Powers, Roots, and Signs

Definition For a real number $a$ and a positive integer $n$, the power $a^n$ is the product of $n$ factors equal to $a$,

The number $a$ is the base and $n$ is the exponent. For real $a$, $b$ and positive integers $m$, $n$ the rules $a^m a^n = a^{m + n}$, $(a^m)^n = a^{mn}$ and $(ab)^n = a^n b^n$ hold.

Definition An exponent applies only to the symbol or parenthesized group immediately before it. Powers are evaluated before multiplication and before negation, so

In the same way $2x^3 = 2 \cdot x^3$, which is different from $(2x)^3 = 8x^3$.

Definition For a nonzero real number $a$ and a positive integer $n$,

With these definitions the rules for integer exponents keep holding for all integers $m$ and $n$, as long as the base is nonzero. For a positive integer $n$ the expression $0^{-n}$ is undefined, since it would mean dividing by $0$. The symbol $0^0$ needs a separate word. In algebra the standard convention is $0^0 = 1$, so that a polynomial written as $a_0 x^0 + a_1 x^1 + a_2 x^2$ still has the value $a_0$ at $x = 0$. This convention says nothing about limits. In calculus a limit of the form $0^0$ is indeterminate, so its value has to be worked out and is not automatically $1$.

Definition For a real number $a \ge 0$, the principal square root $\sqrt{a}$ is the unique real number $b$ with $b \ge 0$ and $b^2 = a$. For $a < 0$ there is no real square root. So $\sqrt{a}$ is always nonnegative, and it is one number, not two.

Definition The absolute value of a real number $x$ is

For every real number $x$,

The reason is that $\lvert x \rvert \ge 0$ and $\lvert x \rvert^2 = x^2$, so $\lvert x \rvert$ is the nonnegative number whose square is $x^2$. The equation $\sqrt{x^2} = x$ holds only for $x \ge 0$.

Definition Let $n \ge 2$ be an integer. If $n$ is even and $a \ge 0$, the real root $\sqrt[n]{a}$ is the unique $b \ge 0$ with $b^n = a$, and for $a < 0$ there is no real even root. If $n$ is odd, every real $a$ has exactly one real $b$ with $b^n = a$, and $\sqrt[n]{a}$ is that number, which has the same sign as $a$. Consequently $\sqrt[n]{x^n} = \lvert x \rvert$ for even $n$ and $\sqrt[n]{x^n} = x$ for odd $n$.

Definition Let $a > 0$, let $m$ be an integer and let $n$ be a positive integer. Then

For a positive base the value does not depend on how the fraction $\frac{m}{n}$ is written, and the usual exponent rules hold for all rational exponents. For the base $0$ one sets $0^{m/n} = 0$ when $\frac{m}{n} > 0$, so that $x^{1/2} = \sqrt{x}$ for every $x \ge 0$.

Negative bases are left out on purpose. Some texts do define $(-8)^{1/3} = \sqrt[3]{-8} = -2$, using an odd denominator in lowest terms, but then the exponent rules break. Since $\frac{1}{3} = \frac{2}{6}$, the rule $a^{m/n} = \sqrt[n]{a^m}$ would also give

which is not $-2$. Calculators and software also disagree about such powers. So in this course a rational exponent always has a base $a > 0$, or $a = 0$ with a positive exponent, and an odd root of a negative number is written with the radical sign, as in $\sqrt[3]{-8}$.