Multiplication via squaring and addition
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Is it possible to define multiplication of two positive integers only using addition and squaring? Of course I have $5 cdot 3 = 5 + 5 + 5$ but I would like something without saying do this $n$-times.
Peano Arithmetic has the following two axioms:
- $x cdot 0 = 0$
- $x cdot y = x cdot (y-1) + x$
So I could also write $3 cdot 5 = 3 cdot (5-1) + 3 = 3^2 + 3 + 3$ but again I don't "know" how often I need to apply the $2$nd axiom.
I tried a few things and noticed that one has:
$$2xy = (x+y)^2-x^2-y^2 text and 4xy = (x+y)^2-(x-y)^2$$
Close to $xy$ but still not what I am looking for. And actually this uses subtraction...
Edit: As clarified in the comments: I am asking how to define multiplication inside the structure $(mathbbN, +, cdot^2)$.
logic
asked Mar 4 at 15:26
SqyuliSqyuli
344111
$endgroup$
|
show 4 more comments
$begingroup$
Is it possible to define multiplication of two positive integers only using addition and squaring? Of course I have $5 cdot 3 = 5 + 5 + 5$ but I would like something without saying do this $n$-times.
Peano Arithmetic has the following two axioms:
- $x cdot 0 = 0$
- $x cdot y = x cdot (y-1) + x$
So I could also write $3 cdot 5 = 3 cdot (5-1) + 3 = 3^2 + 3 + 3$ but again I don't "know" how often I need to apply the $2$nd axiom.
I tried a few things and noticed that one has:
$$2xy = (x+y)^2-x^2-y^2 text and 4xy = (x+y)^2-(x-y)^2$$
Close to $xy$ but still not what I am looking for. And actually this uses subtraction...
Edit: As clarified in the comments: I am asking how to define multiplication inside the structure $(mathbbN, +, cdot^2)$.
logic
asked Mar 4 at 15:26
SqyuliSqyuli
344111
$endgroup$
2
$begingroup$
2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
$endgroup$
– Mauro ALLEGRANZA
Mar 4 at 15:48
2
$begingroup$
In order to define mult with squaring, you have to provide a def of squaring that does not rely of mult...
$endgroup$
– Mauro ALLEGRANZA
Mar 4 at 15:49
2
$begingroup$
Does one need to define squaring? In a simple case one could have a calculator with a squaring button but times button broken, maybe add still works.
$endgroup$
– coffeemath
Mar 4 at 16:06
4
$begingroup$
@MauroALLEGRANZA Or just take squaring as primitive - e.g. ask whether multiplication is definable in the structure $(mathbbN; +, cdot^2)$. That's a perfectly sensible question. (Actually, that specific question is a bit silly - $xy$ is the unique number $alpha$ satisfying $x^2+y^2+alpha+alpha=(x+y)^2$ - so to make the question nontrivial we have to pin down a more limited notion of "define." But my point stands.)
$endgroup$
– Noah Schweber
Mar 4 at 16:30
$begingroup$
@NoahSchweber Acutally that is my original question. I know that multiplication is definable in the structure (mathbbN, +, cdot^2) but not how. It is just written this "weird" because I wanted to avoid talking about structures.
$endgroup$
– Sqyuli
Mar 4 at 16:45
|
show 4 more comments
$begingroup$
Is it possible to define multiplication of two positive integers only using addition and squaring? Of course I have $5 cdot 3 = 5 + 5 + 5$ but I would like something without saying do this $n$-times.
Peano Arithmetic has the following two axioms:
- $x cdot 0 = 0$
- $x cdot y = x cdot (y-1) + x$
So I could also write $3 cdot 5 = 3 cdot (5-1) + 3 = 3^2 + 3 + 3$ but again I don't "know" how often I need to apply the $2$nd axiom.
I tried a few things and noticed that one has:
$$2xy = (x+y)^2-x^2-y^2 text and 4xy = (x+y)^2-(x-y)^2$$
Close to $xy$ but still not what I am looking for. And actually this uses subtraction...
Edit: As clarified in the comments: I am asking how to define multiplication inside the structure $(mathbbN, +, cdot^2)$.
logic
asked Mar 4 at 15:26
SqyuliSqyuli
344111
$endgroup$
Is it possible to define multiplication of two positive integers only using addition and squaring? Of course I have $5 cdot 3 = 5 + 5 + 5$ but I would like something without saying do this $n$-times.
Peano Arithmetic has the following two axioms:
- $x cdot 0 = 0$
- $x cdot y = x cdot (y-1) + x$
So I could also write $3 cdot 5 = 3 cdot (5-1) + 3 = 3^2 + 3 + 3$ but again I don't "know" how often I need to apply the $2$nd axiom.
