Radicals

Motivating question. For $a\ne0$ and $r=\dfrac1n$ ($n\in N$), can $a^{1/n}$ be defined so that $y=a^{1/n}$ satisfies $y^n=a$? This is exactly asking to invert $y=x^n$.

Looking at the graphs of $f(x)=x^{2n}$ (even power) and $g(x)=x^{2n+1}$ (odd power) shows the two cases behave differently: $g$ is one-to-one and onto $R$, so it always has an inverse defined on all of $R$; $f$ is onto $[0,\infty)$ but NOT one-to-one on all of $R$ (both $y$ and $-y$ give the same $f$-value) -- it only becomes one-to-one once restricted to $x\ge0$.

Definition (the $n$th root/radical). (i) For $n$ even and $b>0$, there is a unique $a>0$ with $a^n=b$ (no real root exists if $b<0$, and if $y^n=a$ has a solution $y$, then $-y$ is also a solution). (ii) For $n$ odd and any $b\in R$, there is a unique $a\in R$ with $a^n=b$. In both cases, $a$ is called the $n$th root of $b$, written $b^{1/n}$ or $\sqrt[n]b$; $n=2$ is the square root, $n=3$ the cube root.

Rational exponents. For $r=\dfrac mn$ ($m\in Z,n\in N,\gcd(m,n)=1$) and $a>0$: $a^r=a^{m/n}=(a^{1/n})^m$. The exponent laws of §2.8.1 continue to hold for radicals/rational exponents wherever every individual term involved is defined (e.g. $(-49)^{3/2}$ has no real value, since $x^2=-49$ has no real solution).

Simplifying with rational exponents. $(x^{1/2}y^{-3})^{1/2}=x^{1/4}y^{-3/2}$ for $x,y\ge0$; $\sqrt{x^2-10x+25}=\sqrt{(x-5)^2}=|x-5|$ (never just $x-5$, since the base could be negative). …