## Frequently Asked Questions

#### When is a square root irrational?

The answer is very simple: if the square root a natural number is non-integer, then it is irrational.

From `y`² = `n` ⇒ `y` = √`n` (taking positive root).

Alternatively, (`y`²)^{½} = `n`^{½} ⇒ `y` = `n`^{½}.

Therefore √`n` = n^{½}.

In general, `n`^{1/q} is the `q`th root of `n`, and from this there is a remarkable and general result which we shall prove.

**Theorem**

If `n, q` are natural numbers and `n`^{1/q} (the `q`th root of `n`) is non-integer, then it is irrational.

**Proof**

Given that `n`^{1/q} is non-integer and `n, q` are natural numbers, let us assume that it is rational.

Let `n`^{1/q} = `a/b`, where `a` and `b` are a pair of natural numbers with no common factors; clearly `b` cannot equal one.

Therefore, (`n`^{1/q})^{q} = (`a/b`)^{q}, gives `n` = `a ^{q}/b^{q}`.

As the LHS is integer, the RHS must also be integer. But `a` and `b` have no common factors, so `b`^{q} cannot divide `a`^{q} (see note), unless `b`^{q} = 1, which is a contradiction. Hence there can be no ratio of natural numbers such that `n`^{1/q} is rational and so we prove it must be irrational.

NOTE: The statement, `b`^{q} cannot divide `a`^{q}, is based on the Fundamental Theorem of Arithmetic. For example, if `a` = 5 and `b` = 2 (which have no common factors), then no amount of multiplying 2 by itself (2^{q}) will produce a factor of 5, and so it will never divide into 5^{q}.

**Corollary**

If `n`, `p`, `q` are natural numbers and (`n ^{p}`)

^{1/q}is non-integer then it must be irrational (by the result just proved). As (

`n`)

^{p}^{1/q}=

`n`, it follows that if the result of raising

^{p/q}`n`to any rational power,

`p/q`, is non-integer it must be irrational.