Exponents & Roots

Fold a piece of paper

A sheet of paper is about $0.1$ millimeters thick. Fold it in half and it’s $0.2$; again, $0.4$. Ten folds in, the stack is a full $10$ centimeters. By $23$ folds it would clear a kilometer, and around fold $42$ — if paper allowed it — the stack would reach the Moon. Nothing about one fold feels dramatic; the drama is that each step multiplies instead of adds. Doubling ten times isn’t $10 \times 2$, it’s $2 \times 2 \times \cdots \times 2$ ten times over — and math needs a notation for “multiply this by itself that many times.”

An exponent is repeated multiplication

You met the notation in F1 as the densest shorthand on the priority ladder: in $2^{4}$, the base ($2$) is what gets multiplied and the exponent ($4$) counts the copies: $2^{4} = 2 \times 2 \times 2 \times 2 = 16$. You’ve used it since, too — $60 = 2^{2} \times 3 \times 5$ in F2 leans on it. That “counts the copies” reading is the master key to this unit: every exponent rule is just what happens when you write the copies out in full.

Derive the rules — don’t memorize them

Multiply $2^{3} \cdot 2^{4}$ by unpacking both: $(2 \cdot 2 \cdot 2)(2 \cdot 2 \cdot 2 \cdot 2)$ — seven copies of $2$ in a row, so the answer is $2^{7}$. Copies stack, so same-base products add exponents. Division runs the same movie backwards: $2^{5} \div 2^{2}$ cancels two copies off the top, leaving $2^{3}$ — subtract. And a power of a power makes copies of copies: $(2^{3})^{2} = 2^{3} \cdot 2^{3} = 2^{6}$ — multiply.

Two slips account for most exponent errors, and both come from the notation looking more symmetric than it is. First, $2^{3} \cdot 5^{4}$ looks like it should combine the way $2^{3} \cdot 2^{4}$ does — but write the copies out and there’s nothing to stack: three $2$s and four $5$s share no common base, so the rule simply doesn’t apply. Second, “add or multiply?” blurs under time pressure; the copies decide instantly. Stacking rows of copies adds; copying the whole row multiplies.

It opens on the product rule with $2^{3} \cdot 2^{4}$. Before switching to each other rule, predict the exponent it will produce from the same $3$ and $4$ — the quotient rule should give $-1$ (a small number, not a negative one!), and power-of-a-power $12$.

The pattern behind zero and negative exponents

What could $2^{0}$ mean — zero copies of $2$? The gut answer is $0$, since zero copies feels like nothing. And $2^{-3}$ looks like it should be negative. Both instincts break against a pattern you can verify: step the exponent down by one and the value divides by the base each time — $2^{3}=8$, $2^{2}=4$, $2^{1}=2$. The ladder doesn’t stop there: one more step down forces $2^{0} = 1$ (divide $2$ by $2$), then $2^{-1}=\frac12$, $2^{-2}=\frac14$, $2^{-3}=\frac18$. Zero and negative exponents aren’t a new arbitrary rule — they’re the only values that keep the dividing pattern unbroken. So $a^{0}=1$ for any non-zero $a$, and $a^{-n} = \frac{1}{a^{n}}$: not negative, just small.

The calculator opens on $2^{-3}$ — check it against the ladder. Then try $7^{0}$, and $10^{-2}$: keep that last one in mind, it comes back in scientific notation.

A root is a side length

Squaring’s undo has a picture. A square of side $7$ covers $7^{2} = 49$ unit tiles; the square root asks the reverse — this square covers $49$ tiles: how long is its side? $\sqrt{49} = 7$. Squaring carries a side to its area; the root carries the area back to its side. One honesty note the symbol insists on: $(-7)^{2} = 49$ too, but $\sqrt{\phantom{6}}$ always means the non-negative answer — $\sqrt{49}$ is $7$, never $-7$. (The equation $x^{2} = 49$ has two solutions, but its $\pm$ comes from the equation, not the symbol — a distinction that does real work in the quadratics modules.)

