improve this answer
























                                                                    up vote
                                                                    1
                                                                    down vote













                                                                    If you multiply a number by 3 twice, then you get a number that is 9 times the original number. If you multiply a number by 3 no times, you get a number that is 1 times the original number. This also works for showing 3^1 = 3.



                                                                    This might be more clear with prime factorization. For instance, compare the facorizations of 12 and 15. The factorization of 12 has two 2's, one 3, and no 5's. 15 has no 2's, one 3, and one 5.



                                                                    You can also show it with blocks: a cube has 3^3 blocks, a square has 3^2 blocks, a line has 3 blocks, and a single block is 3^0=1.






                                                                    share|cite|improve this answer






















                                                                      up vote
                                                                      1
                                                                      down vote










                                                                      up vote
                                                                      1
                                                                      down vote









                                                                      If you multiply a number by 3 twice, then you get a number that is 9 times the original number. If you multiply a number by 3 no times, you get a number that is 1 times the original number. This also works for showing 3^1 = 3.



                                                                      This might be more clear with prime factorization. For instance, compare the facorizations of 12 and 15. The factorization of 12 has two 2's, one 3, and no 5's. 15 has no 2's, one 3, and one 5.



                                                                      You can also show it with blocks: a cube has 3^3 blocks, a square has 3^2 blocks, a line has 3 blocks, and a single block is 3^0=1.






                                                                      share|cite|improve this answer












                                                                      If you multiply a number by 3 twice, then you get a number that is 9 times the original number. If you multiply a number by 3 no times, you get a number that is 1 times the original number. This also works for showing 3^1 = 3.



                                                                      This might be more clear with prime factorization. For instance, compare the facorizations of 12 and 15. The factorization of 12 has two 2's, one 3, and no 5's. 15 has no 2's, one 3, and one 5.



                                                                      You can also show it with blocks: a cube has 3^3 blocks, a square has 3^2 blocks, a line has 3 blocks, and a single block is 3^0=1.







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                                                                      answered 11 hours ago









                                                                      Acccumulation

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                                                                          Below I explain in simple terms the (abstract) algebraic motivation behind the uniform extension of the power laws from positive powers to zero and negative powers. Most of the post is elementary, so if you encounter unfamiliar terms you can safely skip past them.



                                                                          $a^m+n = a^m a^n, $ i.e. $p(m!+!n) = p(m)p(n),$ for $,p(k) = a^kin Bbb Rbackslash 0,$ shows that powers $,a^Bbb N_+$ under multiplication have the same algebraic (semigroup) structure as the positive naturals $Bbb N_+$ under addition. If we enlarge $Bbb N_+$ to a monoid $Bbb N$ or group $Bbb Z$ by adjoining a neutral element $0$ along with additive inverses (negative integers) then there is a unique way of extending this structure preserving (hom) power map, namely:



                                                                          $$p(n) = p(0+n) = p(0)p(n) ,Rightarrow, p(0) = 1, rm i.e. a^0= 1$$



                                                                          $$1 = p(0) = p(-n+n) = p(-n)p(n),Rightarrow, p(-n) = p(n)^-1, rm i.e. a^-n = (a^n)^-1$$



                                                                          The fact that it proves very handy to use $0$ and negative integers when deducing facts about positive integers transfers to the isomorphic structure of powers $a^Bbb Z.,$ Because the power map on $,Bbb Z,$ is an extension of that on positive powers we are guaranteed that proofs about positive powers remain true even if the proof uses negative or zero powers, just as for proofs about positive integers that use negative integers and zero.



                                                                          For example using negative integers allows us to concisely state Bezout's lemma for the gcd, i.e. that $,gcd(m,n) = j m + k n,$ for some integers $j,k$ (which may be negative). In particular if $S$ is a group of integers (i.e. closed under subtraction) and it contains two coprime positives then it contains their gcd $= 1$. When translated to the isomorphic power form this says that if a set of powers is closed under division and it contains powers $a^m, a^n$ for coprime $m,n$ then it contains $a^1 = a,,$ e.g. see here where $a^m, a^n$ are integer matrices with determinant $= 1$. However, if we are restricted to positive powers and cancellation (vs. division) then analogous proofs may become much more cumbersome, and the key algebraic structure may become highly obfuscated by the manipulations needed to keep all powers positive, e.g. see here on proving $,a^m = b^m, a^n = b^n,Rightarrow, a = b,$ for integers $a,b,,$ Such enlargments to richer structures with $0$ and inverses allows us to work with objects in simpler forms that better highlight fundamental algebraic structure (here cyclic groups or principal ideals)



