Prove that root 2 Is Irrational.

Given: root 2.

To find: We have to prove that √2 is irrational.

Solution: 

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 √2  is rational.

So, we can find integers a and b (≠ 0) such that √2 = $\frac{a}{b}$.

Where a and b are co-prime.

⇒    (√2)2 = $\frac{a}{b}$2

⇒    2 = $\ \frac{a^{2}}{b^{2}}$

⇒    2𝑏2 = 𝑎2

Therefore, 5 divides 𝑎2.

Now, by Theorem 1.3, it follows that 2 divides a.

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

⇒    𝑎2 = 4𝑐2

⇒    2𝑏2 = 4𝑐2         (Using, 2𝑏2 = 𝑎2)

⇒    𝑏2 = 2𝑐2

Therefore, 2 divides 𝑏2.

Now, by Theorem 1.3, it follows that 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 has arisen because of our incorrect assumption that √2 is rational.

So, we conclude that √2 is irrational.

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Updated on: 10-Oct-2022

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