Exponents & Roots

Powers: one more floor up the tower

We've been building a tower. Counting gave us $+1$ repeated. Addition repeated is multiplication: $3\times4 = 4+4+4$. Now multiplication repeated is exponentiation:

$a^n = \underbrace{a \cdot a \cdots a}_{n \text{ factors}}.$

So $2^5 = 2\cdot2\cdot2\cdot2\cdot2 = 32$. We call $a$ the base and $n$ the exponent, and $n$ simply counts how many copies of $a$ get multiplied. For now $n$ is a positive integer; by the end of this lesson it'll be any rational, and you'll see why we had no choice.

The exponent laws, proved by counting factors

Here's the part the Federation of Boring Textbook Authors hands you as three magic spells to memorize, no explanation included. They're not spells — they're bookkeeping. Watch how this actually works.

Product law: $a^m \cdot a^n = a^{m+n}$. Why? Count the factors. $a^m$ is $m$ copies of $a$, $a^n$ is $n$ copies; mash them together and you have $m+n$ copies:

$\underbrace{(a\cdots a)}_{m}\cdot\underbrace{(a\cdots a)}_{n} = \underbrace{a\cdots a}_{m+n} = a^{m+n}.$

That's the entire proof. $2^3 \cdot 2^2 = (2\cdot2\cdot2)(2\cdot2) = 2^5$.

Power-of-a-power law: $(a^m)^n = a^{mn}$. The outer exponent says "$n$ copies of $a^m$," and each copy contributes $m$ factors, so total factors $= m\cdot n$:

$(a^m)^n = \underbrace{a^m \cdots a^m}_{n} = a^{mn}.$

$(2^3)^2 = 2^3\cdot2^3 = 2^6 = 64$. Check: $64 = 2^6$. Yes.

Quotient law: $\dfrac{a^m}{a^n} = a^{m-n}$ (for $a\ne 0$). Dividing cancels matched factors: $\frac{a^5}{a^2} = \frac{a\cdot a\cdot a\cdot a\cdot a}{a\cdot a} = a^3$. Three survive, $5-2=3$.

Counting factors. That's the whole damn thing. Every exponent law, every time.

$a^0 = 1$ and $a^{-n} = 1/a^n$ are FORCED

Now the part I genuinely love — I have actually shouted about this at beakers. "What is $a^0$?" You can't multiply $a$ by itself zero times and count — the picture breaks. So we don't guess. We demand the laws keep working and see what $a^0$ is forced to be.

The product law must hold even when one exponent is $0$:

$a^n \cdot a^0 = a^{n+0} = a^n.$

So $a^0$ is a number that, multiplied by $a^n$, gives back $a^n$ — that's the multiplicative identity from last lesson. Therefore (for $a\ne 0$):

$a^0 = 1.$

Not by decree. By force. Any other value would shatter the product law, and I will not stand for shattered product laws in my lab.

Same trick for negative exponents. What's $a^{-n}$? Demand the product law:

$a^n \cdot a^{-n} = a^{n + (-n)} = a^0 = 1.$

So $a^{-n}$ multiplies $a^n$ to give $1$ — it's the multiplicative inverse of $a^n$ (straight from Fractions & the Rationals). Therefore:

$a^{-n} = \frac{1}{a^n}.$

So $2^{-3} = \frac{1}{2^3} = \frac18$. A negative exponent isn't "subtraction" — it's "reciprocal." The laws forced it, and you're welcome, because that understanding will save your ass in a dozen future problems.

Fractional exponents: $a^{1/2}$ must be $\sqrt{a}$

Push the same lever. What could $a^{1/2}$ mean? Demand the power-of-a-power law:

$\big(a^{1/2}\big)^2 = a^{(1/2)\cdot 2} = a^1 = a.$

So $a^{1/2}$ is a number that, squared, gives $a$. That's exactly what "square root of $a$" means. Therefore:

$a^{1/2} = \sqrt{a}.$

And $a^{1/n} = \sqrt[n]{a}$ for the same reason — it's the number whose $n$th power is $a$. Fractional exponents and roots are the same idea. The notation $a^{1/2}$ is arguably the better one, because it makes the laws automatic: $a^{1/2}\cdot a^{1/2} = a^{1/2+1/2} = a^1 = a$. The radical sign is just tradition — pretty tradition, but tradition. The power notation is the real beast.

Roots as the inverse question

Step back. Squaring is an operation: $x \mapsto x^2$. The square root is its inverse question: $\sqrt{x}$ asks "what number, squared, gives $x$?" — exactly like subtraction asked "what adds back?" and division asked "what multiplies back?". The whole course runs on inverse questions.

But there's a wrinkle. For a positive $x$, two numbers square to it: $3^2 = 9$ and $(-3)^2 = 9$. So which one is $\sqrt9$? By convention, $\sqrt{x}$ means the nonnegative one. $\sqrt9 = 3$, full stop — not $\pm3$. (When you want both, you write $\pm\sqrt9$.) This is a choice, made so that $\sqrt{\ }$ is an unambiguous answer rather than two answers. Convention, clearly flagged — not a law.

Graph the power and the root side by side; drag the x-probe and watch $\sqrt{x}$ undo $x^2$ on the right half:

Simplifying radicals, and the landmine

Roots play nicely with products, for nonnegative $a, b$:

$\sqrt{ab} = \sqrt{a}\cdot\sqrt{b}.$

(Both square to $ab$, and both are nonnegative, so they're equal.) This lets you simplify: $\sqrt{12} = \sqrt{4\cdot3} = \sqrt4\cdot\sqrt3 = 2\sqrt3$. Pull out perfect-square factors.

Now the landmine, and I am completely serious: tattoo this on the inside of your eyelids:

$\sqrt{a+b} \ne \sqrt{a} + \sqrt{b}.$

Roots do not distribute over addition. Test it: $\sqrt{9+16} = \sqrt{25} = 5$, but $\sqrt9 + \sqrt{16} = 3+4 = 7$. Five is not seven. The square root is not linear, and assuming it is will detonate your algebra for years. Products split; sums do not. This error shows up constantly and it's the kind of bullshit that loses points on every exam in this stratum. Remember the difference. Go run the gauntlet.