Abstract Alegebra: Let <G,*> be the reals under....

[Sydney]

Let <G,*> be the group of real numbers under multiplication. Let <H,*> be a subset of G of the form H={a+b(square root of 2), a,b are rational numbers and a,b can not simultaneously be 0} Is H a subgroup of G with respect to multiplication? Justify

 

[Sydney]

The question did not specify but I do not think so.

 

[pka]

All one has to do to show that \(\displaystyle <H,*>\) is a subgroup of \(\displaystyle G\) is show that \(\displaystyle a \in H\quad \& \quad b \in H\quad \Rightarrow \quad ab \in H\quad \& \quad a^{ - 1} \in H.\) That is show that \(\displaystyle H\) close with respect to the operation and the inverse.

 

[Sydney]

I do not think that that is all that he wants. What about the H={a+b(square root of 2)? where a and b are rational numbers and are not simultaneously 0?[/tex]

 

[pka]

I do not think that that is all that he wants. What about the H={a+b(square root of 2)? where a and b are rational numbers and are not simultaneously 0?

What could you possibly mean by that?
This is a standard subgroup problem.'
To prove a subset of a group is itself a subgroup means showing it is closed under operation and inverse. That is all that has to be done.

 

[Sydney]

Thank you for your help. I guess sometimes we try to make things seem so much harder then they really are.