Math, asked by VivekR, 1 year ago

Prove that √5 is irrational!!!.....(╯_╰)​

Answers

Answered by harshdpatel18
4

Answer:

Step-by-step explanation:

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.

Answered by Anonymous
22

Ur answer... ✌️

Let us assume that √5 is a rational number.

we know that the rational numbers are in the form of p/q form where p,q are intezers.

so, √5 = p/q

p = √5q

we know that 'p' is a rational number. so √5 q must be rational since it equals to p

but it doesn't occurs with √5 since its not an integer

therefore, p =/= √5q

this contradicts the fact that √5 is an irrational number

hence our assumption is wrong and √5 is an irrational number

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