Question

prove that √5 is irrational number​

Answer 1

EXPLANATION.

→ Let we assume that √5 be a rational number.

→ Then, √5 is in the form of p/q. where q ≠ 0

→ √5 = p/q

→ √5q = p

→ squaring on both sides we get,

→ ( √5q)² = (p)²

→ 5q² = p² ........ (1)

→ p² is divisible by 5

→ so, p is also divisible by 5

→ Let p = 5c

→ squaring on both sides we get,

→ (p)² = (5c)²

→ p² = 25c² .........(2)

→ Put the value of equation (2) in (1) we get,

→ 5q² = 25c²

→ q² = 5c²

→ q² is divisible by 5

→ so, q is also divisible by 5

→ This p and q has common factor of 5.

→ we have assume that p and q have a

co - prime but they have a common factor

of 5

→ Therefore, √5 is a irrational number.

Answer 2

Answer:

Question

prove that √5 is irrational number

To prove

√5 is an irrational number

Solution

Let us assume√5 a rational number .

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

P/q form = √5

Let's square the number

$5 = \frac{ {p}^{2} }{ {q}^{2} }$

$5 {q}^{2} = \: {p}^{2}$

$\frac{ {p}^{2} }{5} = {q}^{2}$

5 divides p

So, p is a multiple of 5 .

➡️p = 5m

➡️ p² = 25² m

So, from the equation 1 and 2 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 as 5. Therefore contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

Hence proved

Hope it helps