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Base $10$ uses these digits: $\{0,1,2,3,4,5,6,7,8,9\};\;$ base $2$ uses: $\{0,1\};\;$ but what would base $1$ be?


Let's say we define Base $1$ to use: $\{0\}$. Because $10_2$ is equal to $010_2$, would all numbers be equal?

The way I have thought Base 1 might be represented is tally marks, $0_{10}$ would be represented by nothing. So, $5$ in Base 1 would be represented by $00000$? Or we could define Base 1 to use: $\{$|$\}$ and $5$ would be |||||?

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You're exactly right that such a system would be represented by the use of arbitrary tally marks. Such a system is known as a Unary Numeral System (Wikipedia Entry):

The unary numeral system is the bijective base-1 numeral system. It is the simplest numeral system to represent natural numbers: in order to represent a number N, an arbitrarily chosen symbol representing 1 is repeated N times. This system is used in tallying. For example, using the tally mark |, the number 6 is represented as ||||||.

...

There is no explicit symbol representing zero in unary as there is in other traditional bases, so unary is a bijective numeration system with a single digit. If there were a 'zero' symbol, unary would effectively be a binary system. [boldface mine] In a true unary system there is no way to explicitly represent none of something, though simply making no marks represents it implicitly. Even in advanced tallying systems like Roman numerals, there is no zero character; instead the Latin word for "nothing," nullae, is used.

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  • $\begingroup$ for representing 0 you can define a number to be represented by 1+itself symbols. so 0 would be | and 1 would be || etc. $\endgroup$ – ratchet freak Apr 25 '13 at 10:51
  • $\begingroup$ @ratchetfreak Why not just use _ (nothing, math.stackexchange.com doesn't allow me to use ' ')? When we use tally marks, | is 1, || is 2, and _ (nothing) is 0. $\endgroup$ – Justin Apr 25 '13 at 17:25
  • $\begingroup$ @gangqinlaohu see the second quote block in this answer ;) $\endgroup$ – ratchet freak Apr 25 '13 at 17:28
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    $\begingroup$ @ratchetfreak Exactly, 0 is represented by nothing $\endgroup$ – Justin Apr 25 '13 at 17:29
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    $\begingroup$ The bolded statement is not really right. In a base-1 system, it is simply the case that $1 = 10 = 100 = \cdots$ because $1 = 1^2 = 1^3 = \cdots$, so that the $0$ is a redundant symbol. The use of it in the unary system does not make it binary. That statement, as well as the following one that there is no explicit symbol for $0$, is sensible only under the presumption that base-representations of natural numbers should be bijective. There may certainly be a $0$ in a unary system. $\endgroup$ – Andrew Odesky Sep 30 '17 at 19:00
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I would like to expand on Trevor Wilson's answer. Base-$b$ representation of integers is rooted in the fact that, for any non-negative integer $n$, there is a unique representation of $n$ in the form $$n = \sum_{i=0}^\infty a_ib^i$$ where $0 \le a_i < b$. For example, when $b$ is 3, and $n$ is 47, the unique solution has $a_0 = 2, a_1 = 0, a_2 = 2, a_3 = 1, $ and $a_i = 0$ for all $i>3$. The $a_i$ are called the "base-$b$ digits of $n$"; in our example the base-3 digits of 47 are 1202. We say that the sequence of digits is a numeral, and that it represents the number $n$.

The uniqueness property means that each $n$ has exactly one base-$b$ representation. If one requires that the sequence of $a_i$ is eventually zero (that is, that $a_i = 0$ for all sufficiently large $i$) then the converse holds also: each sequence of digits corresponds to exactly one $n$. In fact there are four properties that hold:

  1. Each $n$ has at least one representation
  2. Each $n$ has no more than one representation
  3. Each representation corresponds to at least one $n$
  4. Each representation corresponds to no more than one $n$

It is quite possible to construct representations that lack some of these properties. For example, consider the base-3 representation, but drop the restriction that says that $0\le a_i < 3$. Then property 2 fails: The number 47 has many base-3 representations: 502, for example, or 362, or 1 12 2 (here $a_1 = 12$), or even one (harder to write) where $a_0 = 47$. Each sequence of digits still represents a single $n$, but a particular $n$ might have many representations as a sequence of digits. Sometimes such representations even have some use.

