# $\sqrt[n]{2}$ 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 , 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 , 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) sqrtn2IsIrrationalForNge3proofUsingFermatsLastTheorem 2013-03-22 13:38:32 2013-03-22 13:38:32 matte (1858) matte (1858) 13 matte (1858) Proof msc 11J72