Show that $a(b+c) = ab + ac$ for real numbers $a,b$ and $c$ constructed as Dedekind cuts

How do you show that the distributive law

$$a(b+c) = ab + ac$$

is true for real numbers $a,b$ and $c$? Where the real numbers are constructed as Dedekind cuts.

I was just wondering around, how you could actually prove this. If someone got a reference where they prove this, that would also be great.

I find it a bit odd that in all real analysis book I read about this topic, none of them seems to care to prove this.

• I would expect this to be relatively straight-forward: You need to be familiar with addition and multiplication, as defined in Dedekind cuts. Then you calculate both sides of that expression and see that you end up with the same cut. Have you tried that? Where did you get stuck? – Arthur May 23 '18 at 0:49

Divide into several cases based on the signs of $a, b, c$. But for the all-positive case, it's not so bad. The DC on the left consists of all rationals $x(y)$ with $x \le a$ and $y \le b+c$. Such a rational $y$ can be expressed as $u+v$, with $u, v$ rational, and $u \le b, v \le c$ (because addition on DCs is well-defined). But now look at $$x(u+v) = xu + xv$$ It's a sum of rationals, the first of which is no more than $ab$, and the second of which is more more than $ac$, and hence is a rational that's no more than $ab + ac$, and thus is in the DC representing $ab+ac$. So the DC for the left hand side is a subset of the DC for the right hand side.
• I think you misunderstand the definition of a DC. The DC representing a real number $a$ consists of all rationals that are no greater than $a$. In particular, the DC representing a rational like $7$ consists of all rationals less than or equal to $7$. Now you might say that the rational ALSO has a DC representation, but then you're getting off track. Instead, think of a rational as a pair $(p, q)$ of integers with no common factors, which stands for the rational number we write as $\frac{p}{q}$. – John Hughes May 23 '18 at 11:38