Is there any number that I can't find?

Question

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.

Answer 1

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 $\lim_{n\to\infty} y_i = y$.

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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.