Question

Prove that √5 is irrational.​

Answer 1

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Let us assume, that √5 is rational number.

i.e. √5 = x/y (where, x and y are co-primes)

y√5= x

Squaring both the sides,

we get,

(y√5)^2 = x^2

⇒5y^2 = x^2……………………………….. (1)

Thus, x^2 is divisible by 5, so x is also divisible by 5.

Let us say, x = 5k,

for some value of k and substituting the value of x in equation (1),

we get,

5y^2 = (5k)^2

⇒y^2 = 5k^2

 is divisible by 5 it means y is divisible by 5.

Therefore,

x and y are co-primes.

Since,

our assumption about  is rational is incorrect.

Hence, √5 is irrational number.

Answer 2

Answer:

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 

Step-by-step explanation:

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