Question

Prove that ${\displaystyle \sqrt{5}}$ is an irrational number. 

Answer 1

Let us assume, to the contrary, that ${\displaystyle \sqrt{5}}$ is irrational.

That is, we can find integers a and b (≠0) such that ${\displaystyle \sqrt{5}=\frac{a}{b}}$.

Suppose a and b have a common factor other than 1, then we can divide by the common factor and assume that a and b are co-prime.

Therefore, b${\displaystyle \sqrt{5}}$ = a.

Squaring on both sides 5b² = a²  ...(1)

The above implies that a² is divisible by 5 and also a is divisible by 5.

Therefore, we can write that a = 5k for some integer k.

Substituting in (1) 5b² = (5k)²

5b² = 25k² (or) b² = 5k²

⇒ b² is divisible by 5, which means b is also divisible by 5.

Therefore, a and b have 5 as a common factor.

This contradicts the fact that a and b are coprime.

We arrived at the contradictory statement as above since our assumption ${\displaystyle \sqrt{5}}$ is not correct.

Hence, we can conclude that ${\displaystyle \sqrt{5}}$ is an irrational number.

Answer 2

If possible, let ${\displaystyle \sqrt{5}}$ be rational and let its simplest form be ${\displaystyle \frac{p}{q}}$.

Then p and q are integers having no common factor other than 1, and q ≠ 0.

Now ${\displaystyle \sqrt{5}=\frac{p}{q}}$

⇒ 5 = ${\displaystyle \frac{p^2}{q^2}}$    ...[on squaring both sides]

⇒ 5q² = p²     ...(i)

⇒ 5 divides p² ⇒ 5 divides p    ...[∵ 5 is prime and 5 divides p² ⇒ 5 divides p]

Let p = 5R for some integer R.

Putting p = 5R in (i), we get

5q² = 25R²

⇒ q² = 5R²

⇒ 5 divides q²      ...[∵ 5 divides 5R²]

⇒ 5 divides q        ...[∵ 5 divides q]

Thus, p and q share a common factor of 5. However, this contradicts the fact that p and q share no common component other than 1. Hence, ${\displaystyle \sqrt{5}}$ is irrational.