**Question 1.** **Asked on :**29 June 2021:11:08:46 AM

*-Added by Master Purushottam*

**Answer:**

Let us assume that √5 is a rational number.

∴ we can write it as √5 = pq

Now we can reduce p and q and we get reduced value of p and q as a and b which are co-prime have no common factor other than 1.

So we have √5 = ab

⇒ √5b = a

⇒ 5b^{2} = a^{2} [squaring both sides] ............ (i)

⇒ b^{2} = a^{2}5

5 divides a^{2} so a is also divisible by 5 (theorem 1.3) ..... (ii)

so 5 is a factor of a.

Now taking a = 5c for some integer c

⇒ 5b^{2} = (5c)^{2}

⇒ 5b^{2} = 25 c^{2}

⇒ b^{2} = 5 c^{2}

⇒ c^{2} = b25

5 divides b^{2} so b is also divisible by 5 (theorem 1.3) ...... (iii)

from (ii) and (iii) we get that

a and b are divisible by 5 this implies that 5 is common factor of a and b, while we assume that a and b are co-prime having no common factor other than 1.

This contradiction has risen due to our wrong assumption so √5 cann't be

expressed as pq, Hence √5 is an irrational number.

*-Answered by Master Purushottam* On 29 June 2021:11:10:35 AM

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