Irrationality of Root 2Date: 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/ |
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