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

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!

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## closed as unclear what you're asking by amWhy, Kirill, SDevalapurkar, Hakim, Andres CaicedoJun 24 at 6:21

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It would expect something like this: Given an irrational numbe $w$, then for every $\epsilon \gt 0$ there exist rational numbers $r, s$ with $r < w < s$ and $s - r \lt \epsilon$. –  mvw Jun 23 at 13:36
How is that possible? That is the question. :( –  Noga Tailcutter Jun 23 at 13:39
If you assume that the decimal expansion of real numbers is well defined, you can chop off the irrational at the $n$th spot after the decimal point and use that as your rational number. Then the (absolute value of the) difference will be less than $1/10^{n}$. Since $1/10^n\to 0$ as $n\to\infty$, you're done. –  Stahl Jun 23 at 13:48
Indeed, but that is why I have added that edit. I thought about it, it is not reasonable at all. :( –  Noga Tailcutter Jun 23 at 13:50
I, being the king of everything that is and will ever be, endow you, Noga of StackExchange the right to cut off the tail of an infinite decimal expansion. From this day forth, you shall be known as Noga Tailcutter. –  Asaf Karagila Jun 23 at 14:11

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 $$\lim_{R \to \infty} \sum_{r=0}^R d_r b^r = x.$$

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that does not converge ? –  Tom Collinge Jun 23 at 14:00
He meant does not converge inside $\mathbb{Q}$. Still a problem - because it does show that from every incomplete Cauchy Converging sequence I can create an irrational number. It does not show that the correspondence is 1-1 onto. –  Noga Tailcutter Jun 23 at 14:05
It isn't 1-1, as I wrote: "Two Cauchy sequences define the same number if the sequence of their differences tends to zero." But you don't really need it to be 1-1, just for every real number to have one (at least) Cauchy sequence of rationals converging to it (that is, it needs to be onto). And why is it onto? Well - it is simply the definition of irrational numbers (one possible definition; there are others). –  Yoni Rozenshein Jun 23 at 14:15
Thanks or the accept! I really recommend Hurkyl's answer, too! –  Yoni Rozenshein Jun 23 at 15:04
If you want to split hairs, it doesn't mean that reals can be approximated by rational, since with this definition they don't live in the same set. You still have to talk about the identification of $ℚ$ with a certain dense set of $ℝ$. –  Evpok Jun 24 at 0:01

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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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...

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This is a very good question, because unfortunately every answer so far is quite misleading.

The astonishing answer is that there are indeed irrational numbers which cannot be approximated by known rational numbers !

These are called uncomputable numbers and Chaitin's constant is one example of them. Chaitin's constant is every bit a Real number as 1,2 or 0.5, it is bounded by {0,1} and still it is impossible to calculate even the first digit after the decimal point.

While the existence of a rational number approximation can be shown, it does not follow that an expansion to a known base can be found.

So it cannot be proved because the assumption

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

is at least for a specific sequence of rationals wrong.

EDIT: Why exactly is it downvoted when the comments agree that uncountable numbers exist ?

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Doesn't this assume approximation by a computable sequence of rationals? –  Count Iblis Jun 23 at 18:59
This point is a little bit subtle, because it defines the real numbers in the traditional classical way and then turns around and defines "can be approximated" in an algorithmic way, rather than a classical way. When you stay classical, everything everyone else has said is correct, so saying that they are actually wrong is rather sketchy. –  Ian Jun 23 at 18:59
I rather doubt it. The point is somewhere down there. We can not find it. Does not mean there is NO decimal expansion to it! The only thing you can say is you can not find a decimal expansion to it. That is the same way of telling, find the NEXT irrational number larger than PI, and you can not find it. That number is not computable. Does not mean there is NO decimal expansion! –  Noga Tailcutter Jun 23 at 19:02
Yes, it's true that Chaitin's constant can't be computably approximated by rational numbers. But there exist approximations by rational numbers to any desired degree of accuracy $-$ we just don't know how to find them. (Also, what Ian said.) –  TonyK Jun 23 at 19:02
Well, the difference is quite tricky and I see again that language is quite ambivalent. As the poster already said he believed that an expansion to a given base naturally exists. While you can strictly define the existence of approximations (Dedekind cuts etc.), both the problem definition and the other answers imply that known rational numbers are meant. –  Thorsten S. Jun 23 at 19:09