Prove that √5 is an irrational number
Answers
Answer:
Prove That Root 5 is Irrational by Contradiction Method
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.
Step-by-step explanation:
Answer:
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Step-by-step explanation:
