$\sqrt[n]{2}$ is irrational for $n\ge 3$ (proof using Fermat’s last theorem)
Theorem 1.
If $n\mathrm{\ge}\mathrm{3}$, then $\sqrt[n]{\mathrm{2}}$ 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
${b}^{n}+{b}^{n}$ | $=$ | ${a}^{n}.$ | (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 $n\ge 3$ (proof using Fermat’s last theorem) |
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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 |