# Proof Exercise II

Gold Member

## Homework Statement

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

## Homework 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!

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tiny-tim
Homework Helper
hi reenmachine!

yes, that looks fine

(alternatively, you could factor y2 - x2)

hilbert2
Gold Member
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##.

pasmith
Homework Helper
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 $x^2 < xy < y^2$.

Gold Member
hi reenmachine!

yes, that looks fine

(alternatively, you could factor y2 - x2)
Hi! :)

thank you for taking the time to respond!

Gold Member
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 $x^2 < xy < y^2$.
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!