Some of these properties are more important than others. Property 4, for example, is crucial, because if it doesn't hold, then there is some sequence of digits that might represent two different numbers, and when you see it you don't know what number is being represented. Such a system can't really be called a system for representing numbers.

Similarly, a system which fails to have property 1 has limited usefulness. Such a system can represent some $n$, but not all.

Depending on where and how it fails, a representation might be more or less useful. Fraction notation, for example, is universally used to represent rational numbers. But it fails to have properties 2 and 3! (It fails 2 since each rational number has many representations, say as $\frac12, \frac24, $ or $\frac{120}{240}$. And it fails 3 since $\frac10$ and $\frac00$ do not represent any rational numbers.) But these failures don't prevent it from being useful as a representation of rational numbers. A more serious failure arises if you try to make fractions represent real numbers; then property 1 fails, since there is no fraction representation for the number $\pi$ or $\sqrt2$.

Now let's return to $$n = \sum_{i=0}^\infty a_ib^i.$$ I said that this representation of non-negative integers has all four properties, but I left out an important limitation: the four properties only hold for $b\ge 2$. If $b=1$, the restriction $0\le a_i<b$ degenerates to $a_i=0$, and we can no longer represent any number except 0. So only 0 has a base-1 represenation. As a numeral system, this is completely useless.

If we drop the $0\le a_i<b$ restriction, we get something that hardly resembles a system of representation at all: Each number $n$ now has many base-1 representations For example, one could write 5 as 14, or 32, or 1121.

So, although it is inconsistent, mathematicians, and especially computer scientists, adopt a different meaning for "base-$1$ representation". They abandon $\sum a_ib^i$ completely and agree to represent the number $n$ as a sequence of exactly $n$ ones. For example, $7$ is represented as 1111111. This restores properties 1–4, so it is a sensible representation.

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  • $\begingroup$ Another way of looking at it is that they keep the sum definition, but replace 0 <= ai < b with 0 < ai <= b $\endgroup$ – Joe K Apr 25 '13 at 18:43
  • $\begingroup$ No, because then you can't represent any number: you need the sequence to be eventually zero. You can patch it up, but it's a bit more complicated than you said. $\endgroup$ – MJD Apr 30 '13 at 23:22
  • $\begingroup$ True. I guess that's just the way I've intuitively thought of it, but you're right, it is more complicated to actually make it work. $\endgroup$ – Joe K Apr 30 '13 at 23:59
  • $\begingroup$ The problem you run into is rooted in the fact that you identify symbols (digits) and their meaning (numbers). There is no real reason (as far as I can tell) for disallowing arbitrary $f(a_i)$ instead of $a_i$ in the "value sum", where $f : \Sigma \to \mathbb{N}$ maps digits to numbers -- as long as 1-4 are fulfilled, of course. $\endgroup$ – Raphael Oct 14 '13 at 23:13
  • $\begingroup$ @MJD, So if "base 1" is the wrong term, what's the correct term for numbers like 11111? (to be more accurate, it should be 00000) $\endgroup$ – Pacerier Mar 21 '17 at 15:51
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Yes, the usual answer is that numbers are represented by "tally marks" in base $1$. However, the numeral $0$ might not be the best choice of a tally mark because if $00000$ were interpreted in base $1$ analogously to its interpretation in other bases, then it would be interpreted as $0 \cdot 1^5 + 0\cdot 1^4 + 0 \cdot 1^3 + 0 \cdot 1^2 + 0 \cdot 1^1$, which is $0$ rather than $5$.

