Irrational Numbers and ways to represent all of them

Question

I just started studying "Analysis" and it appears one of the first topics all textbooks goes through, is making the reader understand that there are "gaps" in Q. Therefore, we need numbers that are irrational to fill in the gaps. And those numbers are ones that we can't express in $\frac{p}{q}$ where p and q are integers.

However, it seems like the "construction" of irrational numbers seems odd. Other than most $\sqrt[n]{a}$, $\log_a[b]$, and some other numbers that were "taught" as being irrational, such as $\pi$ and $e$, it does not appear that we have good "tools" to describe the irrational numbers, (i.e. to move between irrational numbers from the rational numbers).

For example, I'm not sure if I can arrive at every irrational number by the expression $\sqrt[n]{a}$ and choosing any $n,a \in N$, or even choosing any $n,a \in Q$. Surely not, or we would have learned $\pi$ equals some form of this in high school. I'm not even sure whether any sum of the expression $\sqrt[a_1]{b_1} + \sqrt[a_2]{b_1} + ... $ can be used to describe all irrational numbers. Again, it seems not likely. (Please tell me if I'm correct or not).

So in this case, although we have filled the "gap" by defining a "negative", that is, all that is not rational is irrational, but we still have trouble easily "describing" all of these numbers.

My question is: is there a way to describe all irrational numbers using operations on rational numbers? If not, is this due to a limitation of the commonly "defined" operations? Again, if not, why is this of no interest to mathematicians?

Answer 1

Good question.

However, it seems like the "construction" of irrational numbers seems odd.

Well, we don't "contruct" the irrational numbers. We construct the rationals and discover it is whoafully inadequate. If we assume numbers lie on a continuum then even though we can get infinitely precise with rational numbers, we can never get a solid continuum. So we don't actually construct the irrationals, we just know they have to be there because the rationals are not enough.

I'm not sure if I can arrive at every irrational number by the expression a−−√n and choosing any n,a∈N, or even choosing any n,a∈Q.

You're right. We can't. Such a number is a solution to a polynomial $a_nx^2 + ..... + a_1x + a_0 =0$ where $a_i$ are rational coefficients. These are called algebraic numbers. These numbers aren't all rational (in fact most aren't) but as there are only countably infinitely many such polynomials and finitely many solutions per polynomial there are only countably many infinite algebraic numbers. $\pi$ is not a solution to any polynomial (with rational coefficients) and so is not algebraic. We call such numbers trancendental and there are uncountably many of them.

is there a way to describe all irrational numbers using operations on rational numbers?

Not really.

If not, is this due to a limitation of the commonly "defined" operations?

I'd say it has more to do with the limitations of countability. Everything we do by finite iterations will reach some countably many points, but if we believe numbers are a continuum, then no such finite iterations will cover all spots in a solid utterly smooth transition.

Again, if not, why is this of no interest to mathematicians?

This is of intense and extreme interest to mathematicians!

Answer 2

My question is: is there a way to describe all irrational numbers using operations on rational numbers?

For almost any reasonable definition of "operation," the answer is no! This is because we need to describe operations using strings in a mathematical language which only has finitely many symbols, so there are countably many such strings, but uncountably many real numbers by Cantor's theorem, so no such descriptions can cover all real numbers.

For example, if your operations include taking roots of polynomials with rational coefficients, you can only get the algebraic numbers, missing the transcendental numbers such as $\pi$ or $e$. If you allow any computable operation you get a much larger set, the computable numbers, which is still countable and so still misses almost all real numbers. You can go even further to the definable numbers but these get tricky to reason about.

This gets into really quite deep philosophical territory: if it's impossible to explicitly name almost any real number then in what sense do most of the real numbers "really exist"? IMO mathematicians do not have a good consensus answer to this question, although that's debatable. At the very least, I think textbooks on real analysis don't do a good job emphasizing to students what a huge philosophical leap it is to believe that "all" of the real numbers, as constructed e.g. by Cauchy sequences or Dedekind cuts, "really exist."

Some mathematicians answer that most of these real numbers in fact don't "really exist"; see, for example, constructivism as a keyword.

Answer 3

We say that the sequence $(u_n)$ is a Cauchy sequence if $$ \forall\varepsilon>0,\exists n_0\in\mathbb{N},\forall p,q\geqslant n_0,|u_p-u_q|<\varepsilon $$ One can show that $(u_n)$ is a Cauchy sequence iff it converges in $\mathbb{R}$, we define the equivalence relation $\mathcal{R}$ on $\mathbb{Q}^{\mathbb{N}}$ by $$(u_n)\mathcal{R}(v_n)\iff \lim\limits_{n\rightarrow +\infty}(u_n-v_n)=0$$ where the definition of limit is in the sense of being a Cauchy sequence (notice that in the definition we can restrain ourselves with $\varepsilon\in\mathbb{Q}^{+*}$ since $\mathbb{R}$ is archimedian). We can then define $\mathbb{R}$ as the set of all the equivalence class of $\mathcal{R}$.

Answer 4

It's not quite an operation, but there are two traditional ways to define the real numbers in terms of the rationals to fill in the "gaps". Tuvasbien mentioned Cauchy sequences in his answer while I was writing this, so I'll tackle the other.

The other is Dedekind cuts. This is the essentially all the subsets $A\subset \mathbb Q$ with the following properties:

• $A\neq\emptyset$
• $A\neq\mathbb Q$
• If $x\in A$ and $y<x$, then $y\in A$
• If $x\in A$, then there is a $y\in A$ such that $y>x$

An example of such a set is the set of all non-positive rationals and all positive rationals whose square is less than $2$. In a nutshell, it turns out that we can define addition and multiplication and order on Dedekind cuts that agree on the rationals but also define new numbers like that example (which we call $\sqrt2$).

Answer 5

If you accept that all real numbers have one or two infinite decimal expansions (some numbers have two expansions coming from $0.9999\ldots=1.0000\ldots$), and that each infinite decimal expansion represents exactly one real number, then you can say that rational numbers have an expansion ending periodically, like $4.7235353535\ldots$, and irrational numbers have an expansion that does not become periodical. This of course is true for any number base $b\geq2$.