prove that √5 is irrational.
Answers
Step-by-step explanation:
Answer
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
On squaring both sides we get,
⇒ {p}^{2}p2 = 5{q}^{2}q2
⇒ {p}^{2}p2 /5 = {q}^{2}q2
Assuming if p was a prime number and p divides
{a}^{2}a2 , then p divides a, where a is any positive integer.
Hence, 5 is a factor of {p}^{2}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}2 /5 = {q}^{2}q2
⇒ 2{5}a^{2}5a2 /5 = {q}^{2}q2
⇒ 5a{}^{2}2 = q{}^{2}2
⇒ a{}^{2}2 = q{}^{2}2 /5
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.