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Binary relations/proofs

  1. Apr 2, 2014 #1
    1. The problem statement, all variables and given/known data
    2bu0x3.png


    3. The attempt at a solution
    I know what the question means and all, and I know a lot of different proofs. But I really don't know which proof to use for this question. I tried starting up a proof, but I don't exactly know how. Would a proof by contradiction work for a)?
     
  2. jcsd
  3. Apr 2, 2014 #2

    Dick

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    Yes, contradiction would work fine. Try it. Though it looks a lot like the definition of injective to me. What's your definition of 'injective'?
     
  4. Apr 2, 2014 #3
    a function is injective iff its inverse is a function
     
  5. Apr 2, 2014 #4

    Dick

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    Ok, then use that definition to construct a proof by contradiction. What's the negation of the right hand side?
     
  6. Apr 2, 2014 #5
    What do you mean by right hand side, by the way, I get proof by contradiction, but I don't understand the logical expression, like if you were to prove A->B then the contradiction would be ~A->(~B^B), How does that apply here?
     
  7. Apr 2, 2014 #6

    Dick

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    Um, think I'm using the wrong word here again. I mean proof by the contrapositive. You can prove A->B by proving ~B->~A. The right hand side is B, i.e. the for all (x1,x2) part.
     
  8. Apr 2, 2014 #7
    so There exists (x1,x2) in A^2 (f(x1) = f(x2)) ^ (x1 != x2)) -> f is not injective
    I am not sure about this part " in A^2" is it not in A^2 or in A^2
     
  9. Apr 3, 2014 #8
    nevermind, I think I proved it, I get what to do now.
     
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