---
title: "Proof that √2 is irrational"
subject: "Discrete Structure"
chapter: 1
key: "semester_4/discrete_structure:proof-sqrt2-irrational"
source: https://www.bctengineer.com/q/semester_4/discrete_structure/proof-sqrt2-irrational
site: BCT Engineering
---

# Proof that √2 is irrational

Discrete Structure · Chapter 1: Logic, Induction and Reasoning

- Asked in: 2072 Asoj, 5 marks; 2068 Bhadra, 5 marks; 2073 Magh, 4 marks (asked 3 times)
- Video: [Proof: Square Root of 2 is Irrational](https://www.youtube.com/watch?v=2FlMkP7jqQo) — Wrath of Math
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/discrete_structure/proof-sqrt2-irrational
- Whole chapter: https://www.bctengineer.com/md/semester_4/discrete_structure/chapter-1

## Question

Prove that √2 is irrational by giving a proof by contradiction.

## Short answer

Assume √2 = a/b with a,b coprime integers. Squaring gives a² = 2b², so a is even, say a = 2k. Then 4k² = 2b² ⇒ b² = 2k², so b is even, contradicting coprimality. Hence √2 is irrational.

## Answer

**Claim.** √2 is irrational.

**Method.** Proof by contradiction. A rational number is a number that can be written as a/b with integers a, b, b ≠ 0. We assume √2 is rational and derive a contradiction.

#### Proof
1. Suppose √2 is rational. Then √2 = a/b where a and b are integers, b ≠ 0, and a/b is in lowest terms, so gcd(a, b) = 1.
2. Squaring both sides: 2 = a²/b², so **a² = 2b²**.
3. Hence a² is even. If a were odd, a² would be odd. So **a is even**; write a = 2k for some integer k.
4. Substitute: (2k)² = 2b², that is 4k² = 2b², so **b² = 2k²**.
5. Hence b² is even, and by the same argument as step 3, **b is even**.
6. Now both a and b are even, so 2 divides both. This contradicts gcd(a, b) = 1.
7. The assumption that √2 is rational is false.

**Therefore √2 is irrational.** ∎

Note: the fact used in step 3 is "if a² is even then a is even". It is proved by contraposition: if a = 2m + 1, then a² = 4m² + 4m + 1 = 2(2m² + 2m) + 1, which is odd.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is assumed at the start of the proof by contradiction?
   A. √2 is irrational
   B. √2 = a/b in lowest terms, with gcd(a, b) = 1
   C. a and b are both even
   D. a² is odd
   Answer: B. We assume √2 is rational and write it in lowest terms; the contradiction later is with gcd(a,b) = 1. (Key point: Assumption)

2. Squaring √2 = a/b gives which useful relation?
   A. a = 2b
   B. a² = b²/2
   C. a² = 2b²
   D. b² = 2a²
   Answer: C. 2 = a²/b² gives a² = 2b², which shows a² is even. (Key point: Squaring step)

3. Why does a² being even imply that a is even?
   A. Because every square is even
   B. If a were odd, a² would be odd (shown by contraposition)
   C. Because a = 2b
   D. Because gcd(a, b) = 1
   Answer: B. If a = 2m+1 then a² = 2(2m²+2m)+1 is odd, so an even a² forces a to be even. (Key point: Even square)

4. Where does the contradiction come from?
   A. a is even but b is odd
   B. a² = 2b² has no integer solution
   C. Both a and b turn out even, contradicting gcd(a, b) = 1
   D. b is zero
   Answer: C. a = 2k gives b² = 2k², so b is also even; both divisible by 2 contradicts lowest terms. (Key point: Contradiction)

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