Prove that is irrational.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rational Numbers

Question

Prove that $\sqrt{2}$ is irrational.

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Let us assume that $\sqrt{2}$ is rational. So, we can find integers p and q ( ≠ 0) such that,

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

Suppose p and q have a common factor other than 1.

Then, we divide by the common factor to get

\sqrt{2} = \frac{a}{b}

, where a and b are co-prime.
So,

b \sqrt{2} = a

Squaring on both sides, we get

2b 2 = a 2

Therefore, 2 divides a²
Now, we know that, if a prime number, p divides a², then p divides a, where a is a positive integer.
Hence, 2 divides a.

So, we can write a = 2c for some integer c.
Substituting for a, we get

2b 2 = 4c 2

\Rightarrow b 2 = 2c 2

This means that 2 divides b² and so, 2 divides b.

Therefore, a and b have at least 2 as a common factor.
But, this contradicts the fact that a and b have no common factors other than 1.

This contradiction is because of our incorrect assumption that

\sqrt{2}

is rational.
So, we can conclude that

\sqrt{2}

is irrational. [Hence proved]