Sqrt2 IrrationalDate: 05/17/99 at 12:53:00 From: G.Arjun Subject: Real Numbers I would like to know the answer to the question, How do you prove that the square root of 2 (or other roots of other real numbers) is irrational? I am not able to understand the argument. Date: 05/18/99 at 15:13:35 From: Doctor Floor Subject: Re: Real Numbers Hello, G. Arjun, Thanks for your question! To prove that sqrt(2) [sqrt means square root] is an irrational number, we have to show that it cannot be written as a/b, where a and b are integers. Such a proof has to be done indirectly: 1. We assume that sqrt(2) could be written as a/b for integer a and b; 2. We show that this leads to a contradiction. So something impossible has to be derived from our assumption. The fact that we derive someting impossible from the fact that sqrt(2) = a/b shows that sqrt(2)=a/b must be false, and we have proven the theorem. Let's assume that sqrt(2) could be written as a/b for integer a and b. We can derive: sqrt(2) = a/b [sqare both sides] 2 = a^2/b^2 [multiply by b^2; ^2 means squared] 2*b^2 = a^2 a^2 = 2*b^2 Now we have two integers in the equation, 2*b^2 and a^2. We can factor these integers into primes, and find: a^2 = 2^m * {product of odd primes} 2*b^2 = 2^n * {product of odd primes} Because a^2 is a square, m has to be even. Because b^2 is a square, and "2*" brings an additional factor 2, n has to be odd. So m and n are not equal. But if a^2 = 2*b^2, then the factorisation into primes of these two has to be equal. So, m and n should be equal. We have found a contradiction. And our assumption, that sqrt(2) = a/b for integers a and b, cannot hold. This proves that sqrt(2) is irrational. I hope this helps you to understand. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ Date: 05/17/99 at 14:42:20 From: Wayne Subject: Square root of two Can you prove to me how the square root of two is irrational? Date: 05/17/99 at 16:20:17 From: Doctor Rob Subject: Re: Square root of two There are a couple of ways to do that. Both are proofs by contradiction; that is, they assume that sqrt(2) is rational, and derive an impossible conclusion from that assumption. Here is one: Assume sqrt(2) = a/b, reduced to lowest terms (that is, a and b have no common factor besides 1). Then 2 = (a/b)^2 = a^2/b^2, so 2*b^2 = a^2. That means that a^2 is even, which implies that a is even. Then write a = 2*c, a^2 = 4*c^2 = 2*b^2, so 2*c^2 = b^2. That means that b^2 is even, which implies that b is even. This means that a and b have the common factor 2, which is a contradiction. Thus no such fraction a/b can exist, and sqrt(2) is irrational. Here's another: Assume sqrt(2) = a/b, with a and b positive integers, and least denominator among all such fractions. Then 2*b^2 = a^2, 2*b^2 - a*b = a^2 - a*b, 4*b^2 > 2*b^2 = a^2 > b^2, b*(2*b-a) = a*(a-b), 2*b > a > b, (2*b-a)/(a-b) = a/b = sqrt(2), b > a - b > 0 and 2*b - a > 0. Now (2*b-a)/(a-b) is a fraction equal to sqrt(2) with positive integer numerator and denominator, but smaller denominator than a/b has. This is a contradiction. Thus no such fraction a/b can exist, and sqrt(2) is irrational. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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