Irrationality of Root 2

Date: 03/26/98 at 13:08:04
From: mike lord
Subject: DIRECT proof that root 2 is irrational

My question is this:

I know that there are many ways to prove that root 2 is irrational.
I've already seen the contradiction proof as well as the geometric 
proof. But I've heard that there is a way to prove that root 2 is 
irrational using a DIRECT method of proof. Could you please help?

Date: 03/26/98 at 14:03:57
From: Doctor Rob
Subject: Re: DIRECT proof that root 2 is irrational

There is a proof using the simple continued fraction expansion of 
sqrt(2):

   sqrt(2) = 1 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + ...))))).

This expansion never ends. The "convergents" are obtained from this
expansion by stopping after any finite number of steps, and evaluating 
the compound fraction as a simple fraction. It is a theorem that the
convergents are closer to the number being expanded than any other 
rational numbers with equal or smaller denominators. Since the 
denominators of the convergents go to infinity as the number of steps 
does, it is clear that sqrt(2) could not be rational. At each step you 
have a proof that sqrt(2) is not a rational number whose denominator 
is smaller than some bound, and that bound eventually exceeds any 
number after a sufficient number of steps.

Does that qualify as a "direct" proof?

There is also a proof by induction (in the form of Fermat's Method of
Descent), but it has a definite indirect flavor:

Suppose sqrt(2) = a/b, reduced to lowest terms. Note that 
1 < sqrt(2) < 2, so 0 < b < a < 2*b.  Then:

           a^2 = 2*b^2
     a^2 - a*b = 2*b^2 - a*b
       a*(a-b) = b*(2*b-a)
           a/b = (2*b-a)/(a-b)
       sqrt(2) = (2*b-a)/(a-b)

Furthermore, 0 < a - b < a, and 0 < 2*b - a < b.  This means that 
there is a smaller numerator and denominator which will work. Repeat 
this. You will get an infinite sequence of smaller and smaller 
positive integer denominators, but any decreasing sequence of positive 
integers must be finite. This contradiction means that no such 
fraction a/b can exist.

This proof can be generalized to prove the irrationality of sqrt(d) 
for any d not a perfect square:

Let c < sqrt(d) = a/b < c + 1, with a, b, c, and d positive integers.

           a^2 = d*b^2
   a^2 - c*a*b = d*b^2 - c*a*b
     a*(a-c*b) = b*(d*b-c*a)
           a/b = (d*b-c*a)/(a-c*b)
       sqrt(d) = (d*b-c*a)/(a-c*b)

Furthermore, 0 < a - c*b < b, and 0 < d*b - c*a < a.  Repeat the above
argument.

-Doctor Rob,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/