I tried a few things and noticed that one has:
$$2xy = (x+y)^2-x^2-y^2 text and 4xy = (x+y)^2-(x-y)^2$$
Close to $xy$ but still not what I am looking for. And actually this uses subtraction...
Edit: As clarified in the comments: I am asking how to define multiplication inside the structure $(mathbbN, +, cdot^2)$.
logic
logic
asked Mar 4 at 15:26
SqyuliSqyuli
344111
asked Mar 4 at 15:26
SqyuliSqyuli
344111
edited Mar 4 at 16:52
Sqyuli
asked Mar 4 at 15:26
SqyuliSqyuli
344111
asked Mar 4 at 15:26
SqyuliSqyuli
344111
asked Mar 4 at 15:26
SqyuliSqyuli
344111
344111
2
$begingroup$
2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
$endgroup$
– Mauro ALLEGRANZA
Mar 4 at 15:48
2
$begingroup$
In order to define mult with squaring, you have to provide a def of squaring that does not rely of mult...
$endgroup$
– Mauro ALLEGRANZA
Mar 4 at 15:49
2
$begingroup$
Does one need to define squaring? In a simple case one could have a calculator with a squaring button but times button broken, maybe add still works.
$endgroup$
– coffeemath
Mar 4 at 16:06
4
$begingroup$
@MauroALLEGRANZA Or just take squaring as primitive - e.g. ask whether multiplication is definable in the structure $(mathbbN; +, cdot^2)$. That's a perfectly sensible question. (Actually, that specific question is a bit silly - $xy$ is the unique number $alpha$ satisfying $x^2+y^2+alpha+alpha=(x+y)^2$ - so to make the question nontrivial we have to pin down a more limited notion of "define." But my point stands.)
$endgroup$
– Noah Schweber
Mar 4 at 16:30
$begingroup$
@NoahSchweber Acutally that is my original question. I know that multiplication is definable in the structure (mathbbN, +, cdot^2) but not how. It is just written this "weird" because I wanted to avoid talking about structures.
$endgroup$
– Sqyuli
Mar 4 at 16:45
|
show 4 more comments
2
$begingroup$
2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
$endgroup$
– Mauro ALLEGRANZA
Mar 4 at 15:48
2
$begingroup$
In order to define mult with squaring, you have to provide a def of squaring that does not rely of mult...
$endgroup$
– Mauro ALLEGRANZA
Mar 4 at 15:49
2
$begingroup$
Does one need to define squaring? In a simple case one could have a calculator with a squaring button but times button broken, maybe add still works.
$endgroup$
– coffeemath
Mar 4 at 16:06
4
$begingroup$
@MauroALLEGRANZA Or just take squaring as primitive - e.g. ask whether multiplication is definable in the structure $(mathbbN; +, cdot^2)$. That's a perfectly sensible question. (Actually, that specific question is a bit silly - $xy$ is the unique number $alpha$ satisfying $x^2+y^2+alpha+alpha=(x+y)^2$ - so to make the question nontrivial we have to pin down a more limited notion of "define." But my point stands.)
$endgroup$
– Noah Schweber
Mar 4 at 16:30
$begingroup$
@NoahSchweber Acutally that is my original question. I know that multiplication is definable in the structure (mathbbN, +, cdot^2) but not how. It is just written this "weird" because I wanted to avoid talking about structures.
$endgroup$
– Sqyuli
Mar 4 at 16:45
2
2
$begingroup$
2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
$endgroup$
– Mauro ALLEGRANZA
Mar 4 at 15:48
$begingroup$
2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
$endgroup$
– Mauro ALLEGRANZA
Mar 4 at 15:48
2
2
$begingroup$
In order to define mult with squaring, you have to provide a def of squaring that does not rely of mult...
$endgroup$
– Mauro ALLEGRANZA
Mar 4 at 15:49
$begingroup$
In order to define mult with squaring, you have to provide a def of squaring that does not rely of mult...
$endgroup$
– Mauro ALLEGRANZA
Mar 4 at 15:49
2
2
$begingroup$
Does one need to define squaring? In a simple case one could have a calculator with a squaring button but times button broken, maybe add still works.
$endgroup$
– coffeemath
Mar 4 at 16:06
$begingroup$
Does one need to define squaring? In a simple case one could have a calculator with a squaring button but times button broken, maybe add still works.
$endgroup$
– coffeemath
Mar 4 at 16:06
4
4
$begingroup$
@MauroALLEGRANZA Or just take squaring as primitive - e.g. ask whether multiplication is definable in the structure $(mathbbN; +, cdot^2)$. That's a perfectly sensible question. (Actually, that specific question is a bit silly - $xy$ is the unique number $alpha$ satisfying $x^2+y^2+alpha+alpha=(x+y)^2$ - so to make the question nontrivial we have to pin down a more limited notion of "define." But my point stands.)
