Question

Prove that √5 is an irrational number.​

Answer 1

Answer:

To prove that √5 is irrational number

Let us assume that √5 is rational

Then √5 = \frac{a}{b}

b

a

(a and b are co primes, with only 1 common factor and b≠0)

⇒ √5 = \frac{a}{b}

b

a

(cross multiply)

⇒ a = √5b

⇒ a² = 5b² -------> α

⇒ 5/a²

(by theorem if p divides q then p can also divide q²)

⇒ 5/a ----> 1

⇒ a = 5c

(squaring on both sides)

⇒ a² = 25c² ----> β

From equations α and β

⇒ 5b² = 25c²

⇒ b² = 5c²

⇒ 5/b²

(again by theorem)

⇒ 5/b-------> 2

we know that a and b are co-primes having only 1 common factor but from 1 and 2 we can that it is wrong.

This contradiction arises because we assumed that √5 is a rational number

∴ our assumption is wrong

∴ √5 is irrational number

Answer 2

Let assume that √5 is a rational number

p/q = √5

p²/q²=5 (squaring both side)

p²=5q² --------------------1

p² is the multiple of 5

so, P is also the multiple of 5

P=5m (putting in equation --1)

(5m²) =5q²

25m²=5q²

q²=5m²

q is also the multiple of 5

P and q are not co-prime

so,our assumption is wrong

Therefore, √5 is an irrational number

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