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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

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Is there a part missing to the question? –  vadim123 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? –  amWhy Apr 29 '13 at 23:36
    
Have you tried proving the contrapositive? –  wj32 Apr 29 '13 at 23:55

4 Answers 4

up vote 14 down vote accepted

Hint : $ y-x=(\sqrt{y}+\sqrt{x})(\sqrt{y}-\sqrt{x})$ Now look at the signs..

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2  
This idea popped into my mind immediately after reading the question. –  ncmathsadist Apr 30 '13 at 0:15

Hint: write it in contrapositive form - if $\sqrt{x}\geq \sqrt{y}$, then $x\geq y$ (assuming $x,y>0$).

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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 $y = b^2$.

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I don't have enough points — can't comment. Mhenni Benghorbal's aid is in green and under green line. Why not prove the question directly?

I started backwards —— from $\sqrt{x} < \sqrt{y} \quad (1)$ $\qquad \color{Green}{\text{this also means } x,y \geq 0 }$

Multiplication of (1) by $\sqrt{x}$ —— $\sqrt{x^2} < \sqrt{yx}$

Multiplication of (1) by $\sqrt{y}$ —— $\qquad \quad \sqrt{xy} < \sqrt{y^2}$

Hence have $\sqrt{x^2} < ... < \sqrt{y^2}.$ This $\implies |x| < |y|$. Now what to do? Thanks.

EDIT— The overhead proves $\sqrt{x} < \sqrt{y} \implies x < y$. But how to prove the original question directly? I reverse my steps overhead —
$0 \leq x < y \implies |x| < |y| \iff \sqrt{x^2} < \sqrt{y^2} $.
But $\quad \quad \quad \qquad \qquad \qquad \qquad \sqrt{x^2} < \sqrt{y^2} \implies??? \quad \sqrt{x} < \sqrt{y} $ ???

$\color{green}{—————————}$

From Mhenni Benghorbal's ---

$|x| < |y| => x < y $ (since x,y are non-negative)

{This proves the converse of what the OP asked, but it's good practice for you. Good job.}

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