$endgroup$
– Noah Schweber
Mar 4 at 16:30
$begingroup$
@MauroALLEGRANZA Or just take squaring as primitive - e.g. ask whether multiplication is definable in the structure $(mathbbN; +, cdot^2)$. That's a perfectly sensible question. (Actually, that specific question is a bit silly - $xy$ is the unique number $alpha$ satisfying $x^2+y^2+alpha+alpha=(x+y)^2$ - so to make the question nontrivial we have to pin down a more limited notion of "define." But my point stands.)
$endgroup$
– Noah Schweber
Mar 4 at 16:30
$begingroup$
@NoahSchweber Acutally that is my original question. I know that multiplication is definable in the structure (mathbbN, +, cdot^2) but not how. It is just written this "weird" because I wanted to avoid talking about structures.
$endgroup$
– Sqyuli
Mar 4 at 16:45
$begingroup$
@NoahSchweber Acutally that is my original question. I know that multiplication is definable in the structure (mathbbN, +, cdot^2) but not how. It is just written this "weird" because I wanted to avoid talking about structures.
$endgroup$
– Sqyuli
Mar 4 at 16:45
|
show 4 more comments
1 Answer
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oldest
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$begingroup$
Per your comment, the precise question you're asking is:
Is multiplication definable in the structure $(mathbbN; +,cdot^2)$?
The answer is yes: we have $z=xcdot y$ iff $z+z+x^2+y^2=(x+y)^2$.
This is a bit unsatisfying; can we do better?
Well, one natural hope would be for a specific term built out of $+$ and $cdot^2$ which gives multiplication. E.g. raising to the fourth power isn't just definable, it's given by the term $(x^2)^2$. So we now ask:
Is the term $xy$ equivalent (in the obvious sense) to a term in the language $+,cdot^2$?
The answer to this new question is no. One way to see this is by taking derivatives. Suppose $t(x,y)$ is a term built out of $+$ and $cdot^2$. Then when we write $partialoverpartial xt(x,y)$ as a fully-cancelled-where-possible sum of monomials, every monomial in which $y$ occurs have even coefficient$^1$. But the monomial $xy$ itself doesn't have this property.$^2$
$^1$This takes proof, but it's a straightforward induction so I'll leave it to the reader.
$^2$OK fine, technically we need to prove that the fully-cancelled-sum-of-monomials form of a polynomial is unique, but meh - I'll leave it to the reader as well. Induction builds character.
answered Mar 4 at 17:08
Noah SchweberNoah Schweber
128k10152294
$endgroup$
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$begingroup$
Per your comment, the precise question you're asking is:
Is multiplication definable in the structure $(mathbbN; +,cdot^2)$?
The answer is yes: we have $z=xcdot y$ iff $z+z+x^2+y^2=(x+y)^2$.
This is a bit unsatisfying; can we do better?
Well, one natural hope would be for a specific term built out of $+$ and $cdot^2$ which gives multiplication. E.g. raising to the fourth power isn't just definable, it's given by the term $(x^2)^2$. So we now ask:
Is the term $xy$ equivalent (in the obvious sense) to a term in the language $+,cdot^2$?
The answer to this new question is no. One way to see this is by taking derivatives. Suppose $t(x,y)$ is a term built out of $+$ and $cdot^2$. Then when we write $partialoverpartial xt(x,y)$ as a fully-cancelled-where-possible sum of monomials, every monomial in which $y$ occurs have even coefficient$^1$. But the monomial $xy$ itself doesn't have this property.$^2$
$^1$This takes proof, but it's a straightforward induction so I'll leave it to the reader.
$^2$OK fine, technically we need to prove that the fully-cancelled-sum-of-monomials form of a polynomial is unique, but meh - I'll leave it to the reader as well. Induction builds character.
answered Mar 4 at 17:08
Noah SchweberNoah Schweber
128k10152294
$endgroup$
add a comment |
$begingroup$
Per your comment, the precise question you're asking is:
Is multiplication definable in the structure $(mathbbN; +,cdot^2)$?
The answer is yes: we have $z=xcdot y$ iff $z+z+x^2+y^2=(x+y)^2$.
This is a bit unsatisfying; can we do better?
Well, one natural hope would be for a specific term built out of $+$ and $cdot^2$ which gives multiplication. E.g. raising to the fourth power isn't just definable, it's given by the term $(x^2)^2$. So we now ask:
Is the term $xy$ equivalent (in the obvious sense) to a term in the language $+,cdot^2$?
The answer to this new question is no. One way to see this is by taking derivatives. Suppose $t(x,y)$ is a term built out of $+$ and $cdot^2$. Then when we write $partialoverpartial xt(x,y)$ as a fully-cancelled-where-possible sum of monomials, every monomial in which $y$ occurs have even coefficient$^1$. But the monomial $xy$ itself doesn't have this property.$^2$
$^1$This takes proof, but it's a straightforward induction so I'll leave it to the reader.
