# Binary relations/proofs

1. Apr 2, 2014

### Panphobia

1. The problem statement, all variables and given/known data

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. Apr 2, 2014

### Dick

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'?

3. Apr 2, 2014

### Panphobia

a function is injective iff its inverse is a function

4. Apr 2, 2014

### Dick

Ok, then use that definition to construct a proof by contradiction. What's the negation of the right hand side?

5. Apr 2, 2014

### Panphobia

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?

6. Apr 2, 2014

### Dick

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.

7. Apr 2, 2014

### Panphobia

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

8. Apr 3, 2014

### Panphobia

nevermind, I think I proved it, I get what to do now.