Irrational Number

Solution: If not, let us assume $\sqrt{2}$ is rational.

Then $\sqrt{2}$ =p/q (q ≠0 , (p,q) ≠ 2)

=> 2 = p²/q²

^=>2 q²= p²

Now 2 q²is divisible by 2, hence p² is divisible by 2

Then p is divisible by 2. Now let p = 2r

2q²= 4 r²

q²= 2 r²

q²is divisible by 2.

q is divisible by 2

It contradicts our assumption that p and q are not divisible by 2.

Solution: If not, let us assume $\sqrt{3}$ is rational.

Then $\sqrt{3}$ =p/q (q ≠0 , (p,q) ≠ 3)

=> 3 =p²/q²

^=>3q²= p²

Now 3q²is divisible by 3, hence p² is divisible by 3

Then p is divisible by 3. Now let p=3r

3q²= 9 r²

q²= 3 r²

q²is divisible by 3.

q is divisible by 3

It contradicts our assumption that p and q are not divisible by 3.

Solution: If not, let us assume $\sqrt{5}$ is rational.

Then $\sqrt{5}$ =p/q (q ≠0 , (p,q) ≠ 5)

=>5 =p²/q²

5q²= p²

Now 5 q²is divisible by 5, hence p² is divisible by 5

Then p is divisible by 5. Now let p=5r

5q²= 25 r²

q²= 5 r²

q²is divisible by 5.

q is divisible by 5

It contradicts our assumption that p and q are not divisible by 5.