A system with base $10$, has $10$ digits


When we try to break down a number such as $4128$ in a system with base 10, the following operation happens:

$$ \begin{align} 4128 &= 4000 + 100 + 20 + 8 \\ &= 4\cdot1000 + 1\cdot100 + 2\cdot10 + 8\\ &= 4\cdot10^3 + 1\cdot10^2 + 2\cdot10^1 + 8\cdot10^0 \end{align} $$

Why do we multiply by the $10$ power and not for example by $9$ or by $8$? Is it because there are $10$ digits?

  • 1
    $\begingroup$ Because you are using base 10. You would multiply by powers of 9 if you wanted to write a number in that base and you would have only 9 digits: 0,1,2,...,8 to do so. $\endgroup$
    – MasB
    Commented Mar 30, 2021 at 14:53
  • 1
    $\begingroup$ Yes. Read the wikipedia page on positional notation. en.wikipedia.org/wiki/Positional_notation $\endgroup$ Commented Mar 30, 2021 at 14:54

2 Answers 2


TL;DR: The reason is the existence and uniqueness of the positional representation for all non-negative integers.

As you have observed, in the positional notation the integer corresponding to the representation $a_m\dots a_1a_0$ where $a_k\in\{0, 1, \dots, n-1\}$ are individual digits is given by the sum $\sum_{k=0}^m a_kb^k$. It turns out that this gives rise to a one-to-one correspondence if and only if $n=b$.

It is easy to see the failure of existence and uniqueness of representations when $n\ne b$. For example, when $n$ is ten and $b$ is nine then the number nine has two representations: $10$ and $9$. On the other hand, if $n$ is ten and $b$ is eleven then number ten does not have any representation at all. Indeed, in this case the highest single-digit number is nine and the smallest number with more than one digit is eleven.

N.B. for fractions, the positional notation does leave some ambiguity.

More precisely, let $S^0(\mathbb{Z}_n)$ denote the set of infinite sequences of elements of $\mathbb{Z}_n=\{0, 1, \dots, n-1\}$ that eventually become zero. Note that valid representations of integers in the base-$n$ notation are in one-to-one coresponence with the elements of $S^0(\mathbb{Z}_n)$. For a non-negative integer $k$, let $a_k\in\mathbb{Z}_n$ denote $k$th element of the sequence $a\in S^0(\mathbb{Z}_n)$. Define the function $f_{n,b}: S^0(\mathbb{Z}_n)\to\mathbb{Z}$ by

$$ f_{n,b}(a) = \sum_{k=0}^\infty a_kb^k.\tag1 $$

With the above definitions, the question becomes: why do we use $(1)$ with $n=b$ to assign values in $\mathbb{Z}$ to representations in $S^0(\mathbb{Z}_n)$. The answer is provided by the following

Theorem. The function $f_{n,b}$ defined by $(1)$ is a bijection if and only if $n=b$.

Proof. First, we show that if $b\ne n$ then $f_{n,b}$ is not a bijection. Note that if $b<n$ then $f_{n,b}((0, 1, 0, \dots)) = f_{n,b}((b, 0, \dots))$ and so $f_{n,b}$ fails to be injective. On the other hand, for $b>n$ we see that if $a_i > 0$ for some $i>0$ then $f_{n,b}(a) \ge b > n$ and if $a_i=0$ for all $i>0$ then $f_{n,b}(a) < n$. Thus, there is no $a\in S^0(\mathbb{Z}_n)$ for which $f_{n,b}(a) = n$ and $f_{n,b}$ fails to be surjective.

Now, we assume $n=b$ and show that $f_{n,n}$ is a bijection. Suppose that $f_{n,n}(a) = f_{n,n}(a')$. Then from $(1)$ we have

$$ \sum_{k=0}^\infty (a_k-a_k')b^k = 0.\tag2 $$

Note that for any non-negative integer $m$ the right-hand side of $(2)$ is divisible by $b^m$. Therefore, the left-hand side is divisible by $b^m$. Thus, $a_k=a_k'$ for all $k<m$. Therefore, $a=a'$ and $f_{n,n}$ is injective.

Finally, let $x\in\mathbb{Z}$ and define $a_0=x\mod n$ so that $x=q_0n+a_0$ for some integer $q_0$. Next, define $a_1 = q_0\mod n$ so that $q_0=q_1n+a_1$ for some integer $q_1$. Continuing this process until $q_m=0$ we obtain

$$ x = q_0n+a_0 = q_1n^2 + a_1n + a_0 = \dots = a_mn^{m+1} + \dots + a_1n + a_0. $$

However, this means that $x=f_{n,n}(a)$ establishing that $f_{n,n}$ is surjective. $\square$


Why do we multiply by the $10$ power and not for example by $9$ or by $8$? Is it because there are $10$ digits?

Well, it's because that's what "base 10" means. The word "base" means "the number whose powers you multiply the digits by." If you multiply the digits by powers of 9, that's base 9. If you multiply the digits by powers of 8, that's base 8.

The reason that there are 10 digits is that that's how many digits you need in order for base 10 to work. You need digits for these numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. You don't need a digit for the number after 9 because we have a way to write that number using the digits we already have. That way of writing that number is, of course, "10."


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