Question

prove that
$\: prove \:that \: \sqrt{5} \: is \: irratation$
5 is irratation
​

Answer 1

Given :

• √5.

To Prove :

• √5 is an irrational number.

Proof :

Let us assume that, the √5 is an rational number and let its simplest form be a/b.

Then,

a and b are integers having no common factor other than 1, and b ≠ 0.

Now,

→ √5 = a/b

→ 5 = a²/b²⠀⠀ [on squaring both sides]

→ 5b² = a² ⠀⠀ ....(1)

→ 5 divides a² ⠀⠀ [∵ 5 divides 5b²]

→ 5 divides a ⠀⠀ [∵ 5 is prime and 5 divides a² → 5 divides a.]

Let a = 5c for some integer c.

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

→ 5b² = a²

→ 5b² = (5c)²

→ 5b² = 25c²

→ b² = 5c²

→ 5 divides b² ⠀⠀ [∵ 5 divides 5c²]

→ 5 divides b ⠀⠀ [∵ 5 is prime and 5 divides b² → 5 divides b.]

Thus, 5 is a common factor of a and b.

But, this contradicts the fact that a and b have no common factor other than 1.

The contradictions arises by assuming that √5 is rational number.

Hence √5 is an irrational number.

Answer 2

$\huge \red{\bf Answer}$

We need to prove that √5 is irrational

$\pink{\rm Proof:}$

Let us assume that √5 is a rational number.

Sp it t 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

ANSWER

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

$\green{ \bf \sqrt{5} \: is \: a \: irrational \: number}$