Prove that $\sqrt{3}$ is an irrational number.

Real Numbers

Irrational Numbers

Official Answer

√3 is irrational, proved by contradiction. Assume $\sqrt{3} = \frac{p}{q}$ with p, q coprime integers, $q \ne 0$ . Squaring, $p^2 = 3q^2$ , so 3 divides p², hence 3 divides p (since 3 is prime). Let $p = 3c$ ; substituting gives $q^2 = 3c^2$ , so 3 divides q too. Thus 3 divides both p and q, contradicting that they are coprime. This contradiction shows the assumption was false, so √3 cannot be expressed as p/q — it is irrational.

irrational numberproof by contradictionco-primeHCFfundamental theorem of arithmeticprime dividesrational number p/q form

Marking Scheme

11 mark: correctly starting the proof by contradiction — assuming √3 is rational, i.e., $\sqrt{3} = \frac{p}{q}$ with p, q co-prime integers, $q \ne 0$ .
21 mark: correctly showing $p^2 = 3q^2$ , concluding 3 divides p, writing $p = 3c$ , and deriving $q^2 = 3c^2$ (hence 3 divides q).
31 mark: stating the contradiction (p and q have a common factor 3, contradicting co-primality) and concluding √3 is irrational.

Hint

Assume $\sqrt{3} = \frac{p}{q}$ in lowest terms, square both sides, show 3 divides p, then 3 divides q too — contradicting that p, q are co-prime.

Quick Oral Answer

I assume √3 is rational, i.e., $\sqrt{3} = \frac{p}{q}$ with p and q co-prime, then show that 3 must divide both p and q, contradicting their co-primality — so √3 cannot be rational, hence it is irrational.

Analysis & Explanation

A classic proof by contradiction using the prime-divisibility lemma, tested almost every year.

Concept

Assume $\sqrt{3} = \frac{p}{q}$ with p, q coprime integers; squaring gives $p^2 = 3q^2$ .
Lemma: if a prime divides $a^2$ , it divides a — hence 3 divides both p and q.

Key Points

This contradicts the coprimality (HCF = 1) assumption, so √3 cannot be rational.

Common Mistakes

Skipping the explicit coprime assumption at the start.
Not stating the final contradiction clearly before concluding.

Exam Tip / Real-world

The same template (only the prime changes) proves √2, √5, √7 irrational — this argument traces back to Euclid's Elements (Book X).

Common Mistakes

1Not stating that p and q are taken to be co-prime (HCF = 1) at the start — this assumption is essential to reach the final contradiction and is often skipped.
2Using the divisibility lemma incorrectly, e.g., applying it to non-prime numbers, or not justifying why 3 divides p from 3 dividing p².
3Ending the proof without explicitly stating the contradiction and the concluding sentence 'Hence √3 is irrational' — losing the concluding mark even if the algebra is correct.

Interesting Facts

The irrationality of $\sqrt{2}$ (and by the same method, $\sqrt{3}$ ) was known to ancient Greek mathematicians and is considered one of the first 'crisis' discoveries in mathematics, reportedly disturbing the Pythagorean belief that all numbers could be expressed as ratios.

The decimal expansion of $\sqrt{3}$ ( $\approx 1.7320508\ldots$ ) is non-terminating and non-recurring, which is the defining property of any irrational number.

CBSE reintroduced direct proofs of irrationality (like $\sqrt{2}, \sqrt{3}, \sqrt{5}$ ) as a core topic after 2019 following updates to the Real Numbers chapter, making this one of the most predictable 3-mark questions in recent board papers.

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Frequently Asked Questions

Why must p and q be co-prime in this proof?

Because any rational number p/q can always be reduced to its simplest (lowest) form where $HCF(p,q)=1$ ; assuming this lowest form is essential to reach a contradiction when we later show both p and q share a common factor of 3.

Does this same proof method work for $\sqrt{2}$ and $\sqrt{5}$ ?

Yes, the identical method works for $\sqrt{2}, \sqrt{5}, \sqrt{7}$ , or √(any prime number) — only the number 3 needs to be replaced by that prime throughout the proof.