# Proving if $x<y$ then $\sqrt{x} < \sqrt{y}$

I am stuck on this homework problem!

Prove that if $x$ and $y$ are real numbers such that $0<x<y$ , then $\sqrt{x}<\sqrt{y}$.

This is in a chapter involving the least upper bound axiom and the Archimedean Principle, but I cannot figure how to use either of those to prove this! Any help would be most appreciated! Thanks

• Is there a part missing to the question? Apr 29 '13 at 23:35
• oops, I think you want to clarify your problem statement in your post...if $x, y > 0?$...and what is it to prove? Apr 29 '13 at 23:36
• Have you tried proving the contrapositive?
– wj32
Apr 29 '13 at 23:55

Hint : $y-x=(\sqrt{y}+\sqrt{x})(\sqrt{y}-\sqrt{x})$ Now look at the signs..
Suppose $0 < a < b$, multiplying by a gives us, $0 < \color{brown}{a^2 < ab}$.
Similarly, multiplying by $b$ gives $0 < \color{green}{ab < b^2}$ so $ab < b^2$. So from the brown and green, we have $0 < a < b \Leftrightarrow 0 < a^2 < b^2$.
Notice that your result is a corollary of this, let $x = a^2$ and $y = b^2$.
Hint: write it in contrapositive form - if $\sqrt{x}\geq \sqrt{y}$, then $x\geq y$ (assuming $x,y>0$).