Question

1. Prove that √5 is irrational.​

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

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Prove that V5 is irrational. We have to prove v5 is irrational Let us assume the opposite, i.e., V5 is rational Hence, v5 can be written in the form - where a and b (b# 0) are co-prime (no common factor other than

Prove that V5 is irrational. We have to prove v5 is irrational Let us assume the opposite, i.e., V5 is rational Hence, v5 can be written in the form - where a and b (b# 0) are co-prime (no common factor other than 1) Hence, V5 = " V5b = a Squaring both sides (V5b)? = a? 5b? = a?