Prove x^2 < y^2

**Noxide** · February 2nd 2011, 04:38 PM

The question:
x,y integers, 0 < x < y ==> x^2< y^2

I used closure of the positive integers to show that x^2 and y^2 are both greater than 0.

Before this question, I proved a few statements involving the transitivity of the < relation. I can prove the statement above quite easily using the induction postulate for the positive integers, but I would like to see a proof involving the transitivity of the order relation as I can't seem to come up with one on my own. Thanks.

**TheEmptySet** · February 2nd 2011, 04:43 PM

Originally Posted by Noxide

The question:
x,y integers, 0 < x < y ==> x^2< y^2

I used closure of the positive integers to show that x^2 and y^2 are both greater than 0.

Before this question, I proved a few statements involving the transitivity of the < relation. I can prove the statement above quite easily using the induction postulate for the positive integers, but I would like to see a proof involving the transitivity of the order relation as I can't seem to come up with one on my own. Thanks.