Question

Prove that √7 is an irrational number.​

Answer 1

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Let us assume that -/7

7 is rational. Then, there exist co-prime positive integers a and b such that

-/7 =b/a

a ⟹a=b/7

7 Squaring on both sides, we get

a /2 =7b^2

Therefore, 2 is also divisible by 7 and

so, b is also divisible by 7.

Therefore, a and b have at least one common factor, i.e., 7.

But, this contradicts the fact that a and b are co-prime.

But, this contradicts the fact that a and b are co-prime.Thus, our supposition is wrong.

Hence, -/7 is irrational.

Answer 2

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Answer:

Given √7

To prove: √7 is an irrational number.

Proof:

Let us assume that √7 is a rational number.

So it t can be expressed in the form p/q where p,q are co-prime integers and q≠0

√7 = p/q

Here p and q are coprime numbers and q ≠ 0

Solving

√7 = p/q

On squaring both the side we get,

=> 7 = (p/q)2

=> 7q2 = p2……………………………..(1)

p2/7 = q2

So 7 divides p and p and p and q are multiple of 7.

⇒ p = 7m

⇒ p² = 49m² ………………………………..(2)

From equations (1) and (2), we get,

7q² = 49m²

⇒ q² = 7m²

⇒ q² is a multiple of 7

⇒ q is a multiple of 7

Hence, p,q have a common factor 7. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√7 is an irrational number.