Chapter 1 · Question 4

Use the Fundamental Theorem of Arithmetic to prove that $\sqrt{2}$ is irrational.

Answer Revealed

Direct Answer:

Assume

\sqrt{2}=\frac{p}{q}

in lowest terms. Then

p^2=2q^2

, so

p^2

is even and hence

p

is even. Let

p=2k

. Then

4k^2=2q^2

, so

q^2=2k^2

, making

q

even. This contradicts that

p

and

q

are coprime. Hence

\sqrt{2}

is irrational.

Simple Explanation

\sqrt2

were rational, both numerator and denominator would turn out even, which is impossible for a fraction in lowest terms.

Exam-Ready Structure

Suppose

\sqrt2=\frac pq

, where

p

and

q

are coprime positive integers. Squaring gives

2=\frac{p^2}{q^2}

, so

p^2=2q^2

. Hence

p^2

is even, so

p

is even. Put

p=2k

. Then

4k^2=2q^2

, so

q^2=2k^2

, and

q

is also even. Thus

p

and

q

have common factor

2

, contradicting the assumption. Therefore

\sqrt2

is irrational.

Key Points

Assume $\sqrt{2}=\frac{p}{q}$ in lowest terms.
Use the NCERT formula or theorem carefully.
Write units and final conclusion where applicable.

Use the Fundamental Theorem of Arithmetic to prove that $\sqrt{2}$ is irrational.