Question

Prove that √5 is irtional number​

Answer 1

Answer:

Let's prove this by the method of contradiction-

Say, √5 is a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.

⇒√5=p/q

⇒5=p²/q² {Squaring both the sides}

⇒5q²=p² (1)

⇒p² is a multiple of 5. {Euclid's Division Lemma}

⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}

⇒p=5m

⇒p²=25m² (2)

From equations (1) and (2), we get,

5q²=25m²

⇒q²=5m²

⇒q² is a multiple of 5. {Euclid's Division Lemma}

⇒q is a multiple of 5.{Fundamental Theorm of Arithmetic}

Hence, p,q have a common factor 5. this contradicts that they are co-primes. Therefore, p/q is not a rational number. This proves that √5 is an irrational number.

For the second query, as we've proved √5 irrational.

Answer 2

We have to prove√5 is irrational

let us assume the opposite,

i.e, √5 is rational

Hence, √5 can be written as in the form a/b

where a and b ( b≠0) are co prime ( no common factor other than 1)

Hence, √5= a/b

√5b=a

squaring on both sides

(√5b)^2= a^2

5b^2=a^2

a^2/5= b^2

Hence , 5 divides a^2

so,

a=5c

putting a=5c

5b^2=5c^2

5b^2=25 b^2

b^2/5= c^2

Hence 5 divides b^2 or 5 divides b also ----------- 2

By equation 1 and 2

5 divides both a and b

Hence, 5 is a factor of a and b

so, a and b have a factor 5

Therefore, and b are not co prime .

So, our assumption is wrong..

By contradiction,

√5 is irrational..

Hope it helps you my friend....