# Relationship between Difference of Two Numbers and Their Square Roots

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 ${ x, y \leq 1 }$?

• Use the mean value theorem away from zero. – lhf Sep 30 '14 at 2:05
• Ach, my math is rusty, but I'll look into that. – Isaac Kleinman Sep 30 '14 at 2:10

## 3 Answers

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 $\ge \frac{1}{4}$.

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 $\sqrt x+\sqrt y\le1$.

$x$ 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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• 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 $|x-y|$). – hardmath Mar 31 '17 at 15:40