The diagonal of the square and the irrationals.

| incommensurability | geometric demonstration | algebraic demonstration |

One of the most important discoveries of the Pythagorean School is without doubt that of the incommensurability of the side and the diagonal of the square. Two segments L and D are commensurable when they have a common submultiple, that is when L and D are multiples of the same segment, H:
L = m H
D = n H
with m and n integers. The Pythagoreans discovered that if L is the side and D the diagonal of a square, this relation is impossible. What their reasoning was is unknown; among the suggested ones, two are particularly simple, one is geometric and the other algebraic.

The first is founded on the fact that if two segments L and D are commensurable, and L<D<2L, then also D–L and 2L–D are commensurable. In fact we have D – L = (n – m) H and 2L – D = (2m – n) H and therefore also D–L and 2L–D have H as submultiple.

Let us now suppose that the side L and the diagonal D of a square are commensurable, and let H be a common submultiple. Divide the angle ABP in two equal parts, and draw from the point E the perpendicular EF to the diagonal. The two triangles ABE and BEF are equal (they are right angled, have same angle in B, and the side BE in common). Thus BF=AB=L, and PF=D-L. The triangle PEF is an isosceles triangle (in fact the angle EPF is 45 degrees), and therefore we have AE=EF=FP=D-L, and EP=L–(D–L)=2L-D.
Let us complete the square EFPG. Since we supposed that the side L and the diagonal D had a common submultiple, and that the side PF=D–L and the diagonal EP=2L–D of the small square will have the same submultiple H.
If we repeat in this square the construction we operated on the previous one, we'll obtain a new square, even smaller, the side and diagonal of which will have again H as submultiple. Continuing in the same manner, we'll obtain increasingly small squares, all having the side and the diagonal with H as common submultiple.
But this is not possible because the side and the diagonal become increasingly small and after a number of repetitions they would become smaller than H, meaning smaller than their submultiple. We have then reached an absurdity, thus the side and the diagonal of a square cannot be commensurable.

A second demonstration, the more algebraic one, is probably the one recalled by Aristotle when he says:
this kind of demonstration, for example, is the one that determines the incommensurability of the diagonal [and of the side of the square], which is founded on the fact that if we suppose that they are commensurable, uneven numbers turn out to be the same as the even ones.
To prepare the field, we observe that the two consecutive sides and the diagonal of a square form a right-angled triangle, and therefore we'll have D² = L² + L² = 2 L².
Suppose now that D and L are commensurable, meaning that L=mH and D=nH. We then have
n² H² = 2 m² H² and thus n² = 2 m².
In the previous equation we could eliminate all the common factors (if there are any) to m and to n and then suppose that m and n are prime numbers, meaning that they don't have common factors. In particular, they cannot both be even.
Now observe that n² is even (since it is equal to 2m² which is even), thus n is even and m is uneven. We can then write n=2p, and therefore, being n²=4p² , from the previous equation we get that 4p²=2m² , and, dividing by 2: 2 p² = m².
If our reasoning is like that one above, m² will then be even, and therefore m is also even. This is nonsensical because m was uneven.
We can look at this result from a slightly different point of view, considering that we demonstrated that the equation n² = 2 m² does not have any solution in integers, meaning that two integers m and n so that n²/m²=2 don't exist. Extracting the square root, we can conclude that no fraction n/m equal to the root of 2 can exist.
The numbers that can be expressed through fractions are called rationals; those that are not equal to any fraction are called irrationals. We can then state that the root of 2 is an irrational number.
With an analogous reasoning it can be demonstrated that the root of 3, 5 and in general of all integers that are not perfect squares are irrational. In other words, the root of an integer number is either an integer (as in the case of 4 or 9) or an irrational.

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