Spivak "root 2 is irrational number" problem

[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

 

[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 positive integers p,q such that (p/q)²=2, and this leads to the contradiction.

 

[PeroK]

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

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.

 

[Alpharup]

Yes...got it