Not every expression written with a \sqrt{\,} sign is actually a surd. \sqrt{25} looks like one, but it equals the integer 5 and carries no irrationality. \sqrt{26}, sitting right next to it, is a surd — there is no way to simplify it away. The distinction matters because JEE questions routinely hide integers behind radical signs to see if you will waste time rationalising something that was rational all along.
This page trains your eye to sort them at a glance.
What counts as a surd
A surd is an irrational root of a rational number. Put differently: \sqrt[n]{a} is a surd when a is rational, n is a positive integer, and the root does not come out to a rational number.
So the test has two parts:
- Is the expression a root of a rational number?
- Is the root itself irrational?
If both answers are yes, you have a surd. If the root simplifies to a rational number — even a whole integer — it is not a surd.
Why: "surd" is a word from the Latin surdus, meaning "silent" or "deaf" — the mathematicians who coined it were translating an Arabic term for "inexpressible," meaning a root that cannot be expressed as a fraction. An expression that simplifies to an integer is fully expressible, so by the original definition, it never qualified.
The recognition recipe
For a square root \sqrt{N} of a positive integer N:
- If N is a perfect square (that is, N = k^2 for some non-negative integer k), then \sqrt{N} = k is an integer, not a surd.
- If N is not a perfect square, then \sqrt{N} is irrational, and it is a surd.
For a cube root \sqrt[3]{N}: replace "perfect square" with "perfect cube" (N = k^3). Same logic for n-th roots: check whether N is a perfect n-th power.
The full result — that \sqrt{N} is either an integer or irrational, with no fractional possibilities in between — is a classical theorem. You can read the short proof in Most Square Roots Are Irrational.
The quiz
Type any positive integer N and any root index n \geq 2, then click Guess: surd or Guess: not a surd. The board scores you and reveals the ladder of perfect n-th powers around N — so you can see why it is or is not a surd. Score is tracked across the session.
The verdicts, one by one
- \sqrt{25} = 5. Not a surd. 25 is a perfect square.
- \sqrt{26}. Surd. 26 lies between the perfect squares 25 and 36, so it is not one itself, and \sqrt{26} is irrational.
- \sqrt{49} = 7. Not a surd. 49 = 7^2.
- \sqrt[3]{27} = 3. Not a surd. 27 = 3^3 is a perfect cube.
- \sqrt[3]{28}. Surd. 28 is not a perfect cube (3^3 = 27 and 4^3 = 64), so the cube root is irrational.
- \sqrt[4]{16} = 2. Not a surd. 16 = 2^4.
- \sqrt{2}. Surd — the archetypal one, and the reason the word exists.
Why it matters in a JEE problem
Imagine a question asks you to rationalise \dfrac{3}{\sqrt{49} + \sqrt{16}}. A second's hesitation and you might reach for the conjugate \sqrt{49} - \sqrt{16} and the difference-of-squares trick. But \sqrt{49} + \sqrt{16} = 7 + 4 = 11, a plain integer. The "fraction" is just 3/11 in disguise. Spotting the non-surds saves the work.
Why: the surd/non-surd distinction is not just terminology. It tells you whether a radical is an object you need to manipulate (a surd) or a value you can evaluate on sight (a non-surd). Treating them the same way is the slowest possible approach to a JEE paper.
A one-line checklist
Before you rationalise, simplify, or square anything:
- Square roots: is the radicand a perfect square? (1, 4, 9, 16, 25, 36, 49, 64, 81, 100, \dots)
- Cube roots: is the radicand a perfect cube? (1, 8, 27, 64, 125, \dots)
- n-th roots: is the radicand a perfect n-th power?
If yes — it is not a surd, and you can evaluate it directly. If no — treat it as a surd and use the rules from Roots and Radicals.
Related: Roots and Radicals · Is Root-4 Irrational Because It's Under a Root Sign? · When You See a Root of a Non-Perfect-Square, Default to Irrational · Spot a Conjugate Surd in a JEE Problem and Rationalise on Sight · What's the Difference Between a Radical and a Surd?