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    $\begingroup$ So... $0=5$ you say? Great! We can all go home early! :-) $\endgroup$ – Asaf Karagila Apr 25 '13 at 16:59
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    $\begingroup$ The trick is not to use the 1-hour clock, but to use $\mathbb{F}_\text{un},$ whose vector spaces are any set. $\endgroup$ – Loki Clock Apr 26 '13 at 0:29
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    $\begingroup$ I've never been quite certain whether $\mathbb{F}_\text{un}$ is a real thing, or someone is just being $\mathbb{F}_\text{unny}$. $\endgroup$ – Trevor Wilson Apr 26 '13 at 1:20
  • $\begingroup$ @TrevorWilson, What about fractions for base 1? 11.111? $\endgroup$ – Pacerier Mar 21 '17 at 15:56
  • $\begingroup$ @Pacerier In base 1, setting aside the question of whether digits greater than 0 (such as 1) should be allowed, I think a dot would not change the value of an expression. So for example, 11.111 would just be another way to write five. Because all the positive and negative integer powers of 1 are equal, the value of a digit in base 1 is independent of its position, whether before or after the dot. This makes the dot superfluous as an indicator of position. $\endgroup$ – Trevor Wilson Apr 3 '17 at 2:30
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There is no base $1$, and no unary number system. Base $b$ requires at least two symbols from $0$ to $b - 1$. Base $b$ does not use the digit $b$. For instance base $2$ does not use the digit $2$. So any system that uses the digit $1$ cannot be base $1$.

Tally marks are typographic representation of integers, but are not a "base", let alone "base 1".

The printed representation of a number in a base has a length which is proportional to the logarithm of the number. In the tally mark system, the length is proportional to the number.

Bases can represent fractions. For instance 1.11 in binary is one and three quarters. This works thanks to the negative powers to the right of the point. If we multiply one and three quarters by two, we can simply move the binary point: 11.1. By golly, this is three and a half, exactly right.

The best that the tally system can do here is to be repaired with some scheme whereby tally marks after a dot represent an enumeration of the countable set of fractions. For instance .1 could mean half, .11 means a third, .111 two thirds, and so on. But this scheme is fundamentally incompatible with what is on the left of the point. Moving the point has no intuitive meaning.

The use of at least two symbols in bases is related to the field of numbers having two elements: an additive identity (zero) and multiplicative identity (one). The naive tally mark system ignores this concept of a field with two identity elements, which is why it runs aground when it comes to representing fractions. It doesn't even represent zero, except perhaps by means of leaving an ambiguous empty space devoid of tallies to represent nothingness.

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    $\begingroup$ "The printed representation of a number in a base has a length which is proportional to the logarithm of the number. In the tally mark system, the length is proportional to the number." Exactly!, in base 1, the representation is proportional the logarithm in base $1 + \epsilon$ of the number, which is effectively linear. In this interpretation base 1 still counts as a base, although a special in that sense. $\endgroup$ – alfC Apr 25 '13 at 5:27
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    $\begingroup$ Your assertion that base 1 not having field identities is causing it to fail to represent fractions is wrong. It's easy to define a base 1 with two symbols, 0 and 1, in the same way other bases are defined, as an infinite string of digits with the number represented being the sum of the base raised to the power of its position multiplied by the digit. It fails to represent fractions, but that's because 1 raised to any integer power is 1, not because of a missing zero element. $\endgroup$ – Michael Shaw Apr 25 '13 at 14:12
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    $\begingroup$ @alfC Sure and we could argue that 1 is a prime number because it's only divisible by 1 and itself. $\endgroup$ – Kaz Apr 25 '13 at 14:40
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    $\begingroup$ @Kaz: We could argue exactly that. We could also argue that 2 is not a prime number because it's even and no other prime number is even. Many older definitions of prime numbers included 1. Mathematical definitions are not engraved on stone and handed to Moses by God, we make them up. We try to make them up to match reality in some way, but nothing says they can't be changed. $\endgroup$ – Michael Shaw Apr 25 '13 at 18:38
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    $\begingroup$ @Kaz: In a previous comment I explained how to define base 1 in a way that very strongly resembles the other bases. It's the obvious way to define base 1. Saying "base 1 is less useful than other bases" makes sense, saying "base 1 doesn't exist" is silly. $\endgroup$ – Michael Shaw Apr 26 '13 at 4:09
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$\displaystyle \sum_{i=0}^k a_i b^i$ with $1 \le a_i \le b$ can represent any positive integer uniquely for fixed $b$ and has the tally mark interpretation when $b=1$.

