Prove that is group is abelian

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Let G be a set of all non-zero real numbers.
And a * b = (ab/2).



We need to show (G,*) is an abelian group.



I know I need to show ab = ba, but that seems so trivial,



Assuming the group is abelian,



a * b = b* a



ab/2 = ba/2
Crossing out 2, we get ab = ba



Is this the correct answer ?



Multiplying both sides inverse of a and b results in 1 = 1 . Am I doing something wrong ? Or am I missing something?




abelian-groups




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  • 1




    Assuming that the group is abelian is not a good start to a proof that the group is abelian.
    – ancientmathematician
    2 days ago






  • 1




    I don't think it's correct to assume the thing you want to prove to be true.
    – poyea
    2 days ago











  • You're missing the fact that to show something is an Abelian group requires proving that it is a group.
    – Lord Shark the Unknown
    2 days ago










  • Then, How should I proceed? The person who taught me this used similar strategy. Can anyone point me in the right direction/
    – Kushal Niroula
    2 days ago










  • First attend to group axioms: associativity requires $a*(b*c)=(a*b)*c$. Then there's identity and inverses...
    – Lord Shark the Unknown
    2 days ago














up vote
-1
down vote

favorite












Let G be a set of all non-zero real numbers.
And a * b = (ab/2).



We need to show (G,*) is an abelian group.



I know I need to show ab = ba, but that seems so trivial,



Assuming the group is abelian,



a * b = b* a



ab/2 = ba/2
Crossing out 2, we get ab = ba



Is this the correct answer ?



Multiplying both sides inverse of a and b results in 1 = 1 . Am I doing something wrong ? Or am I missing something?




abelian-groups




share|cite|improve this question















  • 1




    Assuming that the group is abelian is not a good start to a proof that the group is abelian.
    – ancientmathematician
    2 days ago






  • 1




    I don't think it's correct to assume the thing you want to prove to be true.
    – poyea
    2 days ago











  • You're missing the fact that to show something is an Abelian group requires proving that it is a group.
    – Lord Shark the Unknown
    2 days ago










  • Then, How should I proceed? The person who taught me this used similar strategy. Can anyone point me in the right direction/
    – Kushal Niroula
    2 days ago










  • First attend to group axioms: associativity requires $a*(b*c)=(a*b)*c$. Then there's identity and inverses...
    – Lord Shark the Unknown
    2 days ago












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Let G be a set of all non-zero real numbers.
And a * b = (ab/2).



We need to show (G,*) is an abelian group.



I know I need to show ab = ba, but that seems so trivial,



Assuming the group is abelian,



a * b = b* a



ab/2 = ba/2
Crossing out 2, we get ab = ba



Is this the correct answer ?



Multiplying both sides inverse of a and b results in 1 = 1 . Am I doing something wrong ? Or am I missing something?




abelian-groups




share|cite|improve this question











Let G be a set of all non-zero real numbers.
And a * b = (ab/2).



We need to show (G,*) is an abelian group.



I know I need to show ab = ba, but that seems so trivial,



Assuming the group is abelian,



a * b = b* a



ab/2 = ba/2
Crossing out 2, we get ab = ba



Is this the correct answer ?



Multiplying both sides inverse of a and b results in 1 = 1 . Am I doing something wrong ? Or am I missing something?





abelian-groups





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked 2 days ago









Kushal Niroula

31




31







  • 1




    Assuming that the group is abelian is not a good start to a proof that the group is abelian.
    – ancientmathematician
    2 days ago






  • 1




    I don't think it's correct to assume the thing you want to prove to be true.
    – poyea
    2 days ago











  • You're missing the fact that to show something is an Abelian group requires proving that it is a group.
    – Lord Shark the Unknown
    2 days ago










  • Then, How should I proceed? The person who taught me this used similar strategy. Can anyone point me in the right direction/
    – Kushal Niroula
    2 days ago










  • First attend to group axioms: associativity requires $a*(b*c)=(a*b)*c$. Then there's identity and inverses...
    – Lord Shark the Unknown
    2 days ago












  • 1




    Assuming that the group is abelian is not a good start to a proof that the group is abelian.
    – ancientmathematician
    2 days ago






  • 1




    I don't think it's correct to assume the thing you want to prove to be true.
    – poyea
    2 days ago











  • You're missing the fact that to show something is an Abelian group requires proving that it is a group.
    – Lord Shark the Unknown
    2 days ago










  • Then, How should I proceed? The person who taught me this used similar strategy. Can anyone point me in the right direction/
    – Kushal Niroula
    2 days ago










  • First attend to group axioms: associativity requires $a*(b*c)=(a*b)*c$. Then there's identity and inverses...
    – Lord Shark the Unknown
    2 days ago







1




1




Assuming that the group is abelian is not a good start to a proof that the group is abelian.
– ancientmathematician
2 days ago




Assuming that the group is abelian is not a good start to a proof that the group is abelian.
– ancientmathematician
2 days ago




1




1




I don't think it's correct to assume the thing you want to prove to be true.
– poyea
2 days ago





I don't think it's correct to assume the thing you want to prove to be true.
– poyea
2 days ago













You're missing the fact that to show something is an Abelian group requires proving that it is a group.
– Lord Shark the Unknown
2 days ago




You're missing the fact that to show something is an Abelian group requires proving that it is a group.
– Lord Shark the Unknown
2 days ago












Then, How should I proceed? The person who taught me this used similar strategy. Can anyone point me in the right direction/
– Kushal Niroula
2 days ago




Then, How should I proceed? The person who taught me this used similar strategy. Can anyone point me in the right direction/
– Kushal Niroula
2 days ago












First attend to group axioms: associativity requires $a*(b*c)=(a*b)*c$. Then there's identity and inverses...
– Lord Shark the Unknown
2 days ago




First attend to group axioms: associativity requires $a*(b*c)=(a*b)*c$. Then there's identity and inverses...
– Lord Shark the Unknown
2 days ago










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We have that $$a*b=dfracab2=acdot bcdotdfrac12$$form the other side$$b*a=bcdot acdotdfrac12$$since real numbers with multiplication operator is an Abelian group (If we needed to show that the proof became more complicated) so is this group.






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    up vote
    0
    down vote



    accepted










    We have that $$a*b=dfracab2=acdot bcdotdfrac12$$form the other side$$b*a=bcdot acdotdfrac12$$since real numbers with multiplication operator is an Abelian group (If we needed to show that the proof became more complicated) so is this group.






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      We have that $$a*b=dfracab2=acdot bcdotdfrac12$$form the other side$$b*a=bcdot acdotdfrac12$$since real numbers with multiplication operator is an Abelian group (If we needed to show that the proof became more complicated) so is this group.






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        We have that $$a*b=dfracab2=acdot bcdotdfrac12$$form the other side$$b*a=bcdot acdotdfrac12$$since real numbers with multiplication operator is an Abelian group (If we needed to show that the proof became more complicated) so is this group.






        share|cite|improve this answer













        We have that $$a*b=dfracab2=acdot bcdotdfrac12$$form the other side$$b*a=bcdot acdotdfrac12$$since real numbers with multiplication operator is an Abelian group (If we needed to show that the proof became more complicated) so is this group.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered 2 days ago









        Mostafa Ayaz

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