Power of two Cartesian product of given sets

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Let $A = 1,2$ and $B = a,b,c$, find $(A times B)^2$.




I found $(A times B) = (1,a),(1,b),(1,c),(2,a),(2,b),(2,c)$



But how do I find
$$(1,a),(1,b),(1,c),(2,a),(2,b),(2,c) times (1,a),(1,b),(1,c),(2,a),(2,b),(2,c) $$
?
Is it $(1,a,1,a),(1,a,1,b),(1,a,1,c), ldots$ ?


If I am wrong please show me the correct method.






elementary-set-theory products







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  • By $A*B$, do you mean $Atimes B$?
    – 5xum
    42 mins ago










  • Yeah , A into B.
    – Shehan Tearz
    35 mins ago














up vote
1
down vote

favorite
1













Let $A = 1,2$ and $B = a,b,c$, find $(A times B)^2$.




I found $(A times B) = (1,a),(1,b),(1,c),(2,a),(2,b),(2,c)$



But how do I find
$$(1,a),(1,b),(1,c),(2,a),(2,b),(2,c) times (1,a),(1,b),(1,c),(2,a),(2,b),(2,c) $$
?
Is it $(1,a,1,a),(1,a,1,b),(1,a,1,c), ldots$ ?


If I am wrong please show me the correct method.






elementary-set-theory products







share|cite|improve this question









New contributor




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



















  • By $A*B$, do you mean $Atimes B$?
    – 5xum
    42 mins ago










  • Yeah , A into B.
    – Shehan Tearz
    35 mins ago












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1






Let $A = 1,2$ and $B = a,b,c$, find $(A times B)^2$.




I found $(A times B) = (1,a),(1,b),(1,c),(2,a),(2,b),(2,c)$



But how do I find
$$(1,a),(1,b),(1,c),(2,a),(2,b),(2,c) times (1,a),(1,b),(1,c),(2,a),(2,b),(2,c) $$
?
Is it $(1,a,1,a),(1,a,1,b),(1,a,1,c), ldots$ ?


If I am wrong please show me the correct method.






elementary-set-theory products







share|cite|improve this question









New contributor




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












Let $A = 1,2$ and $B = a,b,c$, find $(A times B)^2$.




I found $(A times B) = (1,a),(1,b),(1,c),(2,a),(2,b),(2,c)$



But how do I find
$$(1,a),(1,b),(1,c),(2,a),(2,b),(2,c) times (1,a),(1,b),(1,c),(2,a),(2,b),(2,c) $$
?
Is it $(1,a,1,a),(1,a,1,b),(1,a,1,c), ldots$ ?


If I am wrong please show me the correct method.






elementary-set-theory products




elementary-set-theory products






share|cite|improve this question









New contributor




Shehan Tearz 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 question









New contributor




Shehan Tearz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 13 mins ago









Surb

36.8k84376




36.8k84376






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asked 43 mins ago









Shehan Tearz

234




234




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New contributor





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






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











  • By $A*B$, do you mean $Atimes B$?
    – 5xum
    42 mins ago










  • Yeah , A into B.
    – Shehan Tearz
    35 mins ago
















  • By $A*B$, do you mean $Atimes B$?
    – 5xum
    42 mins ago










  • Yeah , A into B.
    – Shehan Tearz
    35 mins ago















By $A*B$, do you mean $Atimes B$?
– 5xum
42 mins ago




By $A*B$, do you mean $Atimes B$?
– 5xum
42 mins ago












Yeah , A into B.
– Shehan Tearz
35 mins ago




Yeah , A into B.
– Shehan Tearz
35 mins ago










2 Answers
2






active

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up vote
3
down vote



accepted










You are wrong.



The set



$M:=(1,a),(1,b),(1,c),(2,a),(2,b),(2,c) times (1,a),(1,b),(1,c),(2,a),(2,b), (2,c)$



consists of pairs of pairs.



