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

Real Numbers

Proof of irrationality of $\sqrt{3}$

Official Answer

$\sqrt{3}$ is irrational. Proof by contradiction: assume $\sqrt{3} = a/b$ in lowest terms ( $a, b$ co-prime, $b \ne 0$ ). Squaring gives $3b^2 = a^2$ , so $3$ divides $a^2$ and, since $3$ is prime, $3$ divides $a$ ; write $a = 3c$ . Substituting gives $9c^2 = 3b^2$ , i.e. $b^2 = 3c^2$ , so $3$ divides $b^2$ and hence $3$ divides $b$ . Thus $3$ is a common factor of both $a$ and $b$ , contradicting that $a/b$ was in its lowest (co-prime) form. This contradiction proves the initial assumption false, so $\sqrt{3}$ cannot be expressed as $a/b$ — hence $\sqrt{3}$ is irrational.

proof by contradiction√3 irrationalco-prime a/ba² = 3b²3 divides aprime divides squarecommon factor contradictionFundamental Theorem of Arithmetic

Marking Scheme

11 mark: setting up the contradiction — assume $\sqrt{3} = a/b$ with $a, b$ co-prime integers, $b \ne 0$ , and squaring to obtain $a^2 = 3b^2$ .
21 mark: correctly using the prime-divisibility property to conclude $3$ divides $a$ , writing $a = 3c$ , and deriving $b^2 = 3c^2$ so that $3$ divides $b$ .
31 mark: stating the contradiction ( $3$ is a common factor of $a$ and $b$ , contradicting co-primality) and concluding that $\sqrt{3}$ is irrational.

Hint

Assume $\sqrt{3} = a/b$ in lowest terms, square to get $a^2 = 3b^2$ , use the fact that $3$ (a prime) dividing $a^2$ forces $3$ to divide $a$ , then show $3$ also divides $b$ — a contradiction.

Quick Oral Answer

I assume $\sqrt{3} = a/b$ in lowest terms; squaring gives $a^2 = 3b^2$ , so $3$ divides $a$ , hence $a = 3c$ and then $b^2 = 3c^2$ means $3$ divides $b$ too — but a and b were co-prime, a contradiction, so $\sqrt{3}$ must be irrational.

Analysis & Explanation

This is the standard NCERT proof-by-contradiction, resting on one lemma: if a prime $p$ divides $a^2$ , then $p$ divides $a$ .

Concept

Assume $\sqrt{3}$ is rational, i.e. $\sqrt{3} = a/b$ with $a, b$ co-prime integers; squaring gives $a^2 = 3b^2$ , so $3$ divides $a^2$ .
Since $3$ is prime, $3$ divides $a$ ; write $a = 3c$ , substitute back to get $b^2 = 3c^2$ , so $3$ divides $b$ too.
Now $3$ is a common factor of $a$ and $b$ , contradicting that $a/b$ was in lowest terms — hence $\sqrt{3}$ is irrational.

Common mistakes

Applying the divisibility lemma with a non-prime number (e.g. $4$ divides $36 = 6^2$ but $4$ does not divide $6$ ) — the lemma only works because 3 is prime, and this must be stated.
Forgetting to declare at the start that a and b are co-prime — that is exactly the assumption the final contradiction breaks.

Real-world

The discovery that $\sqrt{2}, \sqrt{3}$ , etc. cannot be written as fractions shattered the ancient Greek belief that 'all is number', reportedly troubling the Pythagoreans deeply.

Common Mistakes

1Using the divisibility property with a composite number (e.g. 'if $3$ divides $a^2$ then... ' stated without noting 3 is prime, or worse applying it to 4 or 6) — the lemma holds only for primes.
2Forgetting to state at the outset that a and b are co-prime (in lowest terms), which removes the very contradiction the proof relies on.
3Writing a numerical/decimal 'proof' (e.g. $\sqrt{3} = 1.732$ ... never repeats) — this is not a valid proof; CBSE requires the algebraic contradiction argument.

Previously Asked

2018Section CQ253 marks

Prove that √3 is an irrational number.

2023Section CQ273 marks

Prove that √3 is irrational.

Interesting Facts

The irrationality of numbers like $\sqrt{2}$ and $\sqrt{3}$ was known to the ancient Greeks around the 5th century BCE; legend says the Pythagorean Hippasus was punished for revealing that such 'incommensurable' magnitudes exist.

$\sqrt{3} \approx 1.7320508\ldots$ is called Theodorus' constant, after Theodorus of Cyrene who proved the irrationality of the square roots of 3, 5, 7, … up to 17.

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Frequently Asked Questions

Why must a and b be co-prime at the start?

The whole proof relies on reaching a contradiction. By assuming a/b is in lowest terms (co-prime), showing that 3 divides both a and b becomes an impossibility, which is exactly the contradiction that proves $\sqrt{3}$ is irrational.

Why does $3$ dividing $a^2$ imply $3$ divides $a$ ?

Because 3 is a prime number. By the Fundamental Theorem of Arithmetic, if a prime $p$ divides $a^2$ , it must appear in the prime factorisation of $a$ , so $p$ divides $a$ . This fails for composite numbers like 4.

Can I prove it just by writing the decimal of $\sqrt{3}$ ?

No. Observing that $\sqrt{3} = 1.732\ldots$ looks non-repeating is not a proof, because you cannot show non-repetition by writing finitely many digits. CBSE requires the logical contradiction argument.

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