Prove that every irrational numbers can be approximated by rational numbers.

Question

Prove that every irrational numbers can be approximated by rational numbers.

How can I do it? Ok, I admit. I heard it, I thought it is to be true. And I was a kid. Now I when I think about it, I really do not know - what to think of it.

It is highly provable that for every $x < y \in \mathbb{Q}$ there exists $g \in \mathbb{R} \setminus \mathbb{Q}$ such that $ x < g < y$. But how does this even remotely relate to what the theorem is all about?

Edit

The way people proved it then ( kid ages ) was circular reasoning at best. Take any base expansion of a real number:- $$ g_b := \sum b^r d_r $$ and now we can obviously cut off the tail. But wait. Who gave us the right to do so? Nobody! Because it implicitly assumes that the we can arbitrarily approximate any number! So, obviously what I was taught is a lie. Please help.

About * reals can be written as decimal expansions* Problem with that argument is that it is an axiom. Axiomatically, the system is as follows:-

1. Integers : Result of Von Nuemmann numbering or Peano - whatever.
2. Rationals : Given integers - the real result of division.
3. Algebraic Numbers : Given finite polynomials, their real roots.
4. Algebraic irrationals : What does not belong to 2 but belong to 3
5. Whatever remains : we can possibly create from infinite expansion of polynomial equation - but we know that won't only work .

Edit 2

I think I have found out how. We can always cut the real line $[0,1]$ into $I_0,I_1,...,I_{b-1}$ 'b' different slices. $$ S = \{ (0,1/b] , (1/b,2/b],... \} $$ As $g$ is irrational, it has to be within one of them. That would mean that $g \in ( i/b , (i+1)/b ) $. That would mean we choose the first digit of the base b expansion, using i. Now, we can again divide the Interval $I_i$ by same construction. That would give us the next digit in base b. As the intervals are getting reduced by $1/b$, we can approximate the left and right side of $g$ arbitrarily. And that shows why base b expansion is possible for any real number. Now that is the same Cauchy sequence ( of of many ) that Yoni was saying.

Thank you all folks!

Answer 1

Regarding your edit, it is circular but not in a way you imagined. Why does a real number have an infinite base expansion? What is a real number anyway?

One way to define real numbers is as a metric completion of the rationals. That is, take any Cauchy sequence of rationals that does not converge. This Cauchy sequence defines a new number, which is irrational. Furthermore, two Cauchy sequences define the same number if the sequence of their differences tends to zero.

This definition justifies talking about infinite base expansions in the first place, and it also immediately follows from the definition that if $x$ has a base expansion $x = \sum_{r=0}^{\infty} d_r b^r$, then $$\lim_{R \to \infty} \sum_{r=0}^R d_r b^r = x.$$

Answer 2

To make rigorous proofs, we first need a rigorous definition. The usual definition of the real numbers are that they form a complete ordered field. In fact, it is the only complete ordered field (up to isomorphism).

One of the things you can prove is that the real numbers satisfy the Archimedean axiom: for every real number $r$, there exists a natural number $n$ such that $n > r$.

We can use this to solve your particular problem: if we want to find a rational number $q$ such that $|q - r| < \epsilon$, then we can

• choose $n$ to be a natural number such that $n > 1/\epsilon$.
• choose $m = \lfloor rn \rfloor$
• choose $q = m/n$

and you can compute

$$ |q-r| = \left| \frac{m}{n} - r \right| = \frac{1}{n} |m - nr| < \frac{1}{n} < \epsilon $$

as desired. If you pay attention to details, you might wonder why $m$ is well-defined: if $r>0$ we can use the Archimedean axiom again to show there is a natural number $t > rn$, and then we can use induction to show that there must be a largest integer amongst the set of integers less than or equal to $rn$. (a similar argument works for $r<0$)

A more direct proof form the usual form of completeness is to consider the set $S$ of all rational numbers less than $r$: one can argue that $r$ must be the least upper bound of $S$, which in turn implies that $S$ must have elements arbitrarily close to $r$.

Answer 3

Consider an irrational number $x$. We can come up with the following function $f$ (where $b$ is the base used, usually obviously $10$) $$f(x,n)=\frac{\left \lfloor b^nx \right \rfloor}{b^n}$$ As $n$ gets bigger, the rational number $f(x,n) $ starts to approximate $x$ better. For example, to approximate $\pi$ in base-$10$ we can choose $n=3$ and get: $$f(\pi , 3)=\frac{\left \lfloor 10^3\pi \right \rfloor}{10^3}=\frac{3141}{1000}=3.141$$ This whole thing of course pressuposes that any real number can be represented as an infinite string of digits of base-$b$ with a point somewhere inbetween, but that's a different story...