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Suppose that we have a computer program, my question is whether a human can type in any real number - say in $[0,10]$ - that she would like to type in a finite amount of time? Suppose that the program allows typing in any basic expressions like sum, multiplication, square-root, limits etc.

in a more formal way: Choose one real number $x\in [0,10]$, is it possible to express $x$ in a way that a computer can understand*?

*can understand = if $y$ is another such number, computer can tell whether $x>y, y>x$ or $x=y$.

EDIT: For example, square-root 2 can be typed as $\sqrt{2}$ and that is okay. As asked in comments, computer does not have an infinite memory, What is imporant is that the program can distinguish any two numbers that are typed in. For example, even though it is not possible to represent $\sqrt{2}$ in a computer, it can understand that $\sqrt{3}$ is bigger than $\sqrt{2}$.

Nonlinear
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2 Answers2

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When you have finite or countable many symbols:

No, when you only allow symbols from a finite (or countable) set of possible symbols like $\lim$, $\sqrt{}$, digits and so on, the set of all possible terms (with finte length) someone can type is countable. But there are uncountable many reals in $[0,10]$. Thus there are reals which are "untypable"...

This is related to definable numbers: There are only countable many expressions to define a number in first order logic but there are uncountable many reals -> There are undefinable numbers. See also the answers and comments to the question Is there an example for an undefinable number? for examples of undefinable/uncomputable numbers.

When you have uncountable many symbols:

Then you can assign to each real number a different symbol which stands for it (like the symbol "$\pi$" stands for the number $\pi$) and voila: Every real can be typed by using its symbol ;-)

(Let's assume that the continuum hypothesis is true as a axiom for this answer: There is no uncountable set with cardinality less then the cardinality of $\mathbb R$)

Update: There are computable numbers $x$ for which you cannot decide whether $x=0$ or $x > 0$. To cite a comment of Robert Israel to this answer:

Consider a predicate $S(n)$ that can be computed for each natural number $n$. Define $x=\sum_{n=1}^\infty 2^{−n}d(n)$, where $d(n)=1$ if $S(n)$ is true and 0 otherwise. Since each $S(n)$ can be computed, $x$ is computable in the sense that arbitrarily good rational approximations of $x$ can be computed. Now it might be that all $S(n)$ happen to be false, but there is no proof (in your favourite consistent formal system) of this fact. Then it is impossible to decide (in that system) whether $x=0$ [or $x > 0$].

Community
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Stephan Kulla
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  • and if an uncountable set of characters is allowed, you can type any real number using single character – z100 Nov 08 '15 at 19:21
  • @z100: You are right! I add your comment to my answer... – Stephan Kulla Nov 08 '15 at 19:24
  • @z100: But I guess we need the continuum hypothesis as an axiom so that there is no uncountable set of symbols with cardinality less then the cardinality of $\mathbb R$ ;-) – Stephan Kulla Nov 08 '15 at 19:31
  • @tampis I did not mean to write every real number down, then you are right there are uncountable of them. I am wondering if there is a reference / paper about the issue you raise, best. – Nonlinear Nov 08 '15 at 19:52
  • @Nonlinear: Which real numbers do you mean? Definable numbers?! Computable numbers?! Algebraic numbers?! Note that in your current question on may choose each $x\in[0,10]$. So I guess you need to concrete your question (or ask a new one)... – Stephan Kulla Nov 08 '15 at 19:57
  • @Nonlinear: Note that there are definable numbers which are not computable. Note also, that there are definable numbers $x$ and $y$ for which you cannot decide whether $x<y$, $x=y$ or $x>y$ (I have no reference for this statement, so ask in a new question for a reference when you need some) – Stephan Kulla Nov 08 '15 at 20:03
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    Consider a predicate $S(n)$ that can be computed for each natural number $n$. Define $x = \sum_{n=1}^\infty 2^{-n} d(n)$, where $d(n) = 1$ if $S(n)$ is true and $0$ otherwise. Since each $S(n)$ can be computed, $x$ is computable in the sense that arbitrarily good rational approximations of $x$ can be computed. Now it might be that all $S(n)$ happen to be false, but there is no proof (in your favourite consistent formal system) of this fact. Then it is impossible to decide (in that system) whether $x = 0$. – Robert Israel Nov 08 '15 at 21:54
  • @RobertIsrael: Thanks ;-) I add your example to the answer... – Stephan Kulla Nov 08 '15 at 22:30
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Any real number $\xi$ you, or a more experienced mathematician, can think of can be defined and described in terms of text and universally accepted formulas, whence: be expressed in TeX-code on less than two ASCI-pages or so. Any computer will accept this as input. Such an input determines $\xi$ as an element of ${\mathbb R}$ once and for all, in other words: to any desired number of decimal places. Of course it may be the case that your pocket calculator will not be sufficient to compute these digits one after the other.

Christian Blatter
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    is there a reference that you know about this result? – Nonlinear Nov 08 '15 at 19:34
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    I think this answer is wrong. See the wikipedia article "Computable number": There are numbers a computer can never calculate (see the article computability for more details). There are also undefinable numbers... There are only countable many possible TeX-articles. How do you want to define with them each of the uncountable many real numbers?! ;-) – Stephan Kulla Nov 08 '15 at 19:41
  • Can you "think of" an undefinable number? – Robert Israel Nov 08 '15 at 19:47
  • @RobertIsrael very interesting point! – Nonlinear Nov 08 '15 at 19:56
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    @RobertIsrael -- I,too, think this answer is wrong, because of its claim about what can be defined in "two ASCII-pages or so". Many reals in $[0,10]$ can of course be defined by explicitly listing their binary digits, and a trillion (say) independent random bits is practically certain to define a number that cannot be expressed in "two ASCII-pages or so". (This supposes that after generating these bits, I would say that I have "thought of" the number.) – r.e.s. Nov 08 '15 at 20:04
  • Noncomputable maybe, but you don't think of particular undefinable numbers. – Robert Israel Nov 08 '15 at 21:57
  • @RobertIsrael I also doubt that one can think of an undefinable number, but maybe it's possible. That's why I made a follow up question: http://math.stackexchange.com/questions/1519858/is-there-an-example-for-an-undefinable-number – Stephan Kulla Nov 08 '15 at 23:08
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    Is this a joke? – suriv Feb 02 '17 at 23:26
  • @suriv: This is meant to be a serious answer to the question posed by the OP. It has nothing to do with the issue of noncomputable numbers. – Christian Blatter Feb 03 '17 at 08:51