Is it possible to multiply a set by a natural number? [closed]

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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.










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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
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 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















8












$begingroup$


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.










share|cite|improve this question











$endgroup$



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
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 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













8












8








8


1



$begingroup$


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.










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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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
If this question can be reworded to fit the rules in the help center, please edit the question.




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
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 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




    $begingroup$
    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










3 Answers
3






active

oldest

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23












$begingroup$

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 $Bbb Z=dots,-2,-1,0,1,2,dots$.






share|cite|improve this answer









$endgroup$




















    7












    $begingroup$

    Yes..you have already defined the operation..a scalar multiplication on a set.






    share|cite|improve this answer









    $endgroup$




















      6












      $begingroup$

      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.)






      share|cite|improve this answer









      $endgroup$



















        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        23












        $begingroup$

        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 $Bbb Z=dots,-2,-1,0,1,2,dots$.






        share|cite|improve this answer









        $endgroup$

















          23












          $begingroup$

          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 $Bbb Z=dots,-2,-1,0,1,2,dots$.






          share|cite|improve this answer









          $endgroup$















            23












            23








            23





            $begingroup$

            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 $Bbb Z=dots,-2,-1,0,1,2,dots$.






            share|cite|improve this answer









            $endgroup$



            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 $Bbb Z=dots,-2,-1,0,1,2,dots$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 8 at 5:04









            John DoeJohn Doe

            11.1k11238




            11.1k11238





















                7












                $begingroup$

                Yes..you have already defined the operation..a scalar multiplication on a set.






                share|cite|improve this answer









                $endgroup$

















                  7












                  $begingroup$

                  Yes..you have already defined the operation..a scalar multiplication on a set.






                  share|cite|improve this answer









                  $endgroup$















                    7












                    7








                    7





                    $begingroup$

                    Yes..you have already defined the operation..a scalar multiplication on a set.






                    share|cite|improve this answer









                    $endgroup$



                    Yes..you have already defined the operation..a scalar multiplication on a set.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 8 at 5:03









                    ershersh

                    320112




                    320112





















                        6












                        $begingroup$

                        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.)






                        share|cite|improve this answer









                        $endgroup$

















                          6












                          $begingroup$

                          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.)






                          share|cite|improve this answer









                          $endgroup$















                            6












                            6








                            6





                            $begingroup$

                            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.)






                            share|cite|improve this answer









                            $endgroup$



                            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.)







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 8 at 10:03









                            ComFreekComFreek

                            5321411




                            5321411












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