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 $k$:

  1. The base case ($k=1$) trivially holds because $\sqrt[1]{n}=n$ is always integer.

  2. 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

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

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