MT 007.02 / SL 266.01
Ideas in Mathematics: The Grammar of Numbers

Text: The irrationality of the square root of 2

These texts derive, with some modification, from
Selections illustrating the history of Greek mathematics. With an English translation by Ivor Thomas.
I. From Thales to Euclid.
(Harvard UP) Cambridge MA, 1991r/1980corr.
0-674-99369-1 [Loeb Classical Library, 335].
p. 110: III. Pythagorean arithmetic. (f) Irrationality of the square root of 2.

Greek text, transliterated without Alexandrian accents but with a circumflex to distinguish êta and ômega from epsilon and omikron.
Pantes gar hoi dia tou adynatou perainontes to men pseudos syllogizontai, to de eks arkhês eks hypotheseôs deiknyousin, hotan adynaton ti symbainêi tês antiphaseôs tetheisês, hoion hoti asymmetros hê diametros dia to ginesthai ta peritta isa tois artiois symmetrou tetheisês.
To men oun isa ginesthai ta peritta tois artiois syllogizetai, to d' assymetron einai tên diametron eks hyoptheseôs deiknysin, epei pseudos symbainei dia tên antiphasin.

[Aristotle: Prior analytics 1.23.41a.26-27]

Translation:
All who argue per impossibile [from impossibility] infer first by syllogism a false conclusion, then they prove the original conclusion by hypothesis whenever something impossible follows from a contradictory assumption, as, for example, that the diagonal [of a square] is incommensurable [with the side] because odd numbers seem equal to odd numbers if it is assumed to be commensurate.
It is inferred by syllogism that odd numbers are equal to even, and it is proved hypothetically that the diagonal is incommensurate, since a false conclusion follows from the contradictory assumption.

Commentary:
It is generally believed that the Pythagoreans were aware of the irrationality of sqrt(2) (Theodorus, for example, when proving the irrationality of numbers began with sqrt(3)), and that Aristotle has indicated the method by which they proved it.
The proof, interpolated in the text of Euclid as 10.117 (Euclid, ed. Heiberg-Menge, 3.408-410), is roughly as follows:

Suppose AC, the diagonal of a square, to be commensurable with its side AB,
and let their ratio in its smallest terms be a:b.
Now AC^2 : AB^2 = a^2 : b^2
and AC^2 = 2AB^2, a^2 = 2b^2.
Hence a^2, and therefore a, is even.
Since a:b is in its lowest terms, it follows that b is odd.

Let a = 2c.
Then 4c^2 = 2b^2,
or b^2 = 2c^2,
so that b^2 is even,
and therefore b is even.

But b was shown to be odd, and is therefore odd and even, which is impossible.
Therefore AC cannot be commensurable with AB.

 
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