Question

Prove that √5 is irrational..​

Answer 1

$\huge\mathfrak\green{Answer:-}$

Inorder To prove that √5 is irrational number 

Let us assume that √5 is rational 

Then √5 = a/b

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

=> √5 = a/b

(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

Answer:

Inorder To prove that √5 is irrational number

Let us assume that √5 is rational

Then √5 = a/b

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

=> √5 = a/b

(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

Explanation:

Hope this answer will help you dear...