Question

Using the Axioms of Ordered Fields, prove that x^2 >0, for all x not equal to 0.

Answer 1

i don't know how they taught u this. but the usual way is this:

there is a subset \(\mathbb{R}^+\subset\mathbb{R}\) called the set of "positive real numbers" such that
P1.- if \(x,y\in\mathbb{R}^+\) then \(x+y\in\mathbb{R}^+\) and \(xy\in\mathbb{R}^+\)
P2.- only one of the following is true: \(x=0\) or \(x\in\mathbb{R}^+\) or \(-x\in\mathbb{R}^+\).

Answer 2

now since \(x\neq0\) we have only two options:

(1) if \(x\in\mathbb{R}^+\) then by P1 \(x\cdot x=x^2\in\mathbb{R}^+\).

(2) if \(-x\in\mathbb{R}^+\) then \((-x)(-x)=x^2\in\mathbb{R}^+\).
in both cases x^2 is positive, so we infer that \(x^2>0\) by the definition of \(a>b\).

Answer 3

yea I am not sure if I also need to prove that the product of two negatives must be positive.