Question

Neha claimed that there does not exist any irrational number between 1 and 2. Raunak claimed that `sqrt(2)` lies between 1 and 2 and `sqrt(2)` is an irrational number. Who do you think is correct? Justify by proving either `sqrt(2)` as an irrational number or otherwise.

Answer 1

Remark is correct.

`sqrt(2)` is an irrational number that lies between 1 and 2.

Justification:

1² = 1, `(sqrt(2))^2 = 2`, 2² = 4

= 1 < 2 < 4

= 1² < 2 < 2²

= `sqrt(1) < sqrt(2) < sqrt(4)`

= `1 < sqrt(2) < 2`

This shows `sqrt(2)` is a digit between 1 and 2.

Proof of irrationality of `sqrt(2)`:

Let us assume `sqrt(2)` is a rational no.

∴ `sqrt(2) = a/b, b ≠ 0` and a and b are co-prime have no common factor other than 1.

Now, `sqrt(2) = a/b`

`(sqrt(2))^2 = (a/b)^2`   ...(Squaring on both sides)

= 2 = `(a^2)/(b^2)`   ...(i)

= b² = `(a^2)/2`

∴ 2 divides a² = 2 divides a.

Let a = 2m, where m is an integer.

= a² = (2m)²   ...(Squaring on both sides)

= a² = 4m²

= 2b² = 4m²    ...(From equation (i))

= `(2b^2)/4` = m²

= m² = `(b^2)/2`

∴ 2 divides b² = 2 divides b.

∴ a and b have at least 2 as a common factor.

But this contradicts the fact that a and b are co-prime.

∴ Our assumption is wrong.

∴ `sqrt(2)` is an irrational no.