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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 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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  1. anonymous
    • 4 years ago
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    i think this is not one to one

  2. anonymous
    • 4 years ago
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    for example \[\phi(2)=\phi(-2)\]

  3. anonymous
    • 4 years ago
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    and the mistake is in this line \[x^2=y^2\iff x=y\] should be \[x^2=y^2\iff x = \pm y\]

  4. anonymous
    • 4 years ago
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    also it is not onto since nothing gets mapped to negative rationals

  5. ChrisS
    • 4 years ago
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    right... I see I was totally off on this now. Thanks!

  6. anonymous
    • 4 years ago
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    yw

  7. ChrisS
    • 4 years ago
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    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 \[<M_{2}(\mathbb{R}),*>\ with <\mathbb{R},*> where\ \phi(A)\ is\ the\ determinant\ of\ matrix\ A\] This one is also not an isomorphism because multiplication of real numbers is commutative, where Matrix multiplication is not.

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

  9. anonymous
    • 4 years ago
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    it is a homomorphism however, because the determinant of the product is the product of the determinants

  10. anonymous
    • 4 years ago
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    that is \[\phi(AB)=\phi(A)\phi(B)\]

  11. anonymous
    • 4 years ago
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    i am assuming the " * " in both cases means multiplication yes?

  12. anonymous
    • 4 years ago
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    in any case it is certainly not injective because an infinite number of matrices can have the same determinant

  13. ChrisS
    • 4 years ago
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    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 :)

  14. anonymous
    • 4 years ago
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    no problem, i have medals to spare ( and i can't seem to redeem them at the package store)

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