In the middle of a problem you see \sqrt{23} or \sqrt{50} or \sqrt{1000} and you need a snap judgement: rational or irrational? The rule is almost embarrassingly simple: if the number under the root is a positive integer that is not a perfect square, the whole thing is irrational. Full stop. You can build most of your rational-vs-irrational reflexes around this one pattern.

This article gives you the rule, the reason it works, the short list of exceptions you have to check, and a couple of examples where the expression simplifies before the test bites.

The pattern

You see a number of the form

\sqrt{N}, \qquad N \in \mathbb{Z}^+

in an expression. Run this two-step check.

  1. Is N a perfect square? That means N = k^2 for some non-negative integer k. (So N \in \{0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, \dots\}.) If yes → \sqrt{N} = k, a rational number (in fact an integer). Done.
  2. If no → \sqrt{N} is irrational. Treat it as irrational in every downstream step.

Why this works: the square root of a positive integer is either an integer or irrational — it can never be a non-integer rational. There are no half-integers or third-integers hiding in the square-root function. Either N is a perfect square and the root is clean, or it is not and the root is one of those "never terminates, never repeats" decimals.

Why no rational can slip in between

Suppose, for contradiction, that \sqrt{N} = p/q where p, q are integers with no common factor and q > 1 (so p/q is a genuine non-integer fraction). Squaring both sides:

N = \dfrac{p^2}{q^2}, \qquad \text{so} \qquad N q^2 = p^2.

The left-hand side has a factor of q^2. The right-hand side is p^2. Since p and q share no common factor, neither do p^2 and q^2 — any prime dividing q^2 divides q, so it does not divide p, so it does not divide p^2.

But then q^2 divides p^2 forces q^2 = 1, i.e. q = 1. So the fraction was not a genuine non-integer fraction after all — it was an integer in disguise, which means N was a perfect square.

Contrapositive: if N is not a perfect square, \sqrt{N} cannot be rational. So it is irrational. That is the theorem.

Quick recognition table

Use this in your head for small N. Anything not in the "perfect square" column is an irrational square root.

Perfect squares up to 144 versus non-perfect-square integersTwo columns. Left column labelled perfect squares contains one, four, nine, sixteen, twenty-five, thirty-six, forty-nine, sixty-four, eighty-one, one hundred, one hundred twenty-one, and one hundred forty-four, with their square roots listed alongside as integers. Right column labelled not perfect squares contains two, three, five, six, seven, eight, ten, eleven, twelve, thirteen, and so on, with a note that every one of their square roots is irrational. Perfect squares — √N is an integer Not perfect squares — √N is irrational √1 = 1 √4 = 2 √9 = 3 √16 = 4 √25 = 5 √36 = 6 √49 = 7 √64 = 8 √81 = 9 √100 = 10 √121 = 11 √144 = 12 √2, √3, √5, √6, √7, √8, √10, √11, √12, √13, √14, √15, √17, √18, √19, …all irrational memorise the left column up to 400 and the right column is automatic
The list on the left is the entire "rational square root" catalogue for integers up to $144$. Every other positive integer under a square-root sign is irrational. Memorise the squares up to $400$ ($= 20^2$) and your recognition speed triples.

The "unless the expression simplifies" clause

The rule says default to irrational unless the expression simplifies. What simplifications are you watching for?

Simplification 1: pull out a perfect-square factor. \sqrt{50} looks non-perfect, but 50 = 25 \cdot 2, so \sqrt{50} = 5\sqrt{2}. It is still irrational (because \sqrt{2} is), but the form is cleaner. The rule doesn't say "rational" — it correctly says irrational — but you should express it in the a\sqrt{b} form before moving on.

Simplification 2: the expression is a root of a fraction. \sqrt{\tfrac{9}{4}} = \tfrac{3}{2} — both 9 and 4 are perfect squares, and the result is a rational. The rule "perfect square → rational" extends to "ratio of perfect squares → rational" for the square root of a rational.

Simplification 3: canceling inside. \sqrt{\tfrac{18}{8}} = \sqrt{\tfrac{9}{4}} = \tfrac{3}{2}. Simplify the fraction before you decide.

Simplification 4: arithmetic collapses the expression. You might meet \sqrt{9 + 16} in geometry. Do the inside first: 9 + 16 = 25, a perfect square, so \sqrt{9 + 16} = 5. The individual \sqrt{9} and \sqrt{16} were rational on their own, so no drama — but the habit of finishing arithmetic under the root before judging is a good one.

The general rule becomes: evaluate the argument of the root as far as you can, reduce to lowest terms if it is a fraction, and then apply the perfect-square test.

When the rule bites: quick examples

Four snap judgements

(a) \sqrt{23}. Is 23 a perfect square? No (4^2 = 16, 5^2 = 25). → Irrational.

(b) \sqrt{196}. Is 196 a perfect square? Check: 14 \times 14 = 196. Yes. → \sqrt{196} = 14, rational.

(c) \sqrt{\tfrac{49}{25}}. Both top and bottom are perfect squares. → \tfrac{7}{5}, rational.

(d) \sqrt{72}. Is 72 a perfect square? 8^2 = 64, 9^2 = 81. No. → Irrational. But simplify: 72 = 36 \cdot 2, so \sqrt{72} = 6\sqrt{2}.

Why this is a recognition shortcut, not a proof

You are not writing a proof each time. You are running a pattern match: "integer under the root, not in my perfect-squares list → irrational." The proof exists (it is the contradiction argument in the earlier section), but you should not redo it in every problem. You should cache it once, and then every time you see \sqrt{23} on a worksheet, you know it is irrational, with no mental work beyond "23 is not in my squares table."

This is a recognition skill. It is not interesting in itself, but it saves you seconds on every problem involving roots, and over the course of a JEE paper those seconds add up to whole questions.

One exam-hall warning

The rule only applies to the square root of a positive integer (or rational with integer top and bottom). It does not say anything about the square root of an already-irrational number. For instance, \sqrt{\pi} is still irrational, but not because of this rule — you need the separate fact that \pi is irrational. Do not over-generalise.

And cube roots, fourth roots, etc. follow an analogous rule (the root of an integer is rational iff the integer is a perfect cube / perfect fourth power / etc.) but you should apply that rule by name, not by default.

Related: Number Systems · Is √4 Irrational Because It's Under a Root Sign? · Why Do We Need Irrational Numbers at All? · Tennenbaum's Picture-Proof That √2 Is Irrational