However, some people believe that a Pythagorean, possibly Hippasus, discovered the existence of irrational numbers in a different way, by considering the mystic pentagram. Since this figure was sacred to the Pythagoreans, they must have been curious about determining its dimensions. And it is not too difficult to imagine that one of them was led to the discovery of irrationals this way. Indeed, if the five diagonals of a regular pentagon are drawn, forming the mystic pentagram, the ratio of the length of a diagonal of the pentagon to the length of a side is the irrational number phi, the golden ratio.

It makes a great exercise for good beginning geometry students to prove, as some early Greek geometer must have, that the ratio mentioned above cannot be a rational number, using what is essentially Fermat's method of infinite descent. I'll sketch an outline of the proof below.

Consider a regular pentagon of side length s and let the length of each diagonal be d. A regular pentagon is formed inside the original one. Let its length be s', and the length of its diagonals be d'. After drawing the diagram, the student needs to make repeated use of the following elementary facts:

(1) The interior angle in a regular pentagon is 3 * 180 / 5 = 108 degrees.

(2) The sum of the angles in a triangle is 180 degrees.

(3) Two sides of a triangle are equal iff the two angles opposite the sides are equal.

Using these facts it becomes apparent that the diagram has 36 degree angles all over the place [ 36 = (180 - 108)/2] and lots of isosceles triangles. Using this information, the following simple relations can be determined:

(4) s' = 2s - d

(5) d' = d -s

If the ratio d:s is rational, then (by scaling) we may assume that d and s are positive

__integers__. But according to formulas (4) and (5), this means that d' and s' are integers too, and obviously from the diagram d' < d and s' < s. Now, we can imagine continuing the process of drawing diagonals and producing smaller and smaller nested pentagons over and over. Each time the length of the side and the length of the diagonal is a smaller positive integer. But after (much) less than s iterations, the length of a side will be less than 1, not an integer. So d and s can not both be integers.

To see where phi arises, use (4) and (5) to write

(6) d'/s' = (d - s)/(2s - d)

Since the original pentagon and the nested one are similar, we can replace the left hand side by d/s, and then by dividing the numerator and denominator of the right hand side by s we obtain a quadratic equation for (d/s). The positive solution is phi = (1 + sqrt(5))/2.

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