How are irrational numbers represented in the Reals?

Question

I'm reading about the construction of the Real numbers and am curious how irrational constants $\pi, e, \sqrt 2 \in \mathbb R$. I understand that each real element contains all the smaller rationals $\in \mathbb Q$. The above constants $c$, when calculated, are non-terminating and slowly converging. Therefore, how is it possible that one of these constants $c \in \mathbb R$ when the rationals below $c$ have yet to be determined? Are constants like these $\in \mathbb R$ without an underlying interpretation (as a set of smaller rationals)? If that is the case, how is order of $\mathbb R$ established?

Answer 1

In all of the cases below, we define $B=\mathbb R\setminus A,$ the complement of $A.$

We can find the Dedekind cut for $\sqrt{2}$ as:

$$A=\left\{q\in\mathbb Q\mid q<0\lor q^2<2\right\}$$

We get that $(A,B=\mathbb R\setminus A)$ is a cut because:

1. $0\in A, 4\in B.$
2. If $a\in A$ and $a’<a,$ then $a’\in A.$
3. If $b\in B$ and $b’>b,$ then $b’\in B.$
4. The hard part: If $a\in A,$ then there is an $a’>a$ such that $a’\in A.$ For $a>0,$ let $a’=\frac{3a+4}{2a+3}.$ It takes a little work to show this $a’$ satisfies the conditions.

We can define $e$ as the Dedekind cut:

$$A=\left\{q\in\mathbb Q\mid \exists N,q<\sum_{n=0}^N\frac1{n!}\right\}$$

We can use this formula for $\pi:$ $$ \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} \left(\frac{1}{2}\right)^{n-1} = {\pi}$$ to define the Dedekind cut for $\pi:$

$$A=\left\{q\in\mathbb Q\,\Big\vert\,\exists N,q< \sum_{n=0}^{N} \frac{(2n)!!}{(2n+1)!!} \left(\frac{1}{2}\right)^{n-1}\right\} $$

For $e$ and $\pi,$ condition (4) is easier, because if $a<\sum_{n=0}^N f(n),$ you can pick $a’=\sum_{n=0}^N f(n)<\sum_{n=1}^{N+1} f(n)$ works.

But for these, proving $B$ is non-empty is a few more steps - you have to prove the sums are bounded above. Not hard - they are bounded by geometric series - but a necessary step.

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But we don’t really need to show these are the Dedekind cuts. You only need to show that the set of all Dedekind cuts have the properties we want for the real line.

For example, we don’t need a Dedekind cut to show that $\sqrt2$ is a real number. We could just as well use a Cauchy sequence of real numbers which bounce around $\sqrt2.$ Theorems show that there must be a real number limit, and hence a Dedekind cut, since real numbers are Dedekind cuts.

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If we want an algorithm for determining whether a rational number belongs to $B$ in these cases, we can only give good ones for $e$ and $\pi$ by proving that $e$ and $\pi$ are irrational (or, if rational, knowing the exact rational value.)

But set definitions don’t require algorithms.

If $a_N$ is a strictly increasing rational sequence and $b_N$ is a strictly decreasing rational sequence such that $a_N<b_N$ and $b_N-a_N\to 0,$ we can define $$A=\{a\in\mathbb Q\mid \exists N, a<a_N\}\\B=\{b\in\mathbb Q\mid \exists N, b>b_N\}$$

This might not be a cut. If it is not a cut, though, we can prove there must be a unique rational number $q$ such that $(A,B\cup\{q\})$ is a cut.

So there is a unique Dedekind cut corresponding to the real limit of these two sequences, even if we don’t know whether the cut is for a rational or an irrational number, or what our specific $B$ looks like.

More generally, if $A_0,B_0$ are non-empty sets of rational numbers such that,

1. $A_0$ does not have a maximal element,
2. $\forall a\in A_0,b\in B_0, a<b,$
3. For all $n\in\mathbb Z^+,$ $\exists a\in A_0,b\in B_0, b-a<\frac1n,$

Then there is a unique cut $(A,B)$ such that $A_0\subseteq A, B_0\subseteq B.$

We can define $$A=\bigcup_{a\in A_0} \{q\mid q<a\}$$ and $B=\mathbb R\setminus A.$

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Indeed, an alternative construction defines a cut with $B$ not containing a minimum element, just as $A$ doesn’t contain a maximum, and requiring $A\cup B$ be missing at most one rational. But that definition feels more complex, so it is less often used.