Question

Why is the square root of 5 an irrational number?

Answer 1

See explanation...

Explanation:

Here's a sketch of a proof by contradiction:

Suppose `sqrt(5) = p/q` for some positive integers `p` and `q`.

Without loss of generality, we may suppose that `p, q` are the smallest such numbers.

Then by definition:

`5 = (p/q)^2 = p^2/q^2`

Multiply both ends by `q^2` to get:

`5 q^2 = p^2`

So `p^2` is divisible by `5`.

Then since `5` is prime, `p` must be divisible by `5` too.

So `p = 5m` for some positive integer `m`.

So we have:

`5 q^2 = p^2 = (5m)^2 = 5*5*m^2`

Divide both ends by `5` to get:

`q^2 = 5 m^2`

Divide both ends by `m^2` to get:

`5 = q^2/m^2 = (q/m)^2`

So `sqrt(5) = q/m`

Now `p > q > m`, so `q, m` is a smaller pair of integers whose quotient is `sqrt(5)`, contradicting our hypothesis.

So our hypothesis that `sqrt(5)` can be represented by `p/q` for some integers `p` and `q` is false. That is, `sqrt(5)` is not rational. That is, `sqrt(5)` is irrational.