Rationalizing Denominators

Multiplying by a disguised form of 1

Every rationalizing step is one idea in disguise, which is to multiply the fraction by $1$. Any nonzero quantity divided by itself equals $1$, so $\frac{\sqrt{2}}{\sqrt{2}} = 1$ and $\frac{2 - \sqrt{3}}{2 - \sqrt{3}} = 1$. Multiplying a number by $1$ leaves its value untouched, which is exactly why the move is allowed. What changes is the form of the fraction, not the number it stands for.

The reason a well-chosen form of $1$ can clear a root comes down to one fact about square roots: a square root times itself gives back the number under it.

So the whole method is to pick the form of $1$ that turns the denominator into something with no radical. For a single root the choice is easy, and for a two-term denominator it is the conjugate.

Clearing a single square root

When the denominator is a single square root, the form of $1$ to use is that same root over itself. To rationalize $\frac{a}{\sqrt{b}}$ for any $b > 0$ (so the denominator isn’t zero to begin with), multiply the numerator and the denominator by $\sqrt{b}$:

The denominator $\sqrt{b}\cdot\sqrt{b}$ became $b$, with the root gone, and the root moved up to the numerator where the convention allows it. Notice that you must multiply the top by $\sqrt{b}$ as well; multiplying only the bottom would change the value.

Clearing a binomial denominator with its conjugate

A denominator with two terms, like $1 + \sqrt{2}$, cannot be fixed by multiplying by a single root: multiplying $1 + \sqrt{2}$ by $\sqrt{2}$ gives $\sqrt{2} + 2$, which still has a root. The move that works uses the difference-of-squares pattern you proved earlier in this chapter.

The conjugate of a two-term expression is the same expression with the sign between its terms flipped. The conjugate of $a + \sqrt{b}$ is $a - \sqrt{b}$, and the conjugate of $a - \sqrt{b}$ is $a + \sqrt{b}$. Multiplying an expression by its conjugate produces a difference of squares, and the square of a square root is rational, so every radical disappears.

For example, the conjugate of $3 + \sqrt{5}$ is $3 - \sqrt{5}$. Multiply the two together and watch the radical vanish:

The two middle terms, $-3\sqrt{5}$ and $+3\sqrt{5}$, are exact opposites, so they cancel, leaving the whole number $4$. That cancellation happens for the same reason every time, which is what the next proof shows in general.

To rationalize $\frac{1}{a + \sqrt{b}}$, multiply the numerator and the denominator by the conjugate $a - \sqrt{b}$:

The denominator is now rational, and as always, the conjugate over itself is just $1$, so the value is unchanged. As in every example in this lesson, take $a$ to be a positive whole number and $b$ to be a whole number that is not a perfect square. That keeps $a - \sqrt{b}$ from ever being zero: if $b$ were a perfect square, $\sqrt{b}$ would already be a whole number, and there would be no radical left to clear in the first place.

When both terms are square roots

The conjugate trick works just as well when both terms are square roots. The product $(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})$ is again a difference of squares:

Both squares collapse to whole numbers, so the denominator becomes the plain difference $a - b$, with no radical remaining. Here $a$ and $b$ are nonnegative whole numbers with $a \neq b$, so the two roots are genuinely different. If $a = b$, the two roots would already be equal, making the original denominator $\sqrt{a} - \sqrt{a} = 0$ before you ever reached for a conjugate.

Common mistakes