Irrationality of $k$-th roots

Samuel G. Moreno, Esther M. García-Caballero

We aim to prove that for positive integers $k$ and $n$, $\sqrt[k]{n}$ is rational only when it is integer (namely, when $n$ is a perfect $k$-th power). Our proof goes by induction on $k$:

The base case ($k=1$) trivially holds because $\sqrt[1]{n}=n$ is always integer.
Now, the induction hypothesis: assuming our claim to hold for $k=j$, we will prove it for $k=j+1$. Set $r=\sqrt[j+1]{n}=a/b$, where the positive integers $a$ and $b$ are relatively prime (${\rm g.c.d.}(a,b)=1$). It is worth noting that ${\rm g.c.d.}(a^j,b^j)=1$, so Bezout's Lemma ensures the existence of integers $c$, $d$ such that $a^j d-b^j c=1$. Since $a^j=b^j r^j$, then

$\begin{align} 0&=(a^j-b^j r^j)(c+d\,r^j)\\ &=a^j c+(a^j d-b^j c)r^j-b^j\,d\,r^{2j}\\ &=a^j c+r^j-b^j\,d\,r^{2j}, \end{align}$

and thus (using that $r^{2j}=r^{j+1}r^{j-1}=n\,r^{j-1}$)

$\begin{align} r^j&=b^j\,d\,r^{2j}-a^j c\\ &=b^j\,d\,n\,\frac{a^{j-1}}{b^{j-1}}-a^j c\\ &=a^{j-1}(n\,b\,d-a c), \end{align}$

so the rational number $r$ is also the $j$-th root of a positive integer (the positivity of $n\,b\,d-a c$ can be verified by multiplying $a^j d-b^j c=1$ by $a/b^j$). By the induction hypothesis $r$ must be integer, and the proof is complete.

Further Comments: Our argument bears some resemblance with the one sketched out in [Floyd] which was, to the best of our knowledge, the first use of Bézout's Lemma for proving irrationality results. Moreover, our proof may be considered as a refinement of Floyd's beautiful (but somewhat informal) proof, in the sense that we do not require to check that $k$ algebraic expression involving $a$, $b$, and $r$ are integer numbers.

References

R. W. Floyd, What else Pythagoras could have done (Letter to the Editor), Amer. Math. Monthly 96 (1989) 67.
Samuel G. Moreno, Esther M. García-Caballero, Irrationality of $k$-th roots, Amer. Math. Monthly 120 (2013) 688.

71923754