Question

Prove that √5 is irrational​

Answer 1

Answer:

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To prove that √5 is irrational number 

Let us assume that √5 is rational 

Then √5 =  

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

⇒ √5 =  

(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:

Let us assume that √5 is rational

It can be expressed in the form p/q where p,q are co-prime integers and q≠0

⇒√5=p/q

On squaring both the sides we get,

⇒5=p²/q²

⇒5q²=p² —————–(i)

p²/5= q²

So 5 divides p

p is a multiple of 5

⇒p=5m

⇒p²=25m² ————-(ii)

From equations (i) and (ii), we get,

5q²=25m²

⇒q²=5m²

⇒q² is a multiple of 5

⇒q is a multiple of 5

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√5 is an irrational number

hope this helps you.