The traditional proof that the square root of 2 is
irrational (attributed to Pythagoras) depends on
understanding facts about the divisibility of the integers.
(It is often covered in calculus courses
and begins by assuming Sqrt[2]=x/y where
x/y is in smallest terms, then concludes that
both x and y are even, a contradiction. See the
Hardy and Wright reference.)
But the proof we're about to see (from the
Landau reference) requires only an
understanding of the ordering of the real numbers!
Proof.
Suppose Sqrt[2] were rational. Then
Sqrt[2]=x/y where x and y are integers and y > 0.
We will show that it is also equal to another fraction
x1/y1, where x1 and y1 are integers, y1 > 0 and y1 < y.
If this were true, then this procedure
could be applied over and over to each resulting fraction.
Then the denominators would
yield an infinite decreasing sequence of
positive integers y > y1 > ...,
which is impossible.
So, suppose Sqrt[2]=x/y,
that is, x^{2} = 2y^{2};
then we show x1 = 2y  x, y1 = x  y works.
By crossmultiplication, it is easy to check that
x/y = (2y  x) / (x  y).
So x1/y1 yields the same fraction as x/y.
Secondly, it must be the case that 0 < y1 < y,
because this is the same as y < x < 2y,
which is equivalent to 1 < (x/y) < 2.
But this is equivalent to 1 < (x/y)^{2} < 4,
and the last statement can be verified
because (x/y)^{2} = 2 by hypothesis.
Thus we have found an equivalent fraction with smaller
denominator, giving the desired contradiction.
Therefore Sqrt[2] must have been irrational, after all.
QED.
Presentation Suggestions:
After presenting this proof, ask students
as homework to prove that Sqrt[N] is irrational
if N is a positive integer and not a perfect square.
Caution them not to prove "too much":
their proof must fail when N is a perfect square!
You may give them a hint to use the analogous equation
(where Sqrt[N] = x/y and k is an integer):
(x/y) = (Ny  kx) / (x  ky)
Subhint: the k to use is k = Floor[Sqrt[N]].
The Math Behind the Fact:
The reasoning about an
infinite sequence of decreasing positive integers
is another form of mathematical induction
(both depend on the fact that any
nonempty subset of the positive integers
has a least element).
This form of reasoning was invented by Fermat
and is called the method of infinite descent.
How to Cite this Page:
Su, Francis E., et al. "Irrationality by Infinite Descent."
Math Fun Facts.
<http://www.math.hmc.edu/funfacts>.
