# Inequality: $x > \pm\sqrt{2}$

$x^2=2$

and $x=+\sqrt2$ and $x=-\sqrt2$.

That's ok.

But when $x^2>2$, (It's my fault. $x^2<2$ (inequality fault) => $x^2>2$ (ok))

$x > \pm\sqrt{2}$ ?

The answer is $x > \sqrt{2}$, and $x<-\sqrt{2}$.

When I was young, maybe I studied $x$ bigger than positive number, and smaller than negative number.

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I don't have time to give a comprehensive answer for you, but I would suggest you look at the following page: sosmath.com/algebra/inequalities/ineq01/ineq01.html. I also fixed the LaTeX for you. If you want to write "the square root of 2" in LaTeX the code is "\sqrt{2}", not "\root{2}". –  Samuel Reid Feb 13 '12 at 4:52
sorry and thank you so much Austin @Austinm , Samuelr . I'll learn to method for write in this site. –  KKH Feb 13 '12 at 5:16
The notation $x > \pm\sqrt{2}$ really doesn't make any sense. What exactly is it supposed to mean? Do you consider that statement to be true for $x=0$, for example? –  Hans Lundmark Feb 13 '12 at 5:41

From $$x^2\gt 2$$ we can conclude that $$\sqrt{x^2}\gt \sqrt{2}.$$ However, the important point to remember is that $\sqrt{x^2}$ is not equal to $x$, it is equal to $|x|$, the absolute value of $x$. That is, we have $$|x|\gt \sqrt{2}.$$

And, by the definition of the absolute value, $|x|\gt\sqrt{2}$ if and only if $x\gt 0$ and $x\gt \sqrt{2}$, or $x\lt 0$ and $-x\gt\sqrt{2}$, which is equivalent to $x\lt-\sqrt{2}$; so $$|x|\gt\sqrt{2}\text{ is equivalent to }x\gt\sqrt{2}\text{ or }x\lt-\sqrt{2}.$$ You either write it as two inequalities, with an "or" connective, or as the single inequality using the absolute value.

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aha, absolute value.. Thank you so much!! I want to be like you –  KKH Feb 13 '12 at 6:57

I think you have a typo: $x^2<2$ implies $-\sqrt2<x<\sqrt2$ which is fine, but the inequality $x^2>2$ means that either $x>\sqrt2$ or $x<-\sqrt2$, not both simultaneously (which is impossible). See for yourself that the graph of $y=x^2$ is above that of $y=2$ only outside the interval $[-\sqrt2,\sqrt2]$:

$\hskip 1.8in$

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@KKH anon got the above picture from wolframalpha.com it may help you in the future! –  john w. Feb 13 '12 at 5:04
Thank you everyone. But my question is that if $x^2>2$, then x>$\sqrt{2}$, $x<\sqrt{2}$. Your picture is great! –  KKH Feb 13 '12 at 5:22
@KKH: The statement if x²>2, then x>√2,x<√2 is both false and not a question. Could you help me to understand what it is you want to know or what exactly needs to be clarified for you? –  anon Feb 13 '12 at 5:30
@anon:sorry anon. $f(x)=x^2$ like your graph. To f(x)>2, $x>\sqrt{2}$, $x<-\sqrt{2}$. How to know it in equation? I don't know the step... –  KKH Feb 13 '12 at 5:33

If $x > \sqrt{2}$, then $x² > 2$. You seem to be having trouble with the difference between "smaller" and "more negative". A number's relative "size" is its absolute value, or its distance from zero.

Since the problem asks for $x² < 2$, the "less than" sign is unaffected and your desired $x$ is $x < \pm\sqrt{2}$.

If $x<-\sqrt{2} \Rightarrow x² < (-\sqrt{2})² =2.$ If $x < +\sqrt{2} \Rightarrow x² < (+\sqrt{2}) = 2$.

Good luck.

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This is how I always viewed this. Not sure if totally correct.

$x^2 > 2$ is similar to $|x|^2 > 2$, so $|x| > \sqrt{2}$, so $x < -\sqrt{2}$ or $x > \sqrt{2}$

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