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### 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

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

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