Cardinality of set of Dedekind cuts (elementary)

Question

Under the Dedekind construction the irrationals are defined as those cuts $(A,B)$ where $B$ has no least element ($A$ not having a greatest element by definition), for example the $q^2=2$ case. I can see how I can construct a countable number of irrationals that way but I can't see how to get an uncountable number of irrationals without an uncountable number of symbols. If I need an uncountable number of symbols then I don't need the Dedekind construction to start with. What am I missing?

Answer 1

In our enlightened day and age we know that you don't have to name something for it to exist. Did atoms came to existence just when they were named? Was everything afloat until Newton wrote down the laws of gravity?

In mathematics, especially abstract mathematics that deals with infinite sets, we don't have to name something in order for it to exist. Sometimes we can just show that something with the certain property exists, or we can show that it is impossible for something to exist.

The point is that we don't have to define the Dedekind cuts using arithmetic, or some other well-known function. We can prove that there are uncountably many of them. And this means that many of the numbers defined using Dedekind cuts have no definition using algebraic operations on $\Bbb Q$. But why is that a problem?

Answer 2

There is an alternative, geometric interpretation of Dedekind cuts which may be helpful.

View each positive, rational $\frac{n}{m}$ as the integer point in the first quadrant of the plane $\langle n,m \rangle$.

Then each line through the origin determines a unique cut of the plane into a lower set and an upper set - i.e., points below the line and points above the line.

There are uncountably many such cuts since there are uncountably many such lines. Those lines are $y=rx$ for each real number $r$.