$\sqrt[n]{2}$ is irrational for $n\geq 3$ (proof using Fermat’s last theorem)

Theorem 1.

If $n\geq 3$ , then $\sqrt[n]{2}$ is irrational.

The below proof can be seen as an example of a pathological proof. It gives no information to “why” the result holds, or how non-trivial the result is. Yet, assuming Wiles’ proof does not use the above theorem anywhere, it proves the statement. Otherwise, the below proof would be an example of a circular argument.

Proof.

Suppose $\sqrt[n]{2}=a/b$ for some positive integers $a, b$ . It follows that $2=a^{n}/b^{n}$ , or

\displaystyle b^{n}+b^{n}

\displaystyle=

\displaystyle a^{n}.

(1)

We can now apply a recent result of Andrew Wiles [1], which states that there are no non-zero integers $a$ , $b$ satisfying equation (1). Thus $\sqrt[n]{2}$ is irrational. ∎

The above proof is given in [2], where it is attributed to W.H. Schultz.

References

1 A. Wiles, Modular elliptic curves and Fermat’s last theorem, Annals of Mathematics, Volume 141, No. 3 May, 1995, 443-551.
2 W.H. Schultz, An observation, American Mathematical Monthly, Vol. 110, Nr. 5, May 2003. (submitted by R. Ehrenborg).

Title	$\sqrt[n]{2}$ is irrational for $n\geq 3$ (proof using Fermat’s last theorem)
Canonical name	sqrtn2IsIrrationalForNge3proofUsingFermatsLastTheorem
Date of creation	2013-03-22 13:38:32
Last modified on	2013-03-22 13:38:32
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	13
Author	matte (1858)
Entry type	Proof
Classification	msc 11J72

2n is irrational for n≥3 (proof using Fermat’s last theorem)

Theorem 1.

Proof.

References

$\sqrt[n]{2}$ is irrational for $n\geq 3$ (proof using Fermat’s last theorem)