Question

prove that√5 is an irrational number​

Answer 1

Answer:

To prove that √5 is irrational number 

Let us assume that √5 is rational 

Then √5 =  $\frac{a}{b}$

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

⇒ √5 =  $\frac{a}{b}$

(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

Step-by-step explanation:

Hello, my name is vansh jain.

let's say that root 5 is a rational number so It can be expressed in p/q form.

root 5= p/q  

5= p^2/q^2 (squaring both sides)

5q^2 = p^2(1)

p^2 is a multiple of 5

p is also multiple of 5

p=5m

p^2 = 25m^2

from the equation first and second, we get,

5q^2=25m^2 ;  q^2=5m^2

q^2 is a multiple of 5 and q is also a multiple of 5

Hence, p,q have a comman factor 5. this contradict that they are co-prime therefore, p/q is not a rational number proved that root 5 is irrational number.