# Can every irrational number be written in terms of finitely many rational numbers?

Consider the irrational number $\sqrt{2}$. It can be written in terms, i.e., in a closed form expression, of two rational numbers as $2^{\frac{1}{2}}$.

Does it hold in general that every irrational number can be written in terms, i.e., in a closed form expression, of finitely many rational numbers? For irrational numbers like $e$ and $\pi$ this is not immediately clear.

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You need to better define what writing in terms of means. –  Git Gud Apr 7 '14 at 20:08
@GitGud I suppose it's reasonable to assume that each finite sequence of rationals would have at most countably many possible interpretations of what number they represent; then it follows that the set of "representable" reals is countable since the set of finite sequences of rationals is countable. –  Pedro Apr 7 '14 at 20:12
$e$ and $\pi$ are transcendental numbers, which means they do not satisfy any relation of the form $a_nx^n+a_{n-1}x^{n-1} +\cdots + a_1x + a_0 = 0$. (note that for example $\sqrt2$ does satisfy a relation of this form, specifically $x^2 - 2 = 0$.) It is known that there is a certain sense in which most real numbers are transcendental. –  MJD Apr 7 '14 at 20:13
Isn't "√2" already a closed form expression? –  Greenstone Walker Apr 8 '14 at 1:05
@GitGud I think Church and Turing would claim that the answer is negative regardless of what definition is used (and, moreover, that any reasonable definition would be equivalent). –  Kyle Strand Apr 8 '14 at 6:13

A little extra detail, you have included a function, square root. Suppose we add in a finite vocabulary of functions that can also be used, logarithm base $e,$ exponential, trigonometry, inverse trig, hyperbolic trig, your favorite list of less "elementary" functions suh as hypergeometric, Lambert W, whatever.

Any expression is still a finite string combining rational numbers and a fixed alphabet of functions. As a result, the outcome is a countable list of numbers. So, still an uncountable set not accounted for.

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We have to be careful that the vocabulary doesn't become too expressive; for example, if we let the number be expressed with the full power of ZFC set theory, then we run into paradoxes if we try to express what it means for a number to be inexpressible. –  user2357112 Apr 7 '14 at 21:49
@user2357112, I had in mind functions that take a small (hypergeometric takes four) arguments and are, say, real analytic, but in any case continuous. That being said, i don't actually know what you mean. –  Will Jagy Apr 7 '14 at 22:20
@WillJagy: I don't know the precise details of the paradox user2357112 intends to construct, but an intuition of the issue is that if your language for describing numbers is too powerful, then you can define "the least positive integer that cannot be defined in fourteen or fewer English words". Oops, I just defined it in fourteen English words. For a more serious result that doesn't rely on English, just arithmetic, see Goedel's First Incompleteness Theorem. –  Steve Jessop Apr 7 '14 at 22:29
@SteveJessop, I see what you mean. i have no intention of allowing English language descriptions or allow any hint of self-reference. All that is permitted is an alphabet of smooth functions, rational entries; so, $$\sinh \left( \frac{e}{\pi} + \sqrt {163} \right) - \log {1729}$$ –  Will Jagy Apr 7 '14 at 22:37
@WillJagy: this is scraping my memory of my degree a bit, but I think that as long as you avoid predicate logic ("there exists" or "for each" quantifiers) you're OK. I may be wrong and there's a system that doesn't have those things and but that's rich enough to create paradox, but even if so I don't believe your simple formulas are it :-) –  Steve Jessop Apr 7 '14 at 22:46

The answer is no. The number of expressions which can be formed out of finite combinations of rational numbers is countable, whereas the number of irrational numbers is uncountable, so it cannot be possible.

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No, by a counting argument that's worth remembering: There are $\mathfrak{c}=2^{\aleph_0}$ real numbers, but the set of finite sets of rational numbers is countable. The proof on this last statement is relatively straightforward: since we can map rational numbers to whole numbers, we can map finite sets of rationals to finite sets of whole numbers. But now, to each finite set of whole numbers $\{a_1, a_2, a_3, \ldots, a_n\}$ — which we can obviously assume is given in increasing order — associate the number $2^{a_1}3^{a_2}5^{a_3}\ldots p_n^{a_n}$ where $p_n$ is the $n$th prime. You should be able to convince yourself that this mapping is one-to-one, and so there can only be countably many finite sets of rationals.

In fact, this argument shows that 'almost all' reals have no finite description — whether in terms of rationals, roots of algebraic equations, or even an English description like 'the tenth $x\gt 0$ where $e^x\sin(x) = 1.5$' — at all.

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Care must be taken once we allow an english description. See mathoverflow.net/questions/44102/…. That said, I don't think there are any problems with the specific english descriptions you have used (but I'm far from an expert). –  Jason DeVito Apr 7 '14 at 20:28

Assuming your closed form follows some general formula, all such irrational numbers are a function of rational numbers. As such, you can enumerate them. You can then use Cantor's diagonal argument to construct an irrational number that does not belong to the original set, hence there are irrational numbers than cannot be written using your scheme.

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