Lecture · 14 min · basic

Exponents and roots

Repeated multiplication, then the inverse

Euler · powers and roots

If $a^{3}$ is $a \cdot a \cdot a$ , then $a^{0}$ cannot be “multiply $a$ zero times.” The laws of exponents force $a^{0} = 1$ (for $a \neq = 0$ ) the same way they force $a^{- 1} = 1/ a$ . Definitions follow the laws, not the other way around.

On the board today

—Use $a^{m} a^{n} = a^{m + n}$ as the generating law.
—See $a^{1/ n}$ as the inverse operation of raising to $n$ .
—Rewrite radicals as rational exponents.

The generating law

a^{m} a^{n} = a^{m + n}, (a^{m})^{n} = a^{mn}, a^{- n} = \frac{1}{a ^{n}}

Why $a^{0} = 1$ : $a^{n} a^{0} = a^{n + 0} = a^{n}$ , so $a^{0}$ is the multiplicative identity. Why $a^{1/2} = a$ : $(a^{1/2})^{2} = a^{1} = a$ , so $a^{1/2}$ is a square root. For $a > 0$ we take the positive one.

Worked example · Simplify a nest

Simplify $\frac{8 ^{2/3}}{4 ^{- 1/2}}$ .

1. $8 = 2^{3}$ , so $8^{2/3} = (2^{3})^{2/3} = 2^{2} = 4$ .
2. $4^{- 1/2} = (2^{2})^{- 1/2} = 2^{- 1} = \frac{1}{2}$ .
$\frac{8 ^{2/3}}{4 ^{- 1/2}} = \frac{4}{1/2} = 8$

The expression equals $8$ .

Pause the lecture

Is $a^{2} = a$ always true for real $a$ ?

Common mistake. $(a + b)^{2}$ is not $a^{2} + b^{2}$ . Expanding gives $a^{2} + 2 ab + b^{2}$ . The missing cross term is the most common algebraic injury in calculus.

Practice this lecture

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