Question

PROVE THAT ROOT 7-ROOT 5 IS AN IRRATIONAL NUMBER

Answer 1

Let us prove that √5 is an irrational no by contradiction
Let us assume that √5 is a rational no
√5=p/q where p and q are integers, coprimes and q isn't equal to 0
√5q=p
Squaring on both sides
5q^2=p^2
It means p^2 is a multiple of 5
then p is also a multiple of 5
p=5r(let r be some integer)
Squaring on both sides
p^2=25r^2
We know that p^2=5q^2
5q^2=25r^2
q^2=25r^2/5
q^2=5r^2
It means q^2 is a multiple of 5
q is also a multiple of 5
But it contradicts tht p and q are co primes
So our assumption that √5 is rational is wrong
So √5 is irrational

In the same way by putting √7 in the place of √5 we can prove that √7 is irrational

Hope is answer helps you!!

Answer 2

Answer:

Step-by-step explanation:

5q^2=p^2

It means p^2 is a multiple of 5

then p is also a multiple of 5

p=5r(let r be some integer)

Squaring on both sides

p^2=25r^2

We know that p^2=5q^2

5q^2=25r^2

q^2=25r^2/5

q^2=5r^2

It means q^2 is a multiple of 5

q is also a multiple of 5

But it contradicts tht p and q are co primes

So our assumption that √5 is rational is wrong

So √5 is irrational