Proof That the Cube Root of 3 is IrrationalDate: 05/22/2000 at 23:16:44 From: Sajesh Singh Subject: Proof of an irrational number Hi, I need help with the following problem: Show that the cube root of 3 is irrational. Thank you for your help Sajesh Singh Date: 05/23/2000 at 02:39:58 From: Doctor Floor Subject: Re: Proof of an irrational number Hi, Sajesh, Thanks for writing. I will prove that the cube root of an integer - such as 3 - is only rational if it is an integer. I will use an indirect proof. To do this, suppose that the cube root of an integer N is rational, but not an integer. We can write this cube root as a/b in such a way that GCD(a,b) = 1 (a and b are coprime) and b > 1. Since GCD(a,b) = 1, we can't simplify the fraction a/b. We derive N = a^3/b^3. But since a and b are coprime, a^3 and b^3 are coprime, too, and thus the fraction a^3/b^3 can't be simplified. And since b > 1 so too b^3 > 1. But that means that a^3/b^3 is not an integer, while N is an integer. And we have a contradiction. If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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