# Proof Exercise II

1. Jul 31, 2013

### reenmachine

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

Suppose $x$ and $y$ are positive real numbers.If $x < y$ , then $x^2 < y^2$.

2. Relevant equations

Assume $x<y$ with $x,y \in ℝ^+$.This implies that $\exists z \in ℝ^+$ such that $x+z=y$.We have $y^2 = (x+z)^2 = x^2 + 2xz + z^2$ , and this proves that if $x < y$ with $x,y \in ℝ^+$ , then $x^2 < y^2$.

any thoughts on that one? Thank you!

Last edited: Jul 31, 2013
2. Jul 31, 2013

### tiny-tim

hi reenmachine!

yes, that looks fine

(alternatively, you could factor y2 - x2)

3. Jul 31, 2013

### hilbert2

You only need to know that if $x>0$, $y>0$ and $x<y$ then $x+y>0$ and $x-y<0$.

Also, if $a>0$ and $b<0$ then $ab<0$.

4. Jul 31, 2013

### pasmith

You need to explain why $2xz + z^2$ is strictly positive. It is obvious, but then so is the result you're trying to prove.

An alternative proof is that if $0 < x < y$ then, since multiplying by a positive number preserves inequalities, we have $x^2 < xy$ on multiplying by $x$ and $xy < y^2$ on multiplying by $y$. Putting these together we have $x^2 < xy < y^2$.

5. Jul 31, 2013

### reenmachine

Hi! :)

thank you for taking the time to respond!

6. Jul 31, 2013

### reenmachine

I thought the fact that $2xz + z^2$ is positive was obvious from the fact x and z are in $R^+$...

I like your way of doing it , pretty good road to the proof.

thank you!