Given this definition of a positive real number: $$\forall x\in\mathbb{R},x\geq 0:\Leftrightarrow\exists y\in\mathbb{R},x=y^2$$ how to prove that the sum of two positive numbers is a positive number? It is fairly easy for the product, but not so for the sum.
EDIT: To add a context to this, let's say we define real numbers as the completion of rationals (but I'm not sure whether it needs an order to be defined on the rationals). Then, I'd like to define positivity on real numbers in order to create a total order with $x\geq y$ iff $x-y\geq 0$.
Perhaps I don't have the whole answer of the edit, but here's how you could show the first part.
If we define the field $\Bbb R$ so that it is totally ordered : see here, then this means, among other things, that
$$ \forall x,y,z \in \Bbb R, \; x \le y \implies x+z \le y+z \qquad (1) $$
$$ \forall a,b,c \in \Bbb R, \; a \le b \; and \; b \le c\implies a \le c \qquad (2) $$
In this case, if we let $ x=0 $ and $ y, z $ be positive real numbers, we get by $(1)$ :
$$ 0 \le y \implies 0+z \le y+z $$
But $0+z=z \; and \; 0 \le z $, therefore we get by $(2)$ :
$$0 \le y+z $$
Therefore the sum of two positive numbers is a positive number.
EDIT :
Alright, so here is a more "geometrical" way I'm trying to give to prove that the sum of two positive real numbers is a positive real number. First, we'll assume that the product of two real numbers with the same sign is a positive real number.
Let $a,b \in \Bbb R$ so that $x=a^2$ and $y=b^2$ are positive real numbers (by definition).
Let's now consider a right triangle $ABC$ where $AC$ is the length of the hypothenus and so that $AB=x, BC=y$.
By the Pythagorean Theorem, we have : $AC^2=AB^2+BC^2=x^2+y^2$.
The fact that $x,y \ge0 $ is not a problem, because $(-w)^2=w^2 \quad \forall \ w \in \Bbb R $.
From now on, if we let $X=x^2, Y=y^2$ and $Z=AC^2$, we get that :
$ \forall \ X,Y \ge 0, \exists \ Z \ge 0 $ so that $X+Y=Z$.
Therefore the sum of two positive real numbers is a positive real number.