Is it possible to multiply a set by a natural number? [closed]
Clash Royale CLAN TAG#URR8PPP
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Say I have a set $S=1, 4, 10, 7$. Could I then multiply $S$ by $3$? Would that then look like $3S=3, 12, 30, 21$? Any help would be really appreciated.
discrete-mathematics elementary-set-theory
$endgroup$
closed as off-topic by BlueRaja - Danny Pflughoeft, DRF, Cesareo, amWhy, Adrian Keister Jan 8 at 19:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – DRF, amWhy, Adrian Keister
|
show 5 more comments
$begingroup$
Say I have a set $S=1, 4, 10, 7$. Could I then multiply $S$ by $3$? Would that then look like $3S=3, 12, 30, 21$? Any help would be really appreciated.
discrete-mathematics elementary-set-theory
$endgroup$
closed as off-topic by BlueRaja - Danny Pflughoeft, DRF, Cesareo, amWhy, Adrian Keister Jan 8 at 19:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – DRF, amWhy, Adrian Keister
12
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Yes you could just define $$na_i=na_i$$
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– clathratus
Jan 8 at 5:26
13
$begingroup$
You can certainly take a set of numbers $S= 1,4,10,7$ and say "Hey, I'm going to multiply every element by $3$ and get the set $3,12,20,21$ and I'm going to call that set $3S$". And you can say "I'm going to refer to that as multiplying a set by a number, any objections? No? Good."
$endgroup$
– fleablood
Jan 8 at 5:37
5
$begingroup$
I'm not sure what is being asked here. You can most certainly do anything you want in maths (assuming logically valid). Are you asking about usefulness of this construction?
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– freakish
Jan 8 at 7:45
22
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@fleablood I do have an objection since $3 times 10 = 30$
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– F.Carette
Jan 8 at 8:46
1
$begingroup$
Hello and welcome to Math Stack exchange. What is your background? Where did you come across this notation or problem? Especially for a problem like this context is very important.
$endgroup$
– DRF
Jan 8 at 14:26
|
show 5 more comments
$begingroup$
Say I have a set $S=1, 4, 10, 7$. Could I then multiply $S$ by $3$? Would that then look like $3S=3, 12, 30, 21$? Any help would be really appreciated.
discrete-mathematics elementary-set-theory
$endgroup$
Say I have a set $S=1, 4, 10, 7$. Could I then multiply $S$ by $3$? Would that then look like $3S=3, 12, 30, 21$? Any help would be really appreciated.
discrete-mathematics elementary-set-theory
discrete-mathematics elementary-set-theory
edited Jan 8 at 5:26
clathratus
3,730333
3,730333
asked Jan 8 at 5:01
Hunter KimuraHunter Kimura
502
502
closed as off-topic by BlueRaja - Danny Pflughoeft, DRF, Cesareo, amWhy, Adrian Keister Jan 8 at 19:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – DRF, amWhy, Adrian Keister
closed as off-topic by BlueRaja - Danny Pflughoeft, DRF, Cesareo, amWhy, Adrian Keister Jan 8 at 19:57
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – DRF, amWhy, Adrian Keister
12
$begingroup$
Yes you could just define $$na_i=na_i$$
$endgroup$
– clathratus
Jan 8 at 5:26
13
$begingroup$
You can certainly take a set of numbers $S= 1,4,10,7$ and say "Hey, I'm going to multiply every element by $3$ and get the set $3,12,20,21$ and I'm going to call that set $3S$". And you can say "I'm going to refer to that as multiplying a set by a number, any objections? No? Good."
$endgroup$
– fleablood
Jan 8 at 5:37
5
$begingroup$
I'm not sure what is being asked here. You can most certainly do anything you want in maths (assuming logically valid). Are you asking about usefulness of this construction?
$endgroup$
– freakish
Jan 8 at 7:45
22
$begingroup$
@fleablood I do have an objection since $3 times 10 = 30$
$endgroup$
– F.Carette
Jan 8 at 8:46
1
$begingroup$
Hello and welcome to Math Stack exchange. What is your background? Where did you come across this notation or problem? Especially for a problem like this context is very important.
