Square root of 2 is irrational
Proof 6 (Statement and figure of A. Bogomolny)
If x = 21/2 were rational, there would exist a quantity s
commensurable both with 1 and x: 1 = sn and x = sm. (It's
the same as assuming that x = m/n and taking s = 1/n.) The
same will be true of their difference x − 1, which is smaller
than x. And the process could continue indefinitely in
contradiction with the existence of a minimal element. The game
Euclid might have played always ends!
JL ExplanationDefinition.1
commensurable: Mathematics Exactly divisible by the same
unit an integral number of times. Used of two quantities.
If x = √2 = m/n, m, n integers, after choosing s = 1/n, we have an isosceles right triangle with sides 1 = n s
and hypotenuse
x = m s. Now follow the Apostol program to
construct an isosceles right triangle with sides (m − n)s = x−1 and hypotenuse t s, where
t s
x − 1
=
m s
n s
,
t
m − n
=
m
n
,
t
=
1
n
(m2 − n m)
= 2 n − m.
Continuing we get a sequence of isosceles right triangles with
sides nj s and hypotenuse mj s; after at most n steps we
reach a contradiction.
Another JL Explanation
If x = √2 = m/n, after choosing s = 1/n, rescale so that
we have an isosceles right triangle with integer sides n and
integer hypotenuse m. Now follow the Apostol program to
construct an isosceles right triangle with integer sides m − n
and hypotenuse t, where
t
m − n
=
m
n
,
t
=
1
n
(m2 − nm)
= 2 n − m.
Continuing we get a sequence of isosceles right triangles with
integer sides and integer hypotenuse; after at most n steps we
reach a contradiction.