This is called bijective numeration. When $b$ is ten you lose the $0$ numeral but gain a numeral for ten ($A$ say). Many decimal numbers look the same when written in bijective base-ten, such as $432$, but others do not, such as $402$ which becomes $3A2$.

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Though many consider a tally system to be a base 1 system, I think this confuses more than it clarifies. A tally system is not a radix system and does not represent numbers using the same mechanism. The tally system is simply a string whose length represents the number (integer) in question while a radix numeral is a shorthand for a power series.

Therefore to imply that each of these distinct systems have a comparable attribute known as their "base" is misleading. Much like claiming that a playing card deck and a computer plotter each has a comparable operation called "draw".

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  • $\begingroup$ The tally mark system is a bijective base 1; and the simpliest explanation is that you don't use 0. In bijective base 10, you have digits 1 through A; after 9 comes A, then 11. All numbers in the positional and bijective base are writen the same as long as the positional number does not contain a zero. 829 = 829, but 809 becomes 7A9. $\endgroup$ – Liggliluff Feb 20 at 22:40
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I'm adding an answer because I don't really agree with any of the answers here. An expression like $6272$ represents, by convention, the number

$$6\cdot b^3+2\cdot b^2 + 7\cdot b^1 + 2\cdot b^0$$

Now in base-1 it just happens to be the case that $1 = 10 = 100 = \cdots$ just because $1 = 1^2 = 1^3 = \cdots$. Your tallying is just writing in base-1 in the usual way:

$$ 11111 = 1\cdot 1^4 + 1\cdot 1^3+1\cdot 1^2 + 1\cdot 1^1 + 1\cdot 1^0 $$

Things are just a little funny because the symbolism isn't bijective:

$$ 10101 = 1\cdot 1^4 + 0\cdot 1^3+1\cdot 1^2 + 0\cdot 1^1 + 1\cdot 1^0 = 111 $$

Essentially the zeros can always be removed from an expression. Base-1 shares this property in common with base-0, but these two are the only bases with this property. Base-0 is useless, however, because it only manages to represent the number 0. Base-1 can represent any natural number but it's very inefficient. It also fails to represent fractions, which any other base can do which isn't 0 nor 1.

I disagree with the assertions made in the current answer to the question, which claims that there is no explicit way to represent $0$ in a base-1 system (it's just 0!), as well as claiming that adding 0 would somehow "make it binary". These claims are only sensible under the presumption that a base representation of natural numbers should be bijective, which is not necessary to assume.

I think the point of confusion is that for bases other than 0 and 1, one typically likes to choose symbols in such a way that everything is bijective. This is not possible for base-1 and base-0. On the other hand, even for decimal we could allow ourselves to use numbers bigger than 9:

$$ 3002 = 2\cdot 10^3 + 10\cdot 10^2 + 0\cdot 10^1 + 2\cdot 10^0 =2[10]02 $$

and we now see the phenomenon of non-unique representation. Choosing symbols $\{0,1,\ldots,9,10 \}$ for decimal is analogous to using $\{0,1\}$ for base-1.

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  • $\begingroup$ In a positional base 1 system, the only digit you got is 0. Base 2 got digits 0 and 1, and base 1 got digit 0. If you skip digit 0, or use more digits than the value of the base, you are not using a positional base. $\endgroup$ – Liggliluff Feb 20 at 22:42
  • $\begingroup$ Besides failing to represent fractions, arbitrary ordinals can't all be represented as finite sums of powers of 1. $\endgroup$ – Martin Rattigan Feb 23 at 21:19

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