Which Proof Should I Use for Binary Relations and Proofs Homework?

[Panphobia]

Homework Statement

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)?

 

[Dick]

Homework Statement

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)?

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'?

 

[Panphobia]

a function is injective iff its inverse is a function

 

[Dick]

a function is injective iff its inverse is a function

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

 

[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?

 

[Dick]

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?

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.

 

[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

 

[Panphobia]

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