Rationalizing is the habit of multiplying by conjugate-style expressions to remove radicals or expose clean factors. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
Rationalize each term by its conjugate: $\dfrac{1}{\sqrt n + \sqrt{n+1}} \cdot \dfrac{\sqrt{n+1} - \sqrt n}{\sqrt{n+1} - \sqrt n} = \dfrac{\sqrt{n+1} - \sqrt n}{(n+1) - n} = \sqrt{n+1} - \sqrt n$.
2
The sum telescopes: $(\sqrt 2 - \sqrt 1) + (\sqrt 3 - \sqrt 2) + \cdots + (\sqrt{100} - \sqrt{99})$.
3
Only the endpoints survive: $\sqrt{100} - \sqrt 1 = 10 - 1 = 9$.
Answer:
$9$.
Olympiad / Challenge
Rationalize the denominator of $\dfrac{1}{\sqrt 2 + \sqrt 3 - \sqrt 5}$.
1
Group two of the radicals and multiply by the conjugate $\sqrt 2 + \sqrt 3 + \sqrt 5$. The denominator becomes $(\sqrt 2 + \sqrt 3)^2 - (\sqrt 5)^2 = (5 + 2\sqrt 6) - 5 = 2\sqrt 6$.
2
So the expression is $\dfrac{\sqrt 2 + \sqrt 3 + \sqrt 5}{2\sqrt 6}$; rationalize again by $\sqrt 6$: $\dfrac{\sqrt 6(\sqrt 2 + \sqrt 3 + \sqrt 5)}{12}$.
Generalization: the conjugate trick is $a - b = \frac{a^2 - b^2}{a + b}$ in disguise — for a sum of THREE radicals you group and apply it twice, and for cube roots you use $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ to clear $\sqrt[3]{\;}$.
Where it appears: telescoping sums whose terms are $\frac{1}{\sqrt n + \sqrt{n+1}}$ (AIME), simplifying limits of the form $\frac{0}{0}$ by rationalizing the numerator, and denesting radicals like $\sqrt{a + 2\sqrt b}$.
Pitfall: the conjugate of $\sqrt a - \sqrt b$ must flip ONLY the sign between the radicals, giving $\sqrt a + \sqrt b$ — and you must multiply BOTH numerator and denominator by it (multiplying by $1$). Also confirm $a \ne b$, or the rationalized denominator $a - b$ becomes $0$.
Spot the Signal
Look for problems where the key step is multiplying by conjugate-style expressions to remove radicals or expose clean factors.
You can describe the hard part as multiplying by conjugate-style expressions to remove radicals or expose clean factors, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by identify the expression or equation that calls for rationalizing, then rewrite the givens around it.
Name the relation that makes Rationalizing legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Do not use Rationalizing just because the surface looks familiar; verify the required condition first.
Applying Rationalizing because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Let rationalizing reveal the algebraic structure; then compute only what remains.
Try it on:
Practice a contest problem where the key step is multiplying by conjugate-style expressions to remove radicals or expose clean factors.