Question

Prove √5 is irrational

Answer 1

This irrationality proof for the square root of 5 uses Fermat's method of infinite descent: Suppose that √5 is rational, and express it in lowest possible terms (i.e., as a fully reduced fraction) as mn for natural numbers m and n. Then √5 can be expressed in lower terms as 5n − 2mm − 2n, which is a contradiction.

Answer 2

HI

Proof:

Let us assume to the contrary that √5 is rational

Therefore, √5 = a/b , where a and b are coprime integers and b ≠ 0

Squaring on both sides,

5 = a²/b²

5b² = a² ......(1)

Since, a² is divisible by 5,

Therefore, a is divisible by 5

Let a = 5c and substitute in eq(1)

5b² = (5c)²

5b² = 25c²

b² = 5c²

Since, b² is divisible by 5,

Therefore, b is divisible by 5

➡️ a and b have at least one common factor i.e. 5

This contradicts the fact that a and b are coprime.

➡️ This contradiction has arisen due to our incorrect assumption that √5 is rational.

Therefore, we conclude that √5 is irrational.

Hence Proved !

Hope it proved to be beneficial....