Some areas close into whole-number sides, and those are worth knowing by sight: $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144$ — the perfect squares, the landmarks this whole section navigates by.

Trap the root between landmarks

$50$ is not on that list — but it sits just past $49 = 7^{2}$ and well below $64 = 8^{2}$, and a bigger area always needs a longer side. So $\sqrt{50}$ is trapped: bigger than $7$, smaller than $8$, barely past $7$ (it’s $7.07\ldots$). Find the landmarks on either side, read off the sandwich — that’s all root estimation is, and it’s enough to sanity-check any answer with a radical in it.

Before dragging: which two integers trap $\sqrt{90}$, and which is it closer to? Then hunt $\sqrt{2}$ — the most famous root of all. Then set the target to $49$ and feel the handle click onto a landmark. For any other target the hunt never truly ends: $7.07^{2}$ is a hair small, $7.08^{2}$ a hair big, and no decimal that stops ever squares to exactly $50$. Hold that thought — it gets a name two sections down.

Pulling squares out of a root

When the number under the root isn’t a perfect square, you can still undo the square part way. By power of a product, $(\sqrt{a} \cdot \sqrt{b})^{2} = (\sqrt{a})^{2}(\sqrt{b})^{2} = ab$ — so $\sqrt{a} \cdot \sqrt{b}$ is a non-negative number that squares to $ab$, which is the very definition of $\sqrt{ab}$. That’s the product rule for radicals, $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$, and read right-to-left it lets a root split across a product. Split off a perfect square, and that piece walks out whole:

You don’t have to spot the largest square first: pull out $4$ instead — $\sqrt{48} = 2\sqrt{12}$ — and $12$ still hides a $4$, so pull again and land on the same $4\sqrt{3}$. Every route reaches the same destination; the largest square just finishes in one pull. (No landmark jumping out? F2’s atoms are the safety net: $48 = 2^{4} \cdot 3$, and every pair of equal primes is a square in disguise.) Why bother, when a calculator says $6.93$? Because $6.93$ is a rounding and $4\sqrt{3}$ is the number — exact, and the language answer choices speak. When the quadratic formula starts handing you raw radicals, this is the skill it leans on. Cube roots play the same game with perfect cubes: $\sqrt[3]{54} = \sqrt[3]{27 \cdot 2} = 3\sqrt[3]{2}$.

Watch the whole move once at the board — lazy pull, hunt, and check — then do it yourself right below.

Simplify $\sqrt{72}$ in one pull — which chip? Then restart and do it the lazy way, starting with $4$, and count the pulls. Then try $\sqrt{30}$, and believe its refusal: no square factor, no simplification. (The √ Roots & Radicals tab has a simplifier that shows the prime-atom route for any number, cube roots included.)

Roots across × and ÷ — but never across + and −

The product rule runs forward, too: $\sqrt{3} \cdot \sqrt{12} = \sqrt{36} = 6$ — two shaggy roots multiplying into a whole number. Division matches: $\dfrac{\sqrt{50}}{\sqrt{2}} = \sqrt{25} = 5$, and a root of a fraction splits into a fraction of roots — $\sqrt{\dfrac{9}{16}} = \dfrac{3}{4}$, while $\sqrt{\dfrac{7}{4}} = \dfrac{\sqrt{7}}{2}$ stays exact with the $7$ still under its root.

Roots do add one way — as counts of a common radical: $2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$, for the same reason $2x + 5x = 7x$.

Variables under the root

Letters obey the same machine, because pairs are squares: $\sqrt{x^{2}} = x$, and $\sqrt{x^{6}} = \sqrt{x^{3} \cdot x^{3}} = x^{3}$ — halve the exponent, one copy out per pair. An odd exponent leaves a single behind, $\sqrt{x^{5}} = x^{2}\sqrt{x}$, and numbers and letters pull out independently: $\sqrt{9x^{2}} = 3x$. One honesty note: all of this assumes $x$ isn’t negative — squaring erases signs, so the root can’t recover one. Here (and in most SAT radical work) variables under a root are taken as non-negative; the fully signed story needs absolute value, and it can wait.

