The set of positive real numbers is closed under addition, multiplication, and division That the sum, product, and quotient of two positive real numbers is also a positive real number seems obvious, but I can't seem to find a satisfactory proof for these facts anywhere.
 A: In order to be able really give a proper proof of this fact, you need a working definition of the real numbers and the surrounding operations you have outlined. What do you mean by a real number, and what do you really mean to add, multiply and divide them?
The two most common definitions of ${\mathbb{R}}$ would be the Dedekind cut construction or the Cauchy construction. Both of these assume we already have a working definition of ${\mathbb{Q}}$ (which you can also define in terms of ${\mathbb{Z}}$, and ${\mathbb{Z}}$ can be defined in terms of ${\mathbb{N}}$, and ${\mathbb{N}}$ can be defined in terms of the Peano axioms). From these definitions you can indeed prove closure under the operations you mention.
A: After some thought, I've come up with a fairly simple answer using some basic calculus. In light of the cursory response from Riemann'sPointyNose I'm a little wary that these arguments may contain circular reasoning, but here they are anyways.
Closure under addition
Let $a,b \in \mathbb{R}_{>0}$. Consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by the rule $f(x) = a + x$. This function is differentiable with $f'(x) = 1$. Since $f'(x) > 0$ for all $x \in \mathbb{R}$, $f$ is strictly increasing on $\mathbb{R}$. Now $f(x) > f(0) = a > 0$ for all $x \in \mathbb{R}_{>0}$. In particular, $f(b) = a + b > 0$, so $a + b \in \mathbb{R}_{>0}$.
Closure under multiplication
This argument is very similar to the last. Let $a,b \in \mathbb{R}_{>0}$. Consider the function $g: \mathbb{R} \to \mathbb{R}$ defined by the rule $g(x) = ax$. This function is differentiable with $g'(x) = a$. Since $g'(x) > 0$ for all $x \in \mathbb{R}$, $g$ is strictly increasing on $\mathbb{R}$. Now $g(x) > g(0) = 0$ for all $x \in \mathbb{R}_{>0}$. In particular, $g(b) = ab > 0$, so $ab \in \mathbb{R}_{>0}$.
Closure under division
Since division by a number is equivalent to multiplication by its reciprocal, it will suffice to show that $\mathbb{R}_{>0}$ is closed under reciprocation. This argument uses contradiction. Let $a \in \mathbb{R}_{>0}$ and suppose for contradiction that $\frac{1}{a} \notin \mathbb{R}_{>0}$. $a \frac{1}{a} = 1$, so it cannot be that $\frac{1}{a} = 0$. Then $\frac{1}{a} \in \mathbb{R}_{<0}$ and we have that $\frac{1}{a} = -b$ for some $b \in \mathbb{R}_{>0}$. Now $a \frac{1}{a} = a(-b) = -ab$. We have already proven that $ab \in \mathbb{R}_{>0}$, so $a \frac{1}{a} = -ab \in \mathbb{R}_{<0}$. But this contradicts the fact that $a \frac{1}{a} = 1$, so we must have $\frac{1}{a} \in \mathbb{R}_{>0}$.
Note: This last argument relies on the assumption that the negative real numbers are exactly the opposites of the positive real numbers. I don't have the knowledge right now to prove this fact.
