Here is a habit worth burning into muscle memory. The moment you see a square root sitting in a denominator — \dfrac{1}{\sqrt{2}}, \dfrac{5}{\sqrt{3}}, \dfrac{1}{\sqrt{5}+\sqrt{2}} — your first move, before you attempt anything else, is rationalise. Multiply top and bottom by the expression that will clear the radical from below. The denominator turns into an honest integer, and every operation you were about to try — comparing, adding, cancelling, applying an identity — becomes available again.

This isn't about cosmetic prettiness. It's about unlocking the problem.

Why a radical in a denominator blocks progress

Try to add \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{3}}. You need a common denominator. The natural choice is \sqrt{2} \cdot \sqrt{3} = \sqrt{6}. So you'd rewrite:

\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{6}} + \frac{\sqrt{2}}{\sqrt{6}} = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{6}}

And now you still have a radical downstairs. You haven't added two fractions — you've smeared the mess around. The clean path is to rationalise each term first:

\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}, \qquad \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}

Now both denominators are integers. A common denominator is 6, and the addition is ordinary fraction arithmetic:

\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{3} = \frac{3\sqrt{2}}{6} + \frac{2\sqrt{3}}{6} = \frac{3\sqrt{2} + 2\sqrt{3}}{6}

Or try comparing \dfrac{1}{\sqrt{2}} with \dfrac{1}{\sqrt{3}}. Which is larger? Staring at the radicals downstairs makes your brain hesitate. Rationalise:

\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \approx \frac{1.414}{2} \approx 0.707
\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \approx \frac{1.732}{3} \approx 0.577

Now it's obvious. The radical in the denominator wasn't hiding any mathematics — it was just obscuring comparison. Clear it, and the answer appears.

The core trick — multiply by the matching form

Two cases. Learn both as reflexes.

Single-radical denominator. Multiply top and bottom by the same radical:

\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}

Why it works: \sqrt{2} \cdot \sqrt{2} = 2. The downstairs radical squares itself into a rational number.

Binomial-radical denominator. Multiply top and bottom by the conjugate — same two terms with the sign between them flipped:

\frac{1}{\sqrt{2} + 1} \cdot \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1

The sign-flip in the conjugate is what makes the \sqrt{\phantom{x}} cross-terms cancel.

Why the conjugate trick works

For any binomial a + b, the conjugate is a - b. Their product is the difference of squares identity:

(a + b)(a - b) = a^2 - b^2

Set a = \sqrt{x}. Then a^2 = x — the radical vanishes. So:

(\sqrt{x} + y)(\sqrt{x} - y) = x - y^2

which is rational (assuming x and y are rational). The same idea works when both terms carry radicals:

(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) = x - y

No radicals left. That's the whole mechanism. The conjugate is the unique partner that, when multiplied in, kills every square-root cross-term through the a^2 - b^2 pattern.

Worked examples

Four quick ones. Each takes less than a line once the habit is in place.

Example 1. Rationalise \dfrac{5}{\sqrt{3}}.

\frac{5}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3}

Example 2. Rationalise \dfrac{1}{\sqrt{5} + \sqrt{2}}. Conjugate is \sqrt{5} - \sqrt{2}:

\frac{1}{\sqrt{5} + \sqrt{2}} \cdot \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} = \frac{\sqrt{5} - \sqrt{2}}{5 - 2} = \frac{\sqrt{5} - \sqrt{2}}{3}

Example 3. Rationalise \dfrac{1}{3 - \sqrt{2}}. Conjugate is 3 + \sqrt{2}:

\frac{1}{3 - \sqrt{2}} \cdot \frac{3 + \sqrt{2}}{3 + \sqrt{2}} = \frac{3 + \sqrt{2}}{9 - 2} = \frac{3 + \sqrt{2}}{7}

Example 4. Simplify \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}. Multiply by \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1} (which is both the conjugate of the denominator and conveniently matches the numerator):

\frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}

A two-radical mess collapses to 2 - \sqrt{3}, a number you can plot on a number line.

Comparing, once rationalised

Which is larger, \dfrac{1}{\sqrt{5} - 2} or \dfrac{1}{\sqrt{5} + 2}? Good luck eyeballing it.