                                                                          This structure preservation principle is a key property that is employed when enlarging algebraic structures such as groups and rings. If the extended structure preserves the laws (axioms) of the base structure then everything we deduce about the base structure using the extended structure remains valid in the base structure. For example, to solve for integer or rational roots of quadratic and cubics we can employ well-known formulas. Even though these formula may employ complex numbers to solve for integer, rational or real roots, those result are valid in these base number systems because the proofs only employed (ring) axioms that remain valid in the base structures, e.g. associative, commutative, distributive laws. These single uniform formulas greatly simplify ancient methods where the quadratic and cubic formulas bifurcated into motley special cases to avoid untrusted "imaginary" or "negative" numbers, as well as (ab)surds (nowadays, using e.g. set-theoretic foundations, we know rigorous methods to construct such extended numbers systems in a way that proves they remain as consistent as the base number system). Further, as above, instead of appealing to ancient heuristics like the Hankel or Peacock Permanence Principle, we can use the axiomatic method to specify precisely what algebraic structure is preserved in extensions (e.g. the (semi)group structure underlying the power laws).






                                                                          share|cite|improve this answer


























                                                                            up vote
                                                                            1
                                                                            down vote













                                                                            Below I explain in simple terms the (abstract) algebraic motivation behind the uniform extension of the power laws from positive powers to zero and negative powers. Most of the post is elementary, so if you encounter unfamiliar terms you can safely skip past them.



                                                                            $a^m+n = a^m a^n, $ i.e. $p(m!+!n) = p(m)p(n),$ for $,p(k) = a^kin Bbb Rbackslash 0,$ shows that powers $,a^Bbb N_+$ under multiplication have the same algebraic (semigroup) structure as the positive naturals $Bbb N_+$ under addition. If we enlarge $Bbb N_+$ to a monoid $Bbb N$ or group $Bbb Z$ by adjoining a neutral element $0$ along with additive inverses (negative integers) then there is a unique way of extending this structure preserving (hom) power map, namely:



                                                                            $$p(n) = p(0+n) = p(0)p(n) ,Rightarrow, p(0) = 1, rm i.e. a^0= 1$$



                                                                            $$1 = p(0) = p(-n+n) = p(-n)p(n),Rightarrow, p(-n) = p(n)^-1, rm i.e. a^-n = (a^n)^-1$$



                                                                            The fact that it proves very handy to use $0$ and negative integers when deducing facts about positive integers transfers to the isomorphic structure of powers $a^Bbb Z.,$ Because the power map on $,Bbb Z,$ is an extension of that on positive powers we are guaranteed that proofs about positive powers remain true even if the proof uses negative or zero powers, just as for proofs about positive integers that use negative integers and zero.



                                                                            For example using negative integers allows us to concisely state Bezout's lemma for the gcd, i.e. that $,gcd(m,n) = j m + k n,$ for some integers $j,k$ (which may be negative). In particular if $S$ is a group of integers (i.e. closed under subtraction) and it contains two coprime positives then it contains their gcd $= 1$. When translated to the isomorphic power form this says that if a set of powers is closed under division and it contains powers $a^m, a^n$ for coprime $m,n$ then it contains $a^1 = a,,$ e.g. see here where $a^m, a^n$ are integer matrices with determinant $= 1$. However, if we are restricted to positive powers and cancellation (vs. division) then analogous proofs may become much more cumbersome, and the key algebraic structure may become highly obfuscated by the manipulations needed to keep all powers positive, e.g. see here on proving $,a^m = b^m, a^n = b^n,Rightarrow, a = b,$ for integers $a,b,,$ Such enlargments to richer structures with $0$ and inverses allows us to work with objects in simpler forms that better highlight fundamental algebraic structure (here cyclic groups or principal ideals)



