Question

Prove that ${\bf{\red{\sqrt{5}}}}$ is rational.​

Answer 1

$\huge{\color{red}{\textsf{\textbf {\underline{ Anѕwєr : }}}}}$

CORRECT QUESTION PROVE THAT ROOT 5 IS IRRATIONAL NUMBER .

Given:

• √5

• We need to prove that √5 is irrational

Proof:

• Let us assume that √5 is a rational number.

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

⇒ √5 = p/q

On squaring both the sides we get,

⇒5 = p²/q²

⇒5q² = p² —————–(i)

p²/5 = q²

So 5 divides p

p is a multiple of 5

⇒ p = 5m

⇒ p² = 25m² ————-(ii)

From equations (i) and (ii), we get,

5q² = 25m²

⇒ q² = 5m²

⇒ q² is a multiple of 5

⇒ q is a multiple of 5

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√5 is an irrational number.

Hence proved

Answer 2

$\huge\sf \pmb{\orange {\underline\green{\underline{\: \: Ꭺ ꪀ \mathfrak ꕶ᭙ꫀя \: \: }}}}$

$\huge\sf{\underline{To \: prove:}}$

$\implies\sf{√5 \: is \: irrational \: number}$

$\huge\sf{\underline{Given:}}$

• $\sf{√5}$
• $\sf{We \: have \: to \: prove \: that \: √5 \: is \: irrational \: number}$

$\huge\sf{\underline{Proof:}}$

• $\sf{Let \: us \: assume \: that \: √5 \: is \: a \: rational \: number}$

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

$\small\implies \sf{√5 = p/q}$

On squaring both sides we get,

$\small\implies \sf{5 = {p}^{2} / {q}^{2} }$

$\small\implies\sf5 {q}^{2} = {p}^{2}-----(i)$

$\sf {p}^{2} /5 = {q}^{2}$

So 5 divides p.

P is a multiple of 5.

$\small\implies\sf{P = 5m}$

$\small\implies\sf{ {p}^{2} = 25 {m}^{2}- ----(ii)}$

From eq. (i) & (ii), we get,

$\small\implies\sf5 {p}^{2} = 25 {m}^{2}$

$\small\implies\sf{ {q}^{2} = 5 {m}^{2}}$

$\small\implies\sf{{q}^{2} \: is \: a \: multiple \: of \: 5}$

$\small\implies\sf{ q \: is \: a \: multiple \: of \: 5}$

Hence, p,q have a common factor 5. This contradicts our assumption that they are co primes. Therefore, p/q is not a rational number.

$\huge\sf{\underline{Hence\: Proved}}$

$\huge\sf{\underline{√5\: is\: an\: irrational\: number}}$