$endgroup$
– DRF
Jan 8 at 14:26
|
show 5 more comments
12
$begingroup$
Yes you could just define $$na_i=na_i$$
$endgroup$
– clathratus
Jan 8 at 5:26
13
$begingroup$
You can certainly take a set of numbers $S= 1,4,10,7$ and say "Hey, I'm going to multiply every element by $3$ and get the set $3,12,20,21$ and I'm going to call that set $3S$". And you can say "I'm going to refer to that as multiplying a set by a number, any objections? No? Good."
$endgroup$
– fleablood
Jan 8 at 5:37
5
$begingroup$
I'm not sure what is being asked here. You can most certainly do anything you want in maths (assuming logically valid). Are you asking about usefulness of this construction?
$endgroup$
– freakish
Jan 8 at 7:45
22
$begingroup$
@fleablood I do have an objection since $3 times 10 = 30$
$endgroup$
– F.Carette
Jan 8 at 8:46
1
$begingroup$
Hello and welcome to Math Stack exchange. What is your background? Where did you come across this notation or problem? Especially for a problem like this context is very important.
$endgroup$
– DRF
Jan 8 at 14:26
12
12
$begingroup$
Yes you could just define $$na_i=na_i$$
$endgroup$
– clathratus
Jan 8 at 5:26
$begingroup$
Yes you could just define $$na_i=na_i$$
$endgroup$
– clathratus
Jan 8 at 5:26
13
13
$begingroup$
You can certainly take a set of numbers $S= 1,4,10,7$ and say "Hey, I'm going to multiply every element by $3$ and get the set $3,12,20,21$ and I'm going to call that set $3S$". And you can say "I'm going to refer to that as multiplying a set by a number, any objections? No? Good."
$endgroup$
– fleablood
Jan 8 at 5:37
$begingroup$
You can certainly take a set of numbers $S= 1,4,10,7$ and say "Hey, I'm going to multiply every element by $3$ and get the set $3,12,20,21$ and I'm going to call that set $3S$". And you can say "I'm going to refer to that as multiplying a set by a number, any objections? No? Good."
$endgroup$
– fleablood
Jan 8 at 5:37
5
5
$begingroup$
I'm not sure what is being asked here. You can most certainly do anything you want in maths (assuming logically valid). Are you asking about usefulness of this construction?
$endgroup$
– freakish
Jan 8 at 7:45
$begingroup$
I'm not sure what is being asked here. You can most certainly do anything you want in maths (assuming logically valid). Are you asking about usefulness of this construction?
$endgroup$
– freakish
Jan 8 at 7:45
22
22
$begingroup$
@fleablood I do have an objection since $3 times 10 = 30$
$endgroup$
– F.Carette
Jan 8 at 8:46
$begingroup$
@fleablood I do have an objection since $3 times 10 = 30$
$endgroup$
– F.Carette
Jan 8 at 8:46
1
1
$begingroup$
Hello and welcome to Math Stack exchange. What is your background? Where did you come across this notation or problem? Especially for a problem like this context is very important.
$endgroup$
– DRF
Jan 8 at 14:26
$begingroup$
Hello and welcome to Math Stack exchange. What is your background? Where did you come across this notation or problem? Especially for a problem like this context is very important.
$endgroup$
– DRF
Jan 8 at 14:26
|
show 5 more comments
3 Answers
3
active
oldest
votes
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Sure, we sometimes for example denote the set of even integers by $2Bbb Z=dots,-4,-2,0,2,4,dots$, while the set of integers is $Bbb Z=dots,-2,-1,0,1,2,dots$.
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add a comment |
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Yes..you have already defined the operation..a scalar multiplication on a set.
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add a comment |
$begingroup$
In fact, this is closely related to images of functions:
Let $f: A to B$ be a function between two sets and $A' subseteq A$, then we define
$$f[A'] := a in A'quad left(= b in B right)$$
Your idea is a specific instance:
Let $A' = 1, 4, 10, 7 subseteq mathbbN =: A$ and $f: mathbbN to mathbbN$ be the operation "multiply by three", then
$$f[A'] = 3, 12, 30, 21$$
The other answer hints at the notation $NA'$, e.g. $3A'$.
