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Is it possible to have a number system with 0 as its base value? I haven't found any explanation as to why only numbers more than 1 is taken as a base value. For example, binary: base value is 2. decimal base value is 10.

Siamsiam
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    How would you represent nonzero numbers in this system? – Dan Jul 18 '22 at 17:13
  • Using 0 and -1? – Siamsiam Jul 18 '22 at 17:15
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    @Siamsiam But how do you represent -1? Or how do you even represent 1 for that matter? – DotCounter Jul 18 '22 at 17:25
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    "number system" and "base" are pretty flexible term. What kind of "system" is the Roman Numeral system and it's base. If you are considering numbers where an integer $x$ is written as $d_m....d_2d_1d_0$ when $x= d_m\times b^m + ... + d_2\times b^2 + d_1\times b + d_0$ and the $d_i $ belong to the finite set of digits, it should be obvious we can't have $b=0$ because then $b^m$ and $d_m\times b^m$ would always be $0$! and we'd have $x =d_0$ and there are only finitely many possible values of $d_0$. (However.... it is possible to have $b$ but a negative value.) – fleablood Jul 18 '22 at 18:10
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    How would that work of $x = a_m\times 0^m + a_{m-1}\times 0^{m-1} + .... + a_2\times 0^2 + a_1\times 0 + a_0 = a_0$? – fleablood Jul 18 '22 at 18:14

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Part of the foundation of a number system (i.e. system of representing integers before and after the decimal point) of base $b$, where $~b \in \Bbb{Z_{\geq 2}},~$ is that the following sequence is strictly increasing, and goes to infinity:

$$b^0, b^1, b^2, \cdots.$$

The above property does not apply to any integers $~< 2.$

user2661923
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Whenever you change the base of your number system, you're establishing the size / magnitude of a group, as well as how many distinct symbols are needed for all numbers to be written down. In base 2, numbers are grouped in 2's. every group of 2 creates a group who's size is 1 magnitude larger. the number 2 (in base 10) is two 1's (1 + 1 = 10) so it makes a single group of 2's. the only number's you use are 1 and 0.

In base 3, you have 0 1 and 2 available as symbols. 1+1+1 = 10, because you now have a single group of 1's, where a group is defined as 3 units. 3 + 3 + 3 (base10) = 10 + 10 + 10 (base3) = 100 (base3)

If you are in base 0 however, you have no symbols to represent any numbers. You also can't group numbers into units because there are 0 units in a group

smcrowley
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  • How about base 1? – Siamsiam Jul 18 '22 at 17:20
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    base 1 has only 1 symbol that can be used. because of this, the number 0 is defined as an empty string, and every number is defined as that number of 1's. 4 (base10) =1111(base1). 6(base10) = 111111 base 1 – smcrowley Jul 18 '22 at 17:22
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    But the base-1 (aka "unary") numeral system is very different from "normal" base-n systems, which have digits ranging from 0 to n-1, use O(log n) space to represent the integer n (as opposed to unary's O(n)), and can have "decimal point" notation using negative powers of the base to represent non-integers (which unary can't because negative powers of 1 are still just 1). So it often makes sense to exclude unary from discussions of "base" notation. – Dan Jul 18 '22 at 18:20
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    I don't think it is usual to talk of 1 as the "base" of the unary number system because unary is not a form of positional notation and has different properties: e.g., discarding the last digit corresponds to division by the base in a positional notation but to subtraction by $1$ in unary. – Rob Arthan Jul 18 '22 at 18:23
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Generally speaking, when people refer to “Base-b” numerals, it means that:

  • The concatenation of the digits $d_m...d_3d_2d_1d_0$ represents the number $\sum_{k=0}^m d_kb^k$.
  • Numerals may optionally be extended by “decimal point” (more properly called “radix point” for $b \ne 10$) notation with $.d_{-1}d_{-2}d_{-3}...d_{-m} = \sum_{k=0}^m d_{-k}b^{-k}$ where the length $m$ may be either finite or infinite. This allows integer bases to represent non-integer real numbers.
  • As a consequence of $b^k$ in the above summations depending on the index $k$, numerals have “place value”, so digits cannot arbitrarily be re-ordered. (For example, $24 \ne 42$.)
  • Also, because the place value grows exponentially, representation of a number $n$ (if it doesn't require “decimal” places) uses $O(\log |n|)$ digits.
  • Normally, each digit is an integer between $0$ and $b-1$. However, this rule may be relaxed to accommodate negative, non-integer, or even complex bases. It is also possible to use negative-valued digits even with a positive base, with balanced ternary being the best-known of these systems.

It is possible to have $b = 1$; that's called unary. But it differs from conventional Base-b numerals in several important ways:

  • The “each digit is an integer between $0$ and $b-1$” rule is not followed, since having the sole digit be 0 would not allow representing any number other than 0. So the digit is taken to have a value of 1 instead.
  • Since $\forall k, 1^k = 1$, digits do not have place value. All of the 1's in the unary numeral 1111 are just 1's, regardless of how many other 1's are after it.
  • As a consequence of the above, radix point notation is useless, because all negative powers of 1 are also just 1, so can't be used to represent non-integers. (You could, however, still use vulgar fraction notation like $\frac{1}{1111}$ for $\frac{1}{4}$.)
  • The unary numeral for the integer n takes $O(n)$ space instead of $O(\log n)$.

Since unary is “special”, it's common to exclude it from discussions of number bases, or algorithms to convert between bases, and require $b \ge 2$.

Base-0, if it existed, would be even stranger than Base-1. Place value is nonexistent, because $0^k = 0$. Unless you rely on the definition $0^0 = 1$, in which case the one's place could have a nonzero value. But since no other places could, it would be useless to have multi-digit numerals. For every number to be one digit, every number would have to have its own unique symbol. Obviously, it's unfeasible to define an infinite number of digits. So Base-0 just can't work.

Dan
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