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Is there a relationship between the difference of two numbers and the difference of their square roots? For example, can we say that

${| \sqrt x - \sqrt y|\leq |x - y|}$ when ${ x, y \geq 1 }$,

but

${| \sqrt x - \sqrt y|\geq |x - y|}$ when ${ x, y \leq 1 }$?

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  • $\begingroup$ Use the mean value theorem away from zero. $\endgroup$ – lhf Sep 30 '14 at 2:05
  • $\begingroup$ Ach, my math is rusty, but I'll look into that. $\endgroup$ – Isaac Kleinman Sep 30 '14 at 2:10
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Unless $x$ and $y$ are both $0$, we have $$|\sqrt{x}-\sqrt{y}|=\frac{|x-y|}{\sqrt{x}+\sqrt{y}}.$$ It follows that if $\sqrt{x}+\sqrt{y}\ge 1$ then we have $|\sqrt{x}-\sqrt{y}|\le |x-y|$, and if $\sqrt{x}+\sqrt{y}\le 1$ then we have $|\sqrt{x}-\sqrt{y}|\ge |x-y|$.

So the right condition is not quite the one in the post. If either $x\ge 1$ or $y\ge 1$, then the first inequality holds. But it also holds for certain $x$ and $y$ both below $1$, for example if both $x$ and $y$ are $\ge \frac{1}{4}$.

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If $x,y>0$ then $$(\sqrt x-\sqrt y)(\sqrt x+\sqrt y)=x-y$$ and so $$|\sqrt x-\sqrt y|=\frac{|x-y|}{|\sqrt x+\sqrt y|}\ .$$ So $|\sqrt x-\sqrt y|\le|x-y|$ if $\sqrt x+\sqrt y\ge1$, and the reverse if $\sqrt x+\sqrt y\le1$.

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$x$ and $n$ are real numbers, and $n$ is the distance between $x$ and (duh) ($x-n$).

$x^2-(x-n)^2=2nx-n^2$

Example: $26^2-22^2$ (without a calculator and you don't want to square $26$ and $22$)$= 2(4)(26)-4^2=192$

P.S. I'm in tenth grade, and I spent an hour figuring this out myself. This works especially well for very large numbers whose square roots are gross, and you don't have a calculator.

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    $\begingroup$ Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. For further information about writing math on this site see e.g. here, here, here and here. $\endgroup$ – user409521 Mar 31 '17 at 14:12
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    $\begingroup$ Congratulations for having investigated this two-year old Question and drawing a valid conclusion. However the Question already had an Accepted Answer, and you have not highlighted what new information you are adding. Indeed you did not directly address the crux of the Question (when we compare $|\sqrt{x} - \sqrt{y}|$ with $|x-y|$). $\endgroup$ – hardmath Mar 31 '17 at 15:40

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