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I have a question that I think is quite weird and I can't find an answer.

Is there any real number that I can't find using only $+,-,\times,\div,$ limits and radicals?

For example, using some series, I can find $\pi$ or $e$. But is there any number that I can't find using only those?

Sorry if I made any mistakes, I don't know how to ask that question or which tags to put.

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    $\begingroup$ The real numbers can be constructed as limits of Cauchy-series with elements in ℚ. $\endgroup$
    – P. Siehr
    Feb 20, 2018 at 12:07
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    $\begingroup$ What about the Chaitin's constant, Crostul? An uncomputable number which is precisely defined (en.wikipedia.org/wiki/Chaitin's_constant) $\endgroup$
    – rafa11111
    Feb 20, 2018 at 12:14
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    $\begingroup$ What does "allowing limits" mean? $\endgroup$
    – Joppy
    Feb 20, 2018 at 12:17
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    $\begingroup$ There are only countably many predicates you could write down in the language of mathematics. So there are many real numbers which are not definable. $\endgroup$ Feb 20, 2018 at 14:32
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    $\begingroup$ Can you say more clearly what you mean by "find"? Pi and e are numbers that match a description, like "the ratio of a circle's circumference to its diameter" or "the amount a bank account with a dollar in it will increase to in a year if 100% interest is compounded continuously", and then we can come up with algorithms that give us approximations to this description. There are descriptions of numbers that we know exist that cannot be found in this way because there is no algorithm that gives us an approximation. Is that what you're after? $\endgroup$ Feb 20, 2018 at 17:10

5 Answers 5

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Allowing limits is very strong.

A real number $y$ has an integer part and a decimal part $a.x_1x_2x_3x_4...$ We can construct this real number as a limit in the following way:

Let $y_1 = a, y_2=a.x_1, y_3 =a.x_1x_2 , y_4 = a.x_1x_2x_3 , y_5=...$ clearly each $y_i$ is a rational number and $\lim_{n\to\infty} y_i = y$.

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    $\begingroup$ Although you're not just allowing limits; you're allowing arbitrary sequence formation as well. $\endgroup$
    – user14972
    Feb 20, 2018 at 13:09
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    $\begingroup$ Perhaps it's worthwhile saying that the sequences $x_i$ and $y_i$ are definable in terms of $y$ alone, namely by induction $y_1 = a = \lfloor y \rfloor$, $x_i = \lfloor (y-y_{i}) / 10^{i} \rfloor$, $y_{i+1} = y_{i} + x_i \cdot 10^{-i}$. The OP can tell us whether he allows mathematical induction and whatever other bits of logic might or might not be implicit in the question (sequences clearly are implicit in the question, given the examples of $\pi$ and $e$). More to the point, the OP can tell us whether we may use the "greatest integer" function $\lfloor \cdot \rfloor$. $\endgroup$
    – Lee Mosher
    Feb 20, 2018 at 14:45
  • $\begingroup$ @LeeMosher One can define a sequence converging to $e$ in a way that, depending on your definitions, uses only the arithmetic operations (in particular, the series definition of $e$ is hypergeometric). The floor function in your formula isn't the biggest problem; the biggest problem is the symbol "$y$", so you're implicitly allowing arbitrary real constants, which makes the problem trivial. $\endgroup$
    – Jack M
    Feb 21, 2018 at 12:27
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Maybe Chaitin's constant is somthing you are looking for.

It is non-computable, so we cannot approximate it to arbitrary precision, even with stronger tools than the ones allowed by you.

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    $\begingroup$ To be more precise, there are many different Chaitin's constants (it depends on the encoding in the definition), and in some of them a few digits can be computed: mathworld.wolfram.com/ChaitinsConstant.html [Sorry if my first version of this comment sounded rude - I accidentally hit enter before I finished it] $\endgroup$ Feb 20, 2018 at 13:00
  • $\begingroup$ Generally, any non-computable number should be counted as a "real number that [one] can't find using only $+,-,\times,\div,$ limits and radicals" (starting from 1, say). And since the real numbers are a lot more than the number of programs/algorithms/expressions/whatever one can write as a finite sequence of a finite number of characters, there are a lot of non-computable numbers. $\endgroup$
    – md2perpe
    Aug 4, 2021 at 10:26
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Every real number $x$ can be written in the form: $$x=n+\sum_{k=0}^\infty x_k\left(\frac{1}{10}\right)^k,\ x_k\in\{0,1,\dots,9\}$$ and we are used to call $n$ in integer part and $x_k$ its decimal digits. So the answer to you question is yes.

Moreover, you can write any real number $x$ in the form: $$x=n+\sum_{k=1}^\infty b_k\left(\frac{1}{2}\right)^k,\ b_k\in\{0,1\}$$ where $b_k$ are its binary digits.

So, you only need two digits and the operations you've mentioned.

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Any real number can be expressed as a limit of rational numbers, and rational numbers can be expressed by dividing two whole numbers.

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    $\begingroup$ But how can you express a sequence of rational numbers? $\endgroup$
    – Jack M
    Feb 20, 2018 at 14:01
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    $\begingroup$ @OveAhlman answered that question $\endgroup$
    – user532929
    Feb 20, 2018 at 14:03
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It depends upon what you mean by "number that I can't find", but there are an infinity of real numbers that you can't define in any way using what you are describing.

Consider the set of numbers you can "define". The definition of such a number is some kind of text you can encode as a binary number, so this set is a countably infinite set.

Because real numbers are uncountable, there is an infinity of real numbers you can't define.

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  • $\begingroup$ Your answer is wrong or flatly misleading. Please read the comments under the question before giving an answer whose mistake has already been explained there. $\endgroup$
    – user21820
    Feb 21, 2018 at 4:24
  • $\begingroup$ Nah, you know what it is, Karlotcha $\endgroup$ Feb 21, 2018 at 10:01

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