Prove that under root 2 is not a rational number.


To prove that first we need to understand Theorem 1.3 : 


Let 𝑝 be a prime number. If 𝑝 divides π‘Ž2, then 𝑝 divides π‘Ž, where π‘Ž is π‘Ž 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 = π‘Ž/𝑏 .

Where, a and b are co-prime.

⇒    (√2)2 = (π‘Ž/𝑏)2

⇒    2 = π‘Ž2/𝑏2

⇒    2𝑏2 = π‘Ž2

Therefore, 5 divides π‘Ž2.

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

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

⇒    π‘Ž2 = 4𝑐2

⇒    2𝑏2 = 4𝑐2         (Using, 2𝑏2 = π‘Ž2)

⇒    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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