# 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.

• You should specify which construction of the real numbers you're working with. For example: Dedekind Cuts or equivalence classes of Cauchy Sequences of rationals, or something else. From there people here can prove these statements for you.
– Gary
Feb 22, 2022 at 0:21

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.

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.

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 $$a + b \in \mathbb{R}_{>0}$$.
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 $$ab \in \mathbb{R}_{>0}$$.
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 $$\frac{1}{a} \in \mathbb{R}_{>0}$$.