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

Question

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.)

Answer 1

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 $\alpha<0$.

Answer 2

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.