# Thread: Prove x^2 < y^2

1. ## Prove x^2 < y^2

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.

2. 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.
Hint:

$\displaystyle x < y \implies (x)x <(x)y \iff x^2 < xy$

Can you finish from here?

3. If we know that $\displaystyle 0<x<y$ then we know that $\displaystyle x^2<xy$ and $\displaystyle xy<y^2$.

4. Haha, I had this on my page for the longest time. Thanks for pointing it out.