Determine whether the given map \[\phi\] is an isomorphism of the first binary structure with the second. If it is not an isomorphism, why not?
\[<\mathbb{Q},*>\ with\ <\mathbb{Q},*>\ where\ \phi(x)=x^{2}\ for\ x \in \mathbb{Q}\]
I have the one-to-one requirement
\[If\ \phi(x)=\phi(y) then\ x^{2}=y^{2}\]
\[Then\ \sqrt{x^{2}}=\sqrt{y^{2}}\]
So x=y
Thus the map is one-to-one.
Now I need to show that it is onto and I need to show
\[\phi(x*y)=\phi(x)*\phi(y)\]

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- anonymous

i think this is not one to one

- anonymous

for example
\[\phi(2)=\phi(-2)\]

- anonymous

and the mistake is in this line
\[x^2=y^2\iff x=y\] should be
\[x^2=y^2\iff x = \pm y\]

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- anonymous

also it is not onto since nothing gets mapped to negative rationals

- ChrisS

right... I see I was totally off on this now. Thanks!

- anonymous

yw

- ChrisS

Hopefully you won't mind verifying one more for me... Same question but different binary structures... I think I have the answer but would like to run it by you if I could..
for this one the binary structure is
\[\ with <\mathbb{R},*> where\ \phi(A)\ is\ the\ determinant\ of\ matrix\ A\]
This one is also not an isomorphism because multiplication of real numbers is commutative, where Matrix multiplication is not.

- anonymous

no it is not an isomorphism because the set of two by two matrices is the same as
\[\mathbb{R^4}\]

- anonymous

it is a homomorphism however, because the determinant of the product is the product of the determinants

- anonymous

that is
\[\phi(AB)=\phi(A)\phi(B)\]

- anonymous

i am assuming the " * " in both cases means multiplication yes?

- anonymous

in any case it is certainly not injective because an infinite number of matrices can have the same determinant

- ChrisS

awesome, thanks again. We haven't got to homomorphisms yet, but I will note this for when we do.
Yes, I didn't see a symbol for using a dot for multiplication. I know it's confusing since * can be a generic binary operation symbol too but I thought it would make sense in the context of the question.
I do appreciate your help though. I'll add a medal to one of the other questions you have answered since I can't give you two here lol.
Thanks a bunch :)

- anonymous

no problem, i have medals to spare ( and i can't seem to redeem them at the package store)

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