# 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.