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.