Prove That √5 Is Irrational

To prove that first, we need to understand Theorem 1.3 : 

Let 𝑝 be a prime number. If 𝑝 divides 𝑎2, then 𝑝 divides 𝑎, where 𝑎 is 𝑎 positive integer.

Now,

Let us assume, to the contrary, that √5  is rational.

So, we can find integers a and b (≠ 0) such that √5 = 𝑎/𝑏.

Where a and b are co-prime.

⇒    (√5)2 =$\left(\frac{a}{b}\right)^{2}$

⇒    5 = $(\frac{a^{2}}{b^{2}})$ 

⇒    5𝑏2 = 𝑎2

Therefore, 5 divides 𝑎2.

Now, by Theorem 1.3, it follows that 5 divides a.

So, we can write a = 5c for some integer c.

⇒    𝑎2 = 25𝑐2

⇒    5𝑏2 = 25𝑐2         (Using, 5𝑏2 = 𝑎2)

⇒    𝑏2 = 5𝑐2

Therefore, 5 divides 𝑏2.

Now, by Theorem 1.3, it follows that 5 divides b.

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

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that √5 is rational.

So, we conclude that √5 is irrational.

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Updated on: 10-Oct-2022

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