Class 10 Real Numbers: Fundamental Theorem of Arithmetic & Proof of Irrationality (Long Questions)

Let us assume, to the contrary, that sqrt(5) is rational.
Then there exist coprime positive integers a and b such that sqrt(5) = a / b.
Squaring both sides: 5 = a^2 / b^2 => 5b^2 = a^2 ... (1)
Since 5 divides 5b^2, 5 divides a^2. By the theorem, if a prime p divides a^2, then p divides a. Therefore, 5 divides a.
So we can write a = 5c for some integer c.
Substituting into (1): 5b^2 = (5c)^2 = 25c^2 => b^2 = 5c^2.
This means 5 divides b^2, so 5 divides b.
Therefore, 5 is a common factor of both a and b.
This contradicts the fact that a and b are coprime.
Hence, our assumption was incorrect. sqrt(5) is irrational.