[Math] Prove that every irrational numbers can be approximated by rational numbers.

analysissoft-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!

Best Answer

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 $r$.

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