# The n-th root of m is either an integer or irrational

There is an intuitive way to see that for any natural numbers $$m$$ and $$n$$, $$\sqrt[n]m$$ will always either be an integer or an irrational number.

Suppose that there was some $$\sqrt[n]m$$ that was a rational number $$\frac{a}{b}$$ that was not an integer. Suppose further that $$\frac ab$$ is written as a reduced fraction, such that the greatest common divisor of $$a$$ and $$b$$ is $$1$$. Then, since $$\frac{a}{b}$$ is not an integer, $$b > 1$$.

Since $$\frac ab = \sqrt[n]m$$, we have conversely that $$(\frac ab)^n = m$$, which is a natural number by our hypothesis. But let’s take a closer look at $$(\frac ab)^n$$. It evaluates to $$\frac{a^n}{b^n}$$, which is still a reduced fraction. Proof of gcd(a^n, b^n) = 1.

But since $$b > 1$$ before, we have that $$b^n > 1$$ as well, meaning that $$(\frac ab)^n$$ cannot be an integer, contradicting the fact that it equals $$m$$, a natural number.

Parents:

• Mathematics

Mathematics is the study of numbers and other ideal objects that can be described by axioms.