14. Radicals & Rational Exponents

Radicals & Rational Exponents

Radicals ($\sqrt{x}$, $\sqrt[3]{x}$) and fractional exponents ($x^{1/2}$, $x^{1/3}$) are two notations for the same operation. Mastering the conversion between them unlocks every exponent rule in ML.

The dual notation

$x^{1/n} = \sqrt[n]{x}$

$x^{m/n} = \left(\sqrt[n]{x}\right)^m = \sqrt[n]{x^m}$

$x^{-n} = \frac{1}{x^n}$

$x^{2/3}$ means "cube root of $x$, squared" or equivalently "$x$ squared, then cube root."

Simplifying radicals

Pull out perfect-power factors. For square roots, factor out perfect squares:

$\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2}$

$\sqrt{48x^3} = \sqrt{16x^2} \cdot \sqrt{3x} = 4x \cdot \sqrt{3x} \quad (x \geq 0)$

For cube roots, factor out perfect cubes. For fourth roots, perfect fourth powers. Same idea.

Rationalizing denominators

A "clean" form has no radicals in the denominator. Multiply numerator and denominator by a factor that eliminates the root:

$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \quad \text{(multiply top and bottom by } \sqrt{3}\text{)}$

$\frac{1}{1 + \sqrt{2}} = \frac{1 - \sqrt{2}}{(1 + \sqrt{2})(1 - \sqrt{2})} = \frac{1 - \sqrt{2}}{1 - 2} = \sqrt{2} - 1 \quad \text{(multiply by the conjugate)}$

The identity $\sqrt{x^2} = |x|$

When $x$ could be negative, $\sqrt{x^2}$ is not $x$: it's $|x|$. The square root is defined to return a non-negative number, so it must strip any sign:

$\sqrt{(-3)^2} = \sqrt{9} = 3 = |-3|$

$\sqrt{x^2} = |x| \quad \text{for any real } x$

This identity is why solving radical equations often introduces extraneous roots: squaring loses sign information that the square root can't recover.

Solving radical equations

1. Isolate the radical on one side.
2. Raise both sides to the power that cancels the radical (square for $\sqrt{\;\;}$, cube for $\sqrt[3]{\;\;}$).
3. Solve the resulting equation.
4. Verify every candidate. Squaring can introduce extraneous roots.

Verify $x = 5$: $\sqrt{25} = 5$. Pass. Verify $x = -3$: $\sqrt{9} = 3$, but $-3 \neq 3$. Reject. Solution: $x = 5$.

The verification is mandatory because squaring is many-to-one: both $5$ and $-5$ square to $25$, so squared equations can't distinguish them.

Exponent rules (memorize)

$x^a \cdot x^b = x^{a + b}$

$\frac{x^a}{x^b} = x^{a - b}$

$(x^a)^b = x^{ab}$

$(xy)^a = x^a \cdot y^a$

$x^0 = 1 \quad (x \neq 0)$

These work the same for fractional exponents. $(x^4)^{1/2} = x^2$. $x^{1/2} \cdot x^{1/3} = x^{5/6}$.

The high-trap distinction: Add exponents when bases are multiplied ($x^a \cdot x^b = x^{a+b}$); multiply exponents when raising a power to a power ($(x^a)^b = x^{ab}$). Example: $x^2 \cdot x^3 = x^5$, but $(x^2)^3 = x^6$. Confusing these two rules is the single most common exponent error in calculus prep.

ML connection: Adam optimizer

Adam scales the learning rate by $1 / \sqrt{\text{running average of squared gradients}}$. That $x^{-1/2}$ in the update rule is doing two algebra operations at once: a square root (the denominator's $\sqrt{\;\;}$) and a reciprocal (the negative exponent). Numerical stability requires adding $\varepsilon$ inside the square root to avoid dividing by zero, the same domain-check discipline from week 13.