                                                                            This structure preservation principle is a key property that is employed when enlarging algebraic structures such as groups and rings. If the extended structure preserves the laws (axioms) of the base structure then everything we deduce about the base structure using the extended structure remains valid in the base structure. For example, to solve for integer or rational roots of quadratic and cubics we can employ well-known formulas. Even though these formula may employ complex numbers to solve for integer, rational or real roots, those result are valid in these base number systems because the proofs only employed (ring) axioms that remain valid in the base structures, e.g. associative, commutative, distributive laws. These single uniform formulas greatly simplify ancient methods where the quadratic and cubic formulas bifurcated into motley special cases to avoid untrusted "imaginary" or "negative" numbers, as well as (ab)surds (nowadays, using e.g. set-theoretic foundations, we know rigorous methods to construct such extended numbers systems in a way that proves they remain as consistent as the base number system). Further, as above, instead of appealing to ancient heuristics like the Hankel or Peacock Permanence Principle, we can use the axiomatic method to specify precisely what algebraic structure is preserved in extensions (e.g. the (semi)group structure underlying the power laws).






                                                                            share|cite|improve this answer
























                                                                              up vote
                                                                              1
                                                                              down vote










                                                                              up vote
                                                                              1
                                                                              down vote









                                                                              Below I explain in simple terms the (abstract) algebraic motivation behind the uniform extension of the power laws from positive powers to zero and negative powers. Most of the post is elementary, so if you encounter unfamiliar terms you can safely skip past them.



                                                                              $a^m+n = a^m a^n, $ i.e. $p(m!+!n) = p(m)p(n),$ for $,p(k) = a^kin Bbb Rbackslash 0,$ shows that powers $,a^Bbb N_+$ under multiplication have the same algebraic (semigroup) structure as the positive naturals $Bbb N_+$ under addition. If we enlarge $Bbb N_+$ to a monoid $Bbb N$ or group $Bbb Z$ by adjoining a neutral element $0$ along with additive inverses (negative integers) then there is a unique way of extending this structure preserving (hom) power map, namely:



                                                                              $$p(n) = p(0+n) = p(0)p(n) ,Rightarrow, p(0) = 1, rm i.e. a^0= 1$$



                                                                              $$1 = p(0) = p(-n+n) = p(-n)p(n),Rightarrow, p(-n) = p(n)^-1, rm i.e. a^-n = (a^n)^-1$$



                                                                              The fact that it proves very handy to use $0$ and negative integers when deducing facts about positive integers transfers to the isomorphic structure of powers $a^Bbb Z.,$ Because the power map on $,Bbb Z,$ is an extension of that on positive powers we are guaranteed that proofs about positive powers remain true even if the proof uses negative or zero powers, just as for proofs about positive integers that use negative integers and zero.



                                                                              For example using negative integers allows us to concisely state Bezout's lemma for the gcd, i.e. that $,gcd(m,n) = j m + k n,$ for some integers $j,k$ (which may be negative). In particular if $S$ is a group of integers (i.e. closed under subtraction) and it contains two coprime positives then it contains their gcd $= 1$. When translated to the isomorphic power form this says that if a set of powers is closed under division and it contains powers $a^m, a^n$ for coprime $m,n$ then it contains $a^1 = a,,$ e.g. see here where $a^m, a^n$ are integer matrices with determinant $= 1$. However, if we are restricted to positive powers and cancellation (vs. division) then analogous proofs may become much more cumbersome, and the key algebraic structure may become highly obfuscated by the manipulations needed to keep all powers positive, e.g. see here on proving $,a^m = b^m, a^n = b^n,Rightarrow, a = b,$ for integers $a,b,,$ Such enlargments to richer structures with $0$ and inverses allows us to work with objects in simpler forms that better highlight fundamental algebraic structure (here cyclic groups or principal ideals)



                                                                              This structure preservation principle is a key property that is employed when enlarging algebraic structures such as groups and rings. If the extended structure preserves the laws (axioms) of the base structure then everything we deduce about the base structure using the extended structure remains valid in the base structure. For example, to solve for integer or rational roots of quadratic and cubics we can employ well-known formulas. Even though these formula may employ complex numbers to solve for integer, rational or real roots, those result are valid in these base number systems because the proofs only employed (ring) axioms that remain valid in the base structures, e.g. associative, commutative, distributive laws. These single uniform formulas greatly simplify ancient methods where the quadratic and cubic formulas bifurcated into motley special cases to avoid untrusted "imaginary" or "negative" numbers, as well as (ab)surds (nowadays, using e.g. set-theoretic foundations, we know rigorous methods to construct such extended numbers systems in a way that proves they remain as consistent as the base number system). Further, as above, instead of appealing to ancient heuristics like the Hankel or Peacock Permanence Principle, we can use the axiomatic method to specify precisely what algebraic structure is preserved in extensions (e.g. the (semi)group structure underlying the power laws).






