Question

Prove that√7 is a irrational number​

Answer 1

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

Answer:

$let \: \sqrt{7} \: is \: a \: rational \: number \: \\ = > \sqrt{7} \: = \frac{p}{q} \: where \: p \: ana \: q \: are \: integers \: and \: q \: is \: not \: equalto \: zero \\ squaring \: on \: both \: sides \: \\ = > ({ \sqrt{7} })^{2} = ({ \frac{p}{q} )}^{2} \\ = > 7 = \frac{ {p}^{2} }{ {q}^{2} } \\ = > 7 {q}^{2} = {p}^{2} \\ now \: {p \: }^{2} and \: {q}^{2} \: are \: integers \: \\ then \: 7 \: is \: also \: a \: inter \: \\ let \: p \: = 7m \\ = > 7 {q}^{2} = {(7m})^{2} \\ = > 7 {q}^{2} = 49 {m}^{2} = > \frac{7 {q}^{2} }{49} = {m}^{2} \\ = > \frac{ {q}^{2} }{7} = {m}^{2} \\ = > {q}^{2} = 7 {m}^{2} \\ now \: p \: q \: 7 \: and \: m \: are \: coprime \\ p \: q \: 7 \: and \: m \: are \: rational number \\ \: \:then \: \: \sqrt{7} \: should \: also \: be \: a \: rationl \: number \\ hence \: \sqrt{7} \: is \: an \: irrational \: number$