# Do inequalities hold under square-root (or exponentiation in general)?

This has been bothering me lately. My proof-skills are rusty (and were never great to begin with). I dimly recall having seen this (or something related to it) in a math course I took a while ago, but I can't remember any details now.

Obviously, if $a < b$ and $a$ and $b$ are reals >= 1 (or equal to 0), then $a^2 < b^2$, and hence $\sqrt a < \sqrt b$ (since the sqrt of a, b are in the same domain and squaring them would preserve the inequality and we know a must be < b to begin with).

What's not obvious to me is if this holds for real numbers in the range 0-1.

And what about the general case -- do inequalities on arbitrary real numbers hold under exponentiation? (Assuming the exponentiation yields defined results -- no square roots of negative numbers, for example.)

## 2 Answers

Let's restrict ourselves to positive numbers for now. If the exponent is positive, then yes, they do preserve inequalities. If the exponent is negative, then it reverses inequalities. If the exponent is zero, then of course everything maps to 1 so this is trivial.

The easiest way to think about it: the function $f(x)=x^\alpha$ is monotone increasing for $x>0$ if $\alpha>0$ and monotone decreasing for $x>0$ if $\alpha<0$.

• Ahh, thank you. This actually makes sense to me! Commented May 7, 2013 at 5:58
• @Cameron Made a few edits. I need to be careful with negative numbers. Commented May 7, 2013 at 5:59
• Right. I don't really care about negative numbers (since I'm taking square roots anyway), but it's good to note. Commented May 7, 2013 at 6:00

If $0\leq a<b$, then $aa\leq ab$ and $ab< bb$, hence $a^2<b^2$, which as you observed also proves $\sqrt a<\sqrt b$.

Similar reasoning applies to taking other positive integer powers, and hence to positive integer roots, thus all positive fractional exponents, and finally all positive exponents by taking limits.

• Ooh, I like this one too. Thanks! Commented May 7, 2013 at 6:03
• How do you go from positive integer roots to positive fractional? Commented Mar 18, 2023 at 0:37
• @user3180 by combining positive integer powers with positive integer roots, $(a^m)^{1/n} = a^{m/n}$. Commented Mar 27, 2023 at 6:49