This little bit of inspiration came while reading Simon Singh's 'Fermat's Last Theorem.' Now that Fermat's Last Theorem has been proven, I have been able to construct an argument to show that in fact, the equation must have a least one irrational value in it's solution. We begin with an ordinary number line of whole integers. By proving Fermat's Last Theorem, it has been shown that any solution to the equation must include values that do not lie on any of the currently marked points on this line. Now we will attempt to find a fully rational solution. Firstly, we divide the current intervals to include all rational numbers, and then locate a solution (m/n) for one of the values x, y or z in the equation (xn + yn = zn, n > 2).
Fig. 24 Number line with rational solutions for Fermat's Last Theorem.
We can now adjust the scale of the number line to place our rational solution in line with an integer. Now, by Fermat's Last Theorem, the other variables in the solution must not be integers. Therefore for them to remain rational, they must be fractions. Let us find one of the other variables (p/q). We can now readjust the scale so that both our elements are in line with integers (this is because they exhibit a property called commensurability i.e. there exists a smaller segment that will divide into both, which is the geometrical equivalent of rationality). This process can then be extended t the third element (say r/s). However now we have found a scale on which all three fractions; m/n, p/q, and r/s, result in integers. This is impossible however because Fermat's Last Theorem has proven that no such solutions exist, and hence the third rational term cannot be possible. So, at least one of x, y or z in the equation of the theorem must be irrational (or imaginary but I won't go there)!
This proof can be established algebraically as well as geometrically. If a solution is fully rational, then (m/p)n + (q/r)n = (s/t)n
mn.rn.tn + qn.pn.tn = sn.pn.rn (multiply through by the product of the bases - pn.rn.tn)
(mrt)n + (qpt)n = (spr)n
But this is impossible since Fermat's Last Theorem stated that xn + yn ≠ zn
∴ One of the terms in the equation is not rational.
This means that the nth root of any rational number that can be expressed as the sum of the nth roots of two rational numbers, must be irrational. For example the cubed root of 2 (13 + 13) is irrational and the cubed root of a quarter (which is twice a half cubed) is irrational and therefore that of four is also irrational, since rationalizing the denominator gives the cubed root of four divided by four and then if the cubed root of four was a rational number than the cubed root of a quarter would be that number divided by four, another rational number! Since the cubed root of a quarter is irrational this cannot be so, and the irrationality of the cubed root of four is established. This can be a very powerful tool for proving the irrationality of the roots of numbers.
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