Question

prove that √5 is irrational​

Answer 1

Answer:

Step-by-step explanation:

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

after cross multiplication

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

Solution:

Let $\sf{\sqrt{5}}$ is a rational number.

So, $\sf{\sqrt{5} = \dfrac{p}{q}\;\;\;\;[where\;p\;and\;q\;are\;co-prime\;numbers]}$

$\sf{\implies \sqrt{5}q=p\;\;\;\;...........(1)}$

Now, squaring both sides

=> 5q² = p²

$\sf{\implies q^{2}=\dfrac{p^{2}}{5}}$

Now, 5 is a factor of p² and so, 5 is also factor of p.

Let, p =5m

Put the value of p from Eq (1), we get

=> $\sf{\sqrt{5}q=5m}$

Now squaring both the sides, we get

=> 5q² = 25m²

=> q² = 25m²/5

=> q² = 5m²

=> q²/5 = m²     .........(2)

Now, 5 is a factor of q² and so, 5 is also factor of q.

We know that p and q are co-prime number having only 1 common factor, but we just proved that p and q are divided by 5.

So, our assumption is wrong.

Hence, $\sf{\sqrt{5}}$ is irrational number.