                                                                              share|cite|improve this answer














                                                                              Below I explain in simple terms the (abstract) algebraic motivation behind the uniform extension of the power laws from positive powers to zero and negative powers. Most of the post is elementary, so if you encounter unfamiliar terms you can safely skip past them.



                                                                              $a^m+n = a^m a^n, $ i.e. $p(m!+!n) = p(m)p(n),$ for $,p(k) = a^kin Bbb Rbackslash 0,$ shows that powers $,a^Bbb N_+$ under multiplication have the same algebraic (semigroup) structure as the positive naturals $Bbb N_+$ under addition. If we enlarge $Bbb N_+$ to a monoid $Bbb N$ or group $Bbb Z$ by adjoining a neutral element $0$ along with additive inverses (negative integers) then there is a unique way of extending this structure preserving (hom) power map, namely:



                                                                              $$p(n) = p(0+n) = p(0)p(n) ,Rightarrow, p(0) = 1, rm i.e. a^0= 1$$



                                                                              $$1 = p(0) = p(-n+n) = p(-n)p(n),Rightarrow, p(-n) = p(n)^-1, rm i.e. a^-n = (a^n)^-1$$



                                                                              The fact that it proves very handy to use $0$ and negative integers when deducing facts about positive integers transfers to the isomorphic structure of powers $a^Bbb Z.,$ Because the power map on $,Bbb Z,$ is an extension of that on positive powers we are guaranteed that proofs about positive powers remain true even if the proof uses negative or zero powers, just as for proofs about positive integers that use negative integers and zero.



                                                                              For example using negative integers allows us to concisely state Bezout's lemma for the gcd, i.e. that $,gcd(m,n) = j m + k n,$ for some integers $j,k$ (which may be negative). In particular if $S$ is a group of integers (i.e. closed under subtraction) and it contains two coprime positives then it contains their gcd $= 1$. When translated to the isomorphic power form this says that if a set of powers is closed under division and it contains powers $a^m, a^n$ for coprime $m,n$ then it contains $a^1 = a,,$ e.g. see here where $a^m, a^n$ are integer matrices with determinant $= 1$. However, if we are restricted to positive powers and cancellation (vs. division) then analogous proofs may become much more cumbersome, and the key algebraic structure may become highly obfuscated by the manipulations needed to keep all powers positive, e.g. see here on proving $,a^m = b^m, a^n = b^n,Rightarrow, a = b,$ for integers $a,b,,$ Such enlargments to richer structures with $0$ and inverses allows us to work with objects in simpler forms that better highlight fundamental algebraic structure (here cyclic groups or principal ideals)



                                                                              This structure preservation principle is a key property that is employed when enlarging algebraic structures such as groups and rings. If the extended structure preserves the laws (axioms) of the base structure then everything we deduce about the base structure using the extended structure remains valid in the base structure. For example, to solve for integer or rational roots of quadratic and cubics we can employ well-known formulas. Even though these formula may employ complex numbers to solve for integer, rational or real roots, those result are valid in these base number systems because the proofs only employed (ring) axioms that remain valid in the base structures, e.g. associative, commutative, distributive laws. These single uniform formulas greatly simplify ancient methods where the quadratic and cubic formulas bifurcated into motley special cases to avoid untrusted "imaginary" or "negative" numbers, as well as (ab)surds (nowadays, using e.g. set-theoretic foundations, we know rigorous methods to construct such extended numbers systems in a way that proves they remain as consistent as the base number system). Further, as above, instead of appealing to ancient heuristics like the Hankel or Peacock Permanence Principle, we can use the axiomatic method to specify precisely what algebraic structure is preserved in extensions (e.g. the (semi)group structure underlying the power laws).







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                                                                              answered 6 hours ago









                                                                              Bill Dubuque

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