Fundamental Theorem of Arithmetic
Date: 10/23/2001 at 14:52:15 From: Josie Subject: Fundamental Theorem of Arithmetic How do I prove that the cube root of 2 is an irrational number using the Fundamental Theorem of Arithmetic? I guess I'm supposed to prove that the cube root of 2 is not a product of primes, but I don't know how to do that or where to start. Josie
Date: 10/23/2001 at 15:30:17 From: Doctor Rob Subject: Re: Fundamental Theorem of Arithmetic Thanks for writing to Ask Dr. Math, Josie. Prove this by contradiction. Assume that 2^(1/3) is rational, and that there are positive integers a and b such that 2^(1/3) = a/b. Then, cubing both sides, 2 = a^3/b^3, 2*b^3 = a^3. Now write down the prime power factorization of a and b. a = 2^m*c, c odd, b = 2^n*d, d odd. a^3 = 2^(3*m)*c^3, c^3 odd, b^3 = 2^(3*n)*d^3, d^3 odd. Then, substituting, 2*b^3 = 2^(3*n+1)*d^3 = N = 2^(3*m)*c^3 = a^3. Then apply the Fundamental Theorem of Arithmetic to N, and derive a contradiction. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.