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Create an injection from {a^2+b^2

  1. Oct 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Is it possible to create an injection from {a^2+b^2 l a,b[itex]\in[/itex]Q} to {{a+b l a,b[itex]\in[/itex]Q}

    2. Relevant equations
    I am not sure about this. The problem is that many combinations of a,b can yield the same number, so how do I tackle this obstacle?


    3. The attempt at a solution
     
  2. jcsd
  3. Oct 1, 2013 #2

    UltrafastPED

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    Consider the definition of injection.
     
  4. Oct 1, 2013 #3
    the definition says:

    x1≠x2 => f(x1)≠f(x2)

    But the problem is as said that many combinations give rise to the same numbers in the set.
     
  5. Oct 1, 2013 #4

    UltrafastPED

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    Then perhaps it is not possible!
     
  6. Oct 1, 2013 #5
    it certainly should be, since the two sets have same cardinality. I just need help with coming up with a well defined function that does it. Problem is you can't really say something like f(a^2+b^2)=lal+lbl because of the earlier stated.
     
  7. Oct 1, 2013 #6
    Is this necessarily true?

    How about x → x?
     
  8. Oct 1, 2013 #7

    jbunniii

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    It might simplify the problem somewhat to observe that ##\{a+b \, | \, a,b\in \mathbb{Q}\}## is simply ##\mathbb{Q}##.
     
  9. Oct 1, 2013 #8
    I've just looked at a couple of your other posts like this one where again you seem to be confused about what you mean by {a^2+b^2 l a,b∈Q}. Can you see that all elements of this set are in ## \mathbb{Q} ##, and that a and b are only relevant in selecting which elements of ## \mathbb{Q} ## are elements of this set? You can't then define a function f(a,b) and expect it to have any meaning in relation to this set without defining how you get the unique pair (a,b) from ## x \in \mathbb{Q} ##.

    Before you try to think about jections (I hate that word) you need to think more clearly about the membership of sets and the domain, co-domain and range of relations between them.
     
    Last edited: Oct 1, 2013
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