Prove that is sqrt2 irrational number. - Mathematics

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Sum

Prove that is `sqrt2` irrational number.

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Solution 1

Let us assume, to the contrary, that `sqrt2` is rational. So, we can find integers r and s (≠ 0) such that `sqrt2="r"/"s"`.

 Suppose r and s not having a common factor other than 1. Then, we divide by the common factor to get `sqrt2="a"/"b"` 

where a and b are coprime.

So,  b`sqrt2` = a.

Squaring on both sides and rearranging, we get 2b2 = a2.

Therefore, 2 divides a2. Now, by Theorem it following that 2 divides a.

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

Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2.

This means that 2 divides b2, and so 2 divides b (again using Theorem with p = 2).

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 `sqrt2` is rational.

So, we conclude that `sqrt2` is irrational.

Solution 2

Let `sqrt(2)` be rational.

∴ `sqrt(2)  = "p"/"q"` where p and q are co-prime integers and, q ≠ 0
Implies that `sqrt(2"q")  = "p"`

2q2 - p2          ...(i) 

⇒ 2 divides p2

⇒ 2 divides p       ...(A)

Let p = 2C for some integer c

p2 = 4c2

⇒ 2q2 = 4C2

⇒ q2 = 2c2

⇒ 2 divides q2

⇒ 2 divides q        ...(B)

From (A) and (B), we get 

∴ 2 is common factor of both p and q. But this contradicts the fact that p and q have no common factor other than 1.

∴  Our supposition is wrong Hence, `sqrt(2)` is an irrational number.

  Is there an error in this question or solution?
2023-2024 (March) Standard (Board Sample Paper)

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