What kind of number is √2?

Zoom out once. You’ve been collecting number families all through foundations: the counting numbers; zero and the negatives completing the integers (F1); ratios of integers — the rationals — whose decimals always end or repeat (F3, F4). $\sqrt{2} = 1.41421356\ldots$ breaks the pattern: its decimal never ends and never repeats, because $\sqrt{2}$ provably isn’t a ratio of integers at all. Numbers like that are irrational — most roots are, which is exactly why $4\sqrt{3}$ beats any decimal: it’s the only way to say the number exactly. Rationals and irrationals together fill the entire number line: the real numbers.

And one question the reals cannot answer: $\sqrt{-9}$. A square is never negative — positive times positive and negative times negative both land positive (F1) — so no real number squares to $-9$. When an equation demands one anyway, the honest verdict is no real solution, wording you’ll use for real work in M3. (Mathematicians did build a still-bigger family where that question has an answer — the complex numbers, grown around a new number $i$ with $i^{2} = -1$ — but that story sits past this course’s edge.)

Scientific notation — taming huge and tiny numbers

The Earth weighs about $5{,}970{,}000{,}000{,}000{,}000{,}000{,}000{,}000$ kilograms, and a water molecule is about $0.000000000275$ meters wide. Both numbers are almost all zeros — the only information is one short string of digits and how far from the decimal point it sits. Scientific notation stores exactly those two facts: $a \times 10^{n}$, where $a$ keeps a single non-zero digit before the point and the power of ten counts the point’s hops (the same hops you counted in F4). $5300 = 5.3 \times 10^{3}$ — point hopped $3$ left. $0.00042 = 4.2 \times 10^{-4}$ — point hopped $4$ right, and there’s your negative exponent meaning small, not negative.

Convert $5300$, then $0.00042$ and check each exponent against your hop count. Then give it the Earth’s mass — type the digits and count the zeros yourself first.

The one thing to remember

An exponent counts copies in a multiplication, and every rule falls out of writing the copies down: stacking adds, canceling subtracts, copying the whole row multiplies — same base only — and stepping the exponent down divides by the base, which is why $a^{0} = 1$ and negative exponents mean small. A root is the side of the area: trap it between perfect-square landmarks to estimate it, pull square factors out to simplify it ($\sqrt{48} = 4\sqrt{3}$), slide it across $\times$ and $\div$ but never across $+$ or $-$ — and when no real number can do the squaring, say so: no real solution.

What an exponent means

An exponent is repeated multiplication: $2^{4} = 2 \times 2 \times 2 \times 2 = 16$. The base is what’s multiplied; the exponent is how many times.

The six exponent rules

Roots

A square root undoes squaring: $\sqrt{49} = 7$ because $7^{2} = 49$ — and the symbol always means the non-negative root. A cube root undoes cubing: $\sqrt[3]{27} = 3$. The perfect squares $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144$ are the landmarks: to estimate, trap the number between two of them ($49 < 50 < 64$, so $7 < \sqrt{50} < 8$).

The radical rules

To simplify a radical, pull out the largest perfect-square factor (largest perfect cube for a cube root): $\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}$.

Number types

A square is never negative, so $\sqrt{-9}$ has no real solution — the honest phrase when an equation asks a square to be negative.

Scientific notation

A compact way to write very big or very small numbers: $a \times 10^{n}$ where $a$ has one non-zero digit before the point. $5300 = 5.3 \times 10^{3}$; $\ 0.00042 = 4.2 \times 10^{-4}$. A positive exponent means a big number (point moved left); a negative exponent means a small one (point moved right).