Rationalise each:

\frac{1}{\sqrt{5} - 2} \cdot \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{\sqrt{5} + 2}{5 - 4} = \sqrt{5} + 2 \approx 4.24
\frac{1}{\sqrt{5} + 2} \cdot \frac{\sqrt{5} - 2}{\sqrt{5} - 2} = \frac{\sqrt{5} - 2}{5 - 4} = \sqrt{5} - 2 \approx 0.24

One is about 4.24, the other about 0.24. The comparison was invisible before rationalising and trivial after. Notice also that each denominator came out to exactly 1 — so the two original ugly fractions were really just \sqrt{5} + 2 and \sqrt{5} - 2 in disguise.

Cube roots and higher — the trick generalises

Cube roots need a slightly different completer. You're aiming to hit \sqrt[3]{\phantom{x}}^3 = \text{rational}, so you need to top up the downstairs to a full cube.

\frac{1}{\sqrt[3]{2}} \cdot \frac{\sqrt[3]{4}}{\sqrt[3]{4}} = \frac{\sqrt[3]{4}}{\sqrt[3]{8}} = \frac{\sqrt[3]{4}}{2}

You multiplied by \sqrt[3]{4} = \sqrt[3]{2^2} because \sqrt[3]{2} \cdot \sqrt[3]{2^2} = \sqrt[3]{2^3} = 2. In general:

\frac{1}{\sqrt[3]{a^2}} \cdot \frac{\sqrt[3]{a}}{\sqrt[3]{a}} = \frac{\sqrt[3]{a}}{a}

The rule: multiply by whatever power is missing from making the inside a perfect cube. For an n-th root, complete to an n-th power.

When NOT to rationalise

Sometimes the radical form is the target form of the answer. If a problem asks for the answer as p + q\sqrt{r}, and your work produces 2 + 3\sqrt{5}, that IS the target form — no further cleaning is needed. The rationalising habit is about clearing radicals from denominators; a clean integer numerator with a radical term on top is a perfectly acceptable final form.

Also, in a middle step where you plan to multiply by another fraction that will cancel things out anyway, rationalising prematurely is wasted arithmetic. Read ahead before you commit.

The "rationalising the numerator" variant

Rare in ordinary algebra, useful in calculus-style limits. Consider:

\frac{\sqrt{x + h} - \sqrt{x}}{h}

Multiply top and bottom by \sqrt{x + h} + \sqrt{x} (the conjugate of the numerator):

\frac{\sqrt{x + h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} = \frac{(x + h) - x}{h(\sqrt{x + h} + \sqrt{x})} = \frac{1}{\sqrt{x + h} + \sqrt{x}}

The numerator became h, which cancelled with the h underneath. A difficult difference-of-roots turned into a clean expression you can take limits of. Same trick, applied upstairs instead of down, used whenever the radicals in the numerator are the thing blocking progress.

Worked pipeline — a full problem

Simplify \dfrac{1 + \sqrt{3}}{2 - \sqrt{3}}.

Step 1. Identify the radical in the denominator. Yes — \sqrt{3} is in the denominator. Rationalise.

Step 2. Multiply top and bottom by the conjugate 2 + \sqrt{3}:

\frac{1 + \sqrt{3}}{2 - \sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}}

Step 3. Expand the numerator:

(1 + \sqrt{3})(2 + \sqrt{3}) = 2 + \sqrt{3} + 2\sqrt{3} + 3 = 5 + 3\sqrt{3}

Step 4. Expand the denominator using a^2 - b^2:

(2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1

Step 5. Assemble:

\frac{5 + 3\sqrt{3}}{1} = 5 + 3\sqrt{3}

A grimy-looking fraction with a radical upstairs and downstairs turns out to be 5 + 3\sqrt{3}. Clean integer denominator. Explicit radical-plus-integer form on top. Done.

The habit in one sentence

See a radical in the denominator, rationalise before any other move. The rest of the problem opens up.

Why this is non-negotiable

Rationalising isn't an optional beautification. It is a prerequisite for three of the most common operations you'll do in Class 9 algebra, board exams, and competitive papers:

It's one of the five or six habits every exam-preparing student should have on full automatic. The moment your eye lands on a \sqrt{\phantom{x}} underneath a fraction bar, your hand should already be writing the conjugate.