Editing Nth root (section)

== Proof of irrationality for non-perfect ''n''th power ''x'' ==
Assume that <math>\sqrt[n]{x}</math> is rational. That is, it can be reduced to a fraction <math>\frac{a}{b}</math>, where {{mvar|a}} and {{mvar|b}} are integers without a common factor.

This means that <math>x = \frac{a^n}{b^n}</math>.

Since ''x'' is an integer, <math>a^n</math>and <math>b^n</math>must share a common factor if <math>b \neq 1</math>. This means that if <math>b \neq 1</math>, <math>\frac{a^n}{b^n}</math> is not in simplest form. Thus ''b'' should equal 1.

Since <math>1^n = 1</math> and <math>\frac{n}{1} = n</math>, <math>\frac{a^n}{b^n} = a^n</math>.

This means that <math>x = a^n</math> and thus, <math>\sqrt[n]{x} = a</math>. This implies that <math>\sqrt[n]{x}</math> is an integer. Since {{mvar|x}} is not a perfect {{mvar|n}}th power, this is impossible. Thus <math>\sqrt[n]{x}</math> is irrational.