$^2$OK fine, technically we need to prove that the fully-cancelled-sum-of-monomials form of a polynomial is unique, but meh - I'll leave it to the reader as well. Induction builds character.
answered Mar 4 at 17:08
Noah SchweberNoah Schweber
128k10152294
$endgroup$
add a comment |
$begingroup$
Per your comment, the precise question you're asking is:
Is multiplication definable in the structure $(mathbbN; +,cdot^2)$?
The answer is yes: we have $z=xcdot y$ iff $z+z+x^2+y^2=(x+y)^2$.
This is a bit unsatisfying; can we do better?
Well, one natural hope would be for a specific term built out of $+$ and $cdot^2$ which gives multiplication. E.g. raising to the fourth power isn't just definable, it's given by the term $(x^2)^2$. So we now ask:
Is the term $xy$ equivalent (in the obvious sense) to a term in the language $+,cdot^2$?
The answer to this new question is no. One way to see this is by taking derivatives. Suppose $t(x,y)$ is a term built out of $+$ and $cdot^2$. Then when we write $partialoverpartial xt(x,y)$ as a fully-cancelled-where-possible sum of monomials, every monomial in which $y$ occurs have even coefficient$^1$. But the monomial $xy$ itself doesn't have this property.$^2$
$^1$This takes proof, but it's a straightforward induction so I'll leave it to the reader.
$^2$OK fine, technically we need to prove that the fully-cancelled-sum-of-monomials form of a polynomial is unique, but meh - I'll leave it to the reader as well. Induction builds character.
answered Mar 4 at 17:08
Noah SchweberNoah Schweber
128k10152294
$endgroup$
Per your comment, the precise question you're asking is:
Is multiplication definable in the structure $(mathbbN; +,cdot^2)$?
The answer is yes: we have $z=xcdot y$ iff $z+z+x^2+y^2=(x+y)^2$.
This is a bit unsatisfying; can we do better?
Well, one natural hope would be for a specific term built out of $+$ and $cdot^2$ which gives multiplication. E.g. raising to the fourth power isn't just definable, it's given by the term $(x^2)^2$. So we now ask:
Is the term $xy$ equivalent (in the obvious sense) to a term in the language $+,cdot^2$?
The answer to this new question is no. One way to see this is by taking derivatives. Suppose $t(x,y)$ is a term built out of $+$ and $cdot^2$. Then when we write $partialoverpartial xt(x,y)$ as a fully-cancelled-where-possible sum of monomials, every monomial in which $y$ occurs have even coefficient$^1$. But the monomial $xy$ itself doesn't have this property.$^2$
$^1$This takes proof, but it's a straightforward induction so I'll leave it to the reader.
$^2$OK fine, technically we need to prove that the fully-cancelled-sum-of-monomials form of a polynomial is unique, but meh - I'll leave it to the reader as well. Induction builds character.
answered Mar 4 at 17:08
Noah SchweberNoah Schweber
128k10152294
edited Mar 4 at 17:25
answered Mar 4 at 17:08
Noah SchweberNoah Schweber
128k10152294
answered Mar 4 at 17:08
Noah SchweberNoah Schweber
128k10152294
answered Mar 4 at 17:08
Noah SchweberNoah Schweber
128k10152294
128k10152294
add a comment |
add a comment |
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2
$begingroup$
2) must be $x cdot s(y)=x cdot y +x$, where $s(x)$ is the successor of $x$.
$endgroup$
– Mauro ALLEGRANZA
Mar 4 at 15:48
2
$begingroup$
In order to define mult with squaring, you have to provide a def of squaring that does not rely of mult...
$endgroup$
– Mauro ALLEGRANZA
Mar 4 at 15:49
2
$begingroup$
Does one need to define squaring? In a simple case one could have a calculator with a squaring button but times button broken, maybe add still works.
$endgroup$
– coffeemath
Mar 4 at 16:06
4
$begingroup$
@MauroALLEGRANZA Or just take squaring as primitive - e.g. ask whether multiplication is definable in the structure $(mathbbN; +, cdot^2)$. That's a perfectly sensible question. (Actually, that specific question is a bit silly - $xy$ is the unique number $alpha$ satisfying $x^2+y^2+alpha+alpha=(x+y)^2$ - so to make the question nontrivial we have to pin down a more limited notion of "define." But my point stands.)
$endgroup$
– Noah Schweber
Mar 4 at 16:30
$begingroup$
@NoahSchweber Acutally that is my original question. I know that multiplication is definable in the structure (mathbbN, +, cdot^2) but not how. It is just written this "weird" because I wanted to avoid talking about structures.
$endgroup$
– Sqyuli
Mar 4 at 16:45