For example $((1,a), (2,b)) in M$.






share|cite|improve this answer
















  • 1




    Why is he wrong?
    – greedoid
    37 mins ago







  • 1




    @greedoid $(1,a,1,a)neq ((1,a),(1,a))$.
    – 5xum
    36 mins ago










  • Ah yes.........
    – greedoid
    34 mins ago

















up vote
2
down vote













You will get $36$ elements $$((1,a),(1,a)), ((1,a),(1,b)),..., ((2,c),(2,c))$$






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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    You are wrong.



    The set



    $M:=(1,a),(1,b),(1,c),(2,a),(2,b),(2,c) times (1,a),(1,b),(1,c),(2,a),(2,b), (2,c)$



    consists of pairs of pairs.



    For example $((1,a), (2,b)) in M$.






    share|cite|improve this answer
















    • 1




      Why is he wrong?
      – greedoid
      37 mins ago







    • 1




      @greedoid $(1,a,1,a)neq ((1,a),(1,a))$.
      – 5xum
      36 mins ago










    • Ah yes.........
      – greedoid
      34 mins ago














    up vote
    3
    down vote



    accepted










    You are wrong.



    The set



    $M:=(1,a),(1,b),(1,c),(2,a),(2,b),(2,c) times (1,a),(1,b),(1,c),(2,a),(2,b), (2,c)$



    consists of pairs of pairs.



    For example $((1,a), (2,b)) in M$.






    share|cite|improve this answer
















    • 1




      Why is he wrong?
      – greedoid
      37 mins ago







    • 1




      @greedoid $(1,a,1,a)neq ((1,a),(1,a))$.
      – 5xum
      36 mins ago










    • Ah yes.........
      – greedoid
      34 mins ago












    up vote
    3
    down vote



    accepted







    up vote
    3
    down vote



    accepted






    You are wrong.



    The set



    $M:=(1,a),(1,b),(1,c),(2,a),(2,b),(2,c) times (1,a),(1,b),(1,c),(2,a),(2,b), (2,c)$



    consists of pairs of pairs.



    For example $((1,a), (2,b)) in M$.






    share|cite|improve this answer












    You are wrong.



    The set



    $M:=(1,a),(1,b),(1,c),(2,a),(2,b),(2,c) times (1,a),(1,b),(1,c),(2,a),(2,b), (2,c)$



    consists of pairs of pairs.



    For example $((1,a), (2,b)) in M$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 39 mins ago









    Fred

    38.8k1238




    38.8k1238







    • 1




      Why is he wrong?
      – greedoid
      37 mins ago







    • 1




      @greedoid $(1,a,1,a)neq ((1,a),(1,a))$.
      – 5xum
      36 mins ago










    • Ah yes.........
      – greedoid
      34 mins ago












    • 1




      Why is he wrong?
      – greedoid
      37 mins ago







    • 1




      @greedoid $(1,a,1,a)neq ((1,a),(1,a))$.
      – 5xum
      36 mins ago










    • Ah yes.........
      – greedoid
      34 mins ago







    1




    1




    Why is he wrong?
    – greedoid
    37 mins ago





    Why is he wrong?
    – greedoid
    37 mins ago





    1




    1




    @greedoid $(1,a,1,a)neq ((1,a),(1,a))$.
    – 5xum
    36 mins ago




    @greedoid $(1,a,1,a)neq ((1,a),(1,a))$.
    – 5xum
    36 mins ago












    Ah yes.........
    – greedoid
    34 mins ago




    Ah yes.........
    – greedoid
    34 mins ago










    up vote
    2
    down vote













    You will get $36$ elements $$((1,a),(1,a)), ((1,a),(1,b)),..., ((2,c),(2,c))$$






    share|cite|improve this answer
























      up vote
      2
      down vote













      You will get $36$ elements $$((1,a),(1,a)), ((1,a),(1,b)),..., ((2,c),(2,c))$$






      share|cite|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        You will get $36$ elements $$((1,a),(1,a)), ((1,a),(1,b)),..., ((2,c),(2,c))$$






        share|cite|improve this answer












        You will get $36$ elements $$((1,a),(1,a)), ((1,a),(1,b)),..., ((2,c),(2,c))$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 15 mins ago









        Mohammad Riazi-Kermani

        32.3k41853




        32.3k41853




















            Shehan Tearz is a new contributor. Be nice, and check out our Code of Conduct.









             

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