Math, asked by Yeetzboi256, 2 days ago

Prove that √5 + 2 is irrational by the method of contradiction Plz answer will mark brainliest

Answers

Answered by priyanshukamble533
1

Step-by-step explanation:

Given: Number 5

To Prove: Root 5 is irrational

Proof: Let us assume that square root 5 is rational. Thus we can write, √5 = p/q, where p, q are the integers, and q is not equal to 0. The integers p and q are coprime numbers thus, HCF (p,q) = 1.

√5 = p/q

⇒ p = √5 q ------- (1)

On squaring both sides we get,

⇒ p2 = 5 q2

⇒ p2/5 = q2 ------- (2)

Assuming if p was a prime number and p divides a2, then p divides a, where a is any positive integer.

Hence, 5 is a factor of p2.

This implies that 5 is a factor of p.

Thus we can write p = 5a (where a is a constant)

Substituting p = 5a in (2), we get

(5a)2/5 = q2

⇒ 25a2/5 = q2

⇒ 5a2 = q2

⇒ a2 = q2/5 ------- (3)

Hence 5 is a factor of q (from 3)

(2) indicates that 5 is a factor of p and (3) indicates that 5 is a factor of q. This contradicts our assumption that √5 = p/q.

Therefore, the square root of 5 is irrational.

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