# Prove x^2 < y^2

• Feb 2nd 2011, 04:38 PM
Noxide
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.
• Feb 2nd 2011, 04:43 PM
TheEmptySet
Quote:

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?
• Feb 2nd 2011, 04:44 PM
Plato
If we know that \$\displaystyle 0<x<y\$ then we know that \$\displaystyle x^2<xy\$ and \$\displaystyle xy<y^2\$.
• Feb 2nd 2011, 07:36 PM
Noxide
Haha, I had this on my page for the longest time. Thanks for pointing it out.