Spivak "root 2 is irrational number" problem

1. Dec 23, 2015

Alpharup

Am using Spivak. Spivak elegantly proves that √2 is irrational. The proof is convincing. For that he takes 2 natural numbers, p and q ( p, q> 0)...and proves it.
He defines irrational number which can't be expressed in m/n form (n is not zero).
Here he defines m and n as integers.
But in the earlier proof, he takes p and q as only natural numbers and not integers. Why?
Is it because of definition, he earlier defined?
ie...√((a)^2)...|a|.
ie....root of a real number a

2. Dec 23, 2015

Erland

Any rational number m/n (m, n integers, n≠0) can, by multiplying numerator and denominator by -1 if necessary, be written with n>0. If then (m/n)2=2, then also (-m/n)2=2. One of m and -m is positive. Hence, there are postive integers p,q such that (p/q)2=2, and this leads to the contradiction.

3. Dec 23, 2015

PeroK

There's essentially no difference between looking for natural numbers or integers. An integer may be negative, but that makes no difference to divisibility properties.

4. Dec 23, 2015

Yes...got it