Question

prove that root 7 is irrational​

Answer 1

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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.

Answer 2

➠ Solution

Let √7 is a rational number.

Then, √7 = a / b ... { where a and b, are the coprime integers and b ≠ 0}

squaring both sides

( √7 ) ² = { a / b } ²

7 b ² = a² .. { 1 }

7 divides a²

7 divides a

a = 7c , for some integers c

a ² = 49 c² ( squaring both sides)

7 b ² = 49 c ² .. { from 1 }

b ² = 7 c ²

7 divides b ²

7 divides b

7 divides both a and b means it contradicts our supposition that a & b are coprime.

Hence, √7 is an irrational number.