Archimediedian property

Archedian property:

It x \in \mathbb{Q}, x > 0. So for any y \in \mathbb{Q} there is n \in \mathbb{N}_0 such that n\cdot x > y

Demonstration: Archimedian property

If y \leq 0 the result is trivial, because it is enough to take n = 1. We assume that y > 0. We want to prove that it exists n \in \mathbb{N}_0 that meets n > \frac{y}{x}

The quotient of rational numbers \frac{y}{x} it will be a rational number \frac{a}{b} with positive integers a and b. So, n > \frac{y}{x}=\frac{a}{b} is equivalent to say n \cdot b > a And that is accomplished by taking n = a + 1 since:

(a + 1)\cdot b = a\cdot b + b \geq a + b \geq a + 1 > a

that is just what we wanted