A community for students.
Here's the question you clicked on:
 0 viewing
ChrisS
 4 years ago
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 onetoone 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 onetoone.
Now I need to show that it is onto and I need to show
\[\phi(x*y)=\phi(x)*\phi(y)\]
ChrisS
 4 years ago
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 onetoone 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 onetoone. Now I need to show that it is onto and I need to show \[\phi(x*y)=\phi(x)*\phi(y)\]

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think this is not one to one

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for example \[\phi(2)=\phi(2)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and the mistake is in this line \[x^2=y^2\iff x=y\] should be \[x^2=y^2\iff x = \pm y\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0also it is not onto since nothing gets mapped to negative rationals

ChrisS
 4 years ago
Best ResponseYou've already chosen the best response.0right... I see I was totally off on this now. Thanks!

ChrisS
 4 years ago
Best ResponseYou've already chosen the best response.0Hopefully 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 \[<M_{2}(\mathbb{R}),*>\ with <\mathbb{R},*> where\ \phi(A)\ is\ the\ determinant\ of\ matrix\ A\] This one is also not an isomorphism because multiplication of real numbers is commutative, where Matrix multiplication is not.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no it is not an isomorphism because the set of two by two matrices is the same as \[\mathbb{R^4}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is a homomorphism however, because the determinant of the product is the product of the determinants

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is \[\phi(AB)=\phi(A)\phi(B)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am assuming the " * " in both cases means multiplication yes?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in any case it is certainly not injective because an infinite number of matrices can have the same determinant

ChrisS
 4 years ago
Best ResponseYou've already chosen the best response.0awesome, 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
 4 years ago
Best ResponseYou've already chosen the best response.0no problem, i have medals to spare ( and i can't seem to redeem them at the package store)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.