Proof that Sqrt(2) is Irrational
Date: 8/29/96 at 12:49:58 From: Tim Subject: Proof that Sqrt(2) is Irrational Dr. Math - For years I've tried to recall the proof that the square root of 2 is irrational. The proof I recall begins with the assumption that the square root of 2 is rational and therefore = a/b. The proof then shows this is impossible. I remember the proof being clever and elegant.
Date: 8/30/96 at 19:17:39 From: Doctor Tom Subject: Re: Proof that Sqrt(2) is Irrational Yup. Assume that it is rational, of the form a/b, where a/b is reduced to lowest terms. So (a/b)*(a/b) = 2, or a^2 = 2b^2. Since a^2 is even, a must be even, say a = 2c so (2c)^2 = 2b^2, or 4c^2 = 2b^2 or 2c^2 = b^2, so b must also be even. So in a/b, both a and b are even, but we assumed we'd reduced the fraction to lowest terms. We've got a contradiction, so sqrt(2) must be irrational. -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 10/29/2005 at 20:41:41 From: Mitch Subject: Your proof of sqrt(2) Dear Dr. Math, I saw the proof on your website for the square root of 2 being irrational. I can follow all the steps, but the proof doesn't seem valid to me. I looked around the web and saw very similar descriptions for this proof and they too all seem invalid to me. All the proofs start with the idea that a rational number can be written as the ratio of two integers, say a/b, and that for any ratio there exists exactly one fully reduced fraction (where no integer greater than 1 exists that can be evenly divided into both the numerator and denominator.) What I see is there may exist a fraction that is not fully reduced. Even if I assumed a non-fully reduced fraction did exist, this does not imply to me there does not exist a non-fully reduced fraction. I would want to see that no fully reduced fraction of integers exists anywhere to see the conclusion. Thanks in advance for any help.
Date: 11/01/2005 at 21:15:47 From: Doctor Rick Subject: Re: Thank you (Your proof of sqrt(2)) Hi, Mitch. The version shown above is very brief, more an outline of the proof, and shouldn't be judged as if it were a formal proof. Here's a little more formal version that addresses your concerns about the ratio a/b being in lowest terms. ======================================================= To prove: The square root of 2 is irrational. In other words, there is no rational number whose square is 2. Proof by contradiction: Begin by assuming that the thesis is false, that is, that there does exist a rational number whose square is 2. By definition of a rational number, that number can be expressed in the form c/d, where c and d are integers, and d is not zero. Moreover, those integers, c and d, have a greatest common divisor, and by dividing each by that GCD, we obtain an equivalent fraction a/b that is in lowest terms: a and b are integers, b is not zero, and a and b are relatively prime (their GCD is 1). Now we have  (a/b)^2 = 2 Multiplying both sides by b^2, we have  a^2 = 2b^2 The right side is even (2 times an integer), therefore a^2 is even. But in order for the square of a number to be even, the number itself must be even. Therefore we can write  a = 2f Using this to replace a in , we obtain  (2f)^2 = 2b^2  4f^2 = 2b^2  2f^2 = b^2 The left side is even, therefore b^2 must be even, and by the same reasoning as before, b must be even. But now we have found that both a and b are even, contradicting the assumption that a and b are relatively prime. Therefore the assumption is incorrect, and there must NOT be a rational number whose square is 2. ======================================================= That's my proof. Admittedly it still leaves out some steps, for instance a proof that if the square of a number is even then the number itself must be even. But I filled in the part that had you confused. -Doctor Rick, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.