Note that in general only $|A'| geq |f[A']|$ holds, e.g. consider $0mathbbZ = 0$. However, with this "function framework", it is easy to state a criterion for when the equality holds:
If $f$ is injective on $A'$, then $|A'| = |f[A']|$.
Or in words: "If you map different elements in $A'$ to different elements in $B$, then certainly, we cannot lose elements due to duplicates in the set $f[A']$."
Can you see why $|2mathbbZ| = |mathbbZ|$? This proves "there are as many even numbers as integers". (Of course, you have to first define the cardinality function $|cdot|$ for arbitrary infinite sets first.)
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Sure, we sometimes for example denote the set of even integers by $2Bbb Z=dots,-4,-2,0,2,4,dots$, while the set of integers is $Bbb Z=dots,-2,-1,0,1,2,dots$.
$endgroup$
add a comment |
$begingroup$
Sure, we sometimes for example denote the set of even integers by $2Bbb Z=dots,-4,-2,0,2,4,dots$, while the set of integers is $Bbb Z=dots,-2,-1,0,1,2,dots$.
$endgroup$
add a comment |
$begingroup$
Sure, we sometimes for example denote the set of even integers by $2Bbb Z=dots,-4,-2,0,2,4,dots$, while the set of integers is $Bbb Z=dots,-2,-1,0,1,2,dots$.
$endgroup$
Sure, we sometimes for example denote the set of even integers by $2Bbb Z=dots,-4,-2,0,2,4,dots$, while the set of integers is $Bbb Z=dots,-2,-1,0,1,2,dots$.
answered Jan 8 at 5:04
John DoeJohn Doe
11.1k11238
11.1k11238
add a comment |
add a comment |
$begingroup$
Yes..you have already defined the operation..a scalar multiplication on a set.
$endgroup$
add a comment |
$begingroup$
Yes..you have already defined the operation..a scalar multiplication on a set.
$endgroup$
add a comment |
$begingroup$
Yes..you have already defined the operation..a scalar multiplication on a set.
$endgroup$
Yes..you have already defined the operation..a scalar multiplication on a set.
answered Jan 8 at 5:03
ershersh
320112
320112
add a comment |
add a comment |
$begingroup$
In fact, this is closely related to images of functions:
Let $f: A to B$ be a function between two sets and $A' subseteq A$, then we define
$$f[A'] := a in A'quad left(= b in B right)$$
Your idea is a specific instance:
Let $A' = 1, 4, 10, 7 subseteq mathbbN =: A$ and $f: mathbbN to mathbbN$ be the operation "multiply by three", then
$$f[A'] = 3, 12, 30, 21$$
The other answer hints at the notation $NA'$, e.g. $3A'$.
Note that in general only $|A'| geq |f[A']|$ holds, e.g. consider $0mathbbZ = 0$. However, with this "function framework", it is easy to state a criterion for when the equality holds:
If $f$ is injective on $A'$, then $|A'| = |f[A']|$.
Or in words: "If you map different elements in $A'$ to different elements in $B$, then certainly, we cannot lose elements due to duplicates in the set $f[A']$."
Can you see why $|2mathbbZ| = |mathbbZ|$? This proves "there are as many even numbers as integers". (Of course, you have to first define the cardinality function $|cdot|$ for arbitrary infinite sets first.)
$endgroup$
add a comment |
$begingroup$
In fact, this is closely related to images of functions:
Let $f: A to B$ be a function between two sets and $A' subseteq A$, then we define
$$f[A'] := a in A'quad left(= b in B right)$$
Your idea is a specific instance:
Let $A' = 1, 4, 10, 7 subseteq mathbbN =: A$ and $f: mathbbN to mathbbN$ be the operation "multiply by three", then
$$f[A'] = 3, 12, 30, 21$$
The other answer hints at the notation $NA'$, e.g. $3A'$.
