system
Given the following post, respond as if you're commenting on Reddit:
human
Can there be a numerical system without repeating decimals ?

For example we use base 10 which has 1/3 as a repeating decimal.
gpt
There is no numerical base in which all numbers have a terminating expansion. (There is also no numerical base in which all numbers have a *unique* expansion, although that is a separate statement.)

Why is the former statement true? Well, for one, with a finite alphabet (digits and decimal point), you can only form countably many finite strings, which would correspond to the terminating expansions. But there are uncountably many real numbers. So no number base can do the job of expressing all numbers with a terminating expansion. 

Here's some more insight about which numbers can be expressed with terminating expansions....

Fix a numerical base *b*, which may be any positive real number (not equal to 1). Can we characterize the numbers that have a terminating expansion in base *b*? Yes. If a number *a* has a terminating expansion in base *b*, then *a* has an expansion of the form

> a = d*_m_*d*_m-1_*...d*_1_*d*_0_*.d*_-1_*d*_-2_*...d*_-n_*

where d*_k_* are the digits of *a*, and these digits must be integers. This means *a* is equal to

> a = d*_m_*b^(m) + d*_m-1_*b^(m-1) + ... + d*_1_*b + d*_0_* + d*_-1_*/b + d*_-2_*/b^(2) + ... + d*_-n_*/b^(n)

Multiplying by b^(n) and rearranging shows that *a* must satisfy an equation of the form:

> p(b) + b^(n)a  = 0

where *p* is some polynomial with integer coefficients. Now if let P(x, y) be the polynomial

> P(x, y) = p(y) + y^(n)x

this means that P(a, b) = 0. That is, there is a polynomial with integer coefficients that vanishes at the point (a, b). This is another way of saying that *a* and *b* are algebraically dependent over the integers. Intuitively, we can obtain either *a* or *b* from the other by a finite number of arithmetic operations with integers (addition, subtraction, division, multiplication, roots).

So we have the following partial characterization. Given a number base *b*, if the number *a* has a terminating expansion in base *b*, then the set {a, b} must be algebraically dependent over the rational numbers **Q**. Equivalently, if {a, b} is algebraically independent over **Q**, then *a* cannot have a terminating expansion in base *b*.

It then follows immediately that there is no number base in which all numbers have a terminating expansion, although this is not quite as obvious. It turns out that the real numbers are a so-called field extension of the rational numbers with transcendence degree the cardinality of the continuum. This tells us a lot more than we need. But, in particular, given a number base *b*, it is always possible to find at least one real number *a* that is algebraically *independent* of *b*, and so all expansions of *a* in base *b* must be non-terminating.

(If the number base *b* is rational, then we can just take *a* = e^(b), which is always algebraically independent of *b*. This follows by the Lindemann-Weierstrass theorem.)