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is irrational for (proof using Fermat's last theorem)
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(Proof)
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Theorem 1 If $n\ge 3$ , 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 \begin{eqnarray} \label{fermat} b^n + b^n &=& a^n. \end{eqnarray}We can now apply a recent result of Andrew Wiles [ 1], which states that there are no non-zero integers $a$ , $b$ satisfying equation ( ![[*] [*]](http://images.planetmath.org:8080/cache/objects/4292/js//usr/share/latex2html/icons/crossref.png) ). Thus $\sqrt[n]{2}$ is irrational. 
The above proof is given in [2], where it is attributed to W.H. Schultz.
- 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).
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" is irrational for (proof using Fermat's last theorem)" is owned by matte. [ full author list (2) ]
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Cross-references: equation, Andrew Wiles, integers, positive, circular argument, theorem, information, pathological, proof, irrational
This is version 10 of is irrational for (proof using Fermat's last theorem), born on 2003-05-23, modified 2006-07-25.
Object id is 4292, canonical name is NthRootOf2IsIrrationalForNge3ProofUsingFermatsLastTheorem.
Accessed 2702 times total.
Classification:
| AMS MSC: | 11J72 (Number theory :: Diophantine approximation, transcendental number theory :: Irrationality; linear independence over a field) |
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Pending Errata and Addenda
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