Note that in general only $|A'| geq |f[A']|$ holds, e.g. consider $0mathbbZ = 0$. However, with this "function framework", it is easy to state a criterion for when the equality holds:
If $f$ is injective on $A'$, then $|A'| = |f[A']|$.
Or in words: "If you map different elements in $A'$ to different elements in $B$, then certainly, we cannot lose elements due to duplicates in the set $f[A']$."
Can you see why $|2mathbbZ| = |mathbbZ|$? This proves "there are as many even numbers as integers". (Of course, you have to first define the cardinality function $|cdot|$ for arbitrary infinite sets first.)
$endgroup$
add a comment |
$begingroup$
In fact, this is closely related to images of functions:
Let $f: A to B$ be a function between two sets and $A' subseteq A$, then we define
$$f[A'] := a in A'quad left(= b in B right)$$
Your idea is a specific instance:
Let $A' = 1, 4, 10, 7 subseteq mathbbN =: A$ and $f: mathbbN to mathbbN$ be the operation "multiply by three", then
$$f[A'] = 3, 12, 30, 21$$
The other answer hints at the notation $NA'$, e.g. $3A'$.
Note that in general only $|A'| geq |f[A']|$ holds, e.g. consider $0mathbbZ = 0$. However, with this "function framework", it is easy to state a criterion for when the equality holds:
If $f$ is injective on $A'$, then $|A'| = |f[A']|$.
Or in words: "If you map different elements in $A'$ to different elements in $B$, then certainly, we cannot lose elements due to duplicates in the set $f[A']$."
Can you see why $|2mathbbZ| = |mathbbZ|$? This proves "there are as many even numbers as integers". (Of course, you have to first define the cardinality function $|cdot|$ for arbitrary infinite sets first.)
$endgroup$
In fact, this is closely related to images of functions:
Let $f: A to B$ be a function between two sets and $A' subseteq A$, then we define
$$f[A'] := a in A'quad left(= b in B right)$$
Your idea is a specific instance:
Let $A' = 1, 4, 10, 7 subseteq mathbbN =: A$ and $f: mathbbN to mathbbN$ be the operation "multiply by three", then
$$f[A'] = 3, 12, 30, 21$$
The other answer hints at the notation $NA'$, e.g. $3A'$.
Note that in general only $|A'| geq |f[A']|$ holds, e.g. consider $0mathbbZ = 0$. However, with this "function framework", it is easy to state a criterion for when the equality holds:
If $f$ is injective on $A'$, then $|A'| = |f[A']|$.
Or in words: "If you map different elements in $A'$ to different elements in $B$, then certainly, we cannot lose elements due to duplicates in the set $f[A']$."
Can you see why $|2mathbbZ| = |mathbbZ|$? This proves "there are as many even numbers as integers". (Of course, you have to first define the cardinality function $|cdot|$ for arbitrary infinite sets first.)
answered Jan 8 at 10:03
ComFreekComFreek
5321411
5321411
add a comment |
add a comment |
12
$begingroup$
Yes you could just define $$na_i=na_i$$
$endgroup$
– clathratus
Jan 8 at 5:26
13
$begingroup$
You can certainly take a set of numbers $S= 1,4,10,7$ and say "Hey, I'm going to multiply every element by $3$ and get the set $3,12,20,21$ and I'm going to call that set $3S$". And you can say "I'm going to refer to that as multiplying a set by a number, any objections? No? Good."
$endgroup$
– fleablood
Jan 8 at 5:37
5
$begingroup$
I'm not sure what is being asked here. You can most certainly do anything you want in maths (assuming logically valid). Are you asking about usefulness of this construction?
$endgroup$
– freakish
Jan 8 at 7:45
22
$begingroup$
@fleablood I do have an objection since $3 times 10 = 30$
$endgroup$
– F.Carette
Jan 8 at 8:46
1
$begingroup$
Hello and welcome to Math Stack exchange. What is your background? Where did you come across this notation or problem? Especially for a problem like this context is very important.
$endgroup$
– DRF
Jan 8 at 14:26