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Homework Help: Algebra + Surjectivity

  1. Mar 19, 2006 #1
    (**under addition** not multiplication)
    in my question I am trying to prove that "the set of all integers (Z)" mapped to "the set of all rational numbers (Q)" is isomorphic
    ie. Z --> Q

    through out my work I have shown that yes it is injective, and it is a homomorphism, but I have not shown it is or isnt surjective.
    I think that it is not surjective, bucause the cardnality of Z is less than the cardnality of Q.
    ie. |Q| > |Z| (the set Q is much bigger than the set Z)

    so my question to you is, am i correct? and if you feel different about this, please explain to me what you think.
    thank you
  2. jcsd
  3. Mar 19, 2006 #2


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    I'm very confused by your presentation; I don't know what you're trying to do.

    But I can correct an error: |Q| and |Z| have the same cardinality. One way to see this is through one of Cantor's diagonal arguments.
  4. Mar 19, 2006 #3
    really, they have the same cardinality.... cause if they do then it is clearly surjective..because and injective map that has the same card is sujective
  5. Mar 19, 2006 #4
    and i have no idea what cantors diagnal arguments are..
  6. Mar 19, 2006 #5
    im am trying to show that Z is isomorphic to Q
    basically its bijective and its a homomorphism
  7. Mar 19, 2006 #6
    But what on earth is "it"? You have not given any particular function (the only kind of object that could reasonably be bijective and a homomorphism...) in any of your posts.
    Last edited: Mar 19, 2006
  8. Mar 19, 2006 #7
    there is no function... it is just the two maps
    the question is
    Are the additive groups Z and Q isomorphic.

    thats it..
    maybe one can say that
    y is mapped to y'/m
  9. Mar 19, 2006 #8


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    Assuming you meant "because an injective map between sets of the same cardinality is surjective", then this is only a theorem about finite sets.

    The classical example is that the set of even integers:

    ... -4, -2, 0, 2, 4, ...

    is clearly bijective with the set of integers

    ... -2, -1, 0, 1, 2, ...

    and yet, we have an obvious injective map from the even integers to the integers that is not surjective.

    Maps (of groups) are functions.
    Last edited: Mar 19, 2006
  10. Mar 19, 2006 #9
    oh yes your right for finite sets..!! .darn...
    well back to the drawing board
  11. Mar 19, 2006 #10
    Um, "map" is synonymous with "function". And what two maps? You claim to have proved that something ("it") is injective and a homomorphism, what precisely is "it"?
  12. Mar 19, 2006 #11
    ok sorry, this is not wourth it.. im wrong i the first place i screwed up..
    i meant two groups.. one map
    Z-->Q.. but clearly I was wrong from the beginning
  13. Mar 19, 2006 #12
    I'm only trying to help by attempting to get you to clarify what it is you've done!

    Suppose f: Q -> Z is a homomorphism. Could there possibly exist a rational number x such that f(x) = 1?
  14. Mar 19, 2006 #13


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    Suppose you have a homomorphism f : G -> H. If G is abelian, what can you say of f(G)? Now suppose you have a homomorphism f : Z -> Q. There is some property of Z that, like being abelian, carries over to f(Z). But this property is something Q doesn't have, so you know that f cannot be surjective. What this means is that no homomorphism from the integers to the rationals is surjective, meaning that the two cannot be isomorphic. So what is this key property?
  15. Mar 19, 2006 #14
    hmmmm.. im not sure what carries over to f(Z), but i do know that if Q did not include 0 and if it was under multiplication not addition then it would be commutative/abelian.
  16. Mar 19, 2006 #15


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    Q is an abelian group under addition, and Qx is an abelian group under multiplication. Keep thinking about what carries over to f(Z). Here's a hint: is it possible for there to be a homomorphism f from the integers to the rationals such that f(3) = 8/19, f(4) = 9/19? Why or why not? Or how about this: if I tell you that f(3) = 8/19, how many possibilities are there for f(4)? Are there none? Is there a unique possibility? If so, what is it? Are there many possibilities? If so, how many and what are they?
  17. Mar 20, 2006 #16

    matt grime

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    I'm not sure what you're getting at AKG, so I hope I"m not just repeating what you're doing.

    n=1+1..+1 (n 1s added up).

    Nuff said.
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