How can we represent $10$ in a decimal system?

Question

This may sound like a silly question to begin with but I'm having problems finding a proper answer.

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The question is generally targeting numeral systems of any base, but for simplicity, I will demonstrate the problem on the representation of natural numbers in a decimal system.

In the decimal system, there are ten basic symbols (also known as digits $0-9$), which are corresponding to the first ten natural numbers including zero.

In order to generate the rest of the numbers, the value of each symbol is multiplied by $10$ raised to the power of the symbol's position, and the results are summed together.

For example, the number $496$ is equivalent to $4\cdot10^2+9\cdot10^1+6\cdot10^0$.

More generally, the number ${d_{n-1}}{d_{n-2}}\dots{d_{1}}{d_{0}}$ is equivalent to $\sum\limits_{i=0}^{n-1}d_i\cdot10^i$.

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My question here onwards is very simple:

The number $10$ is equivalent to $1\cdot10^1+0\cdot10^0$.

This seems like a recursive definition, where $10$ is used in order to represent itself.

So how exactly do we "allow ourselves" to use this system in order to represent $10$?

More generally, how can we actually use a base-B system in order to represent B?

Unless I have missed the actual definition of a numeral system (given here), it seems that an additional symbol has to be added just for the sake of representing the base of the system itself.

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Thank you.

Answer 1

Following André Nicolas's suggestion, if you don't like using “$10$” you can replace it with a symbol like “t”. Or you could replace the troublesome expression “$d_i\cdot 10^i$” with “the product of $d_i$ and the $i$'th power of ten”.

The point here is that you are confusing the number $10$ with its representation as a numeral. The definition you quoted defines a certain set of notations for numbers. To describe and implement the definitions, you need to be able to calculate, but you don't already need to understand the notation you are defining. You have confused the issue by using the same notation when you describe the calculations define the numerals, but this is not necessary. You could perform the calculations on an abacus, for example, or describe them in Peano arithmetic as Git Gud suggests.

Answer 2

The sum of powers of $10$ is not the definition of the decimal system. It is a consequence.

What we learn in school is:

The decimal system consists of words in $\{0,1,2,...,9\}$ such that $0$ is never the left most letter.

The successor of $0$ is defined to be $1$. The successor of $1$ is defined to be $2$. ...

The successor of $99...9$ is defined to be the word $100...0$.

Finally, the successor of $a_1...a_{n-1}a_n99...9$, with $a_n\neq9$ is defined to be $a_1...a_{n-1}b00...0$, where $b$ is the successor of $a_n$.

In particular $10$ was just defined to be the successor of $9$.

A posteriori you check (by induction) that $$a_na_{n-1}...a_1a_0=a_n\cdot 10^n+...+a_0.$$

You need, first to define what is $\cdot$ and $10^k$ and prove by induction some of their properties.

The proof: $0=0$. That's the first step. Assume that $$a_na_{n-1}...a_1a_0=a_n\cdot 10^n+...+a_0.$$ If $a_0\neq9$ then $$a_na_{n-1}...a_1a_0+1=a_n\cdot 10^n+...+a_0+1=a_n\cdot 10^n+...+(a_0+1).$$ Assume $a_0=9$ and let $k\leq n$ be the largest such that $a_0=...=a_k=9$. Then $$\begin{align}a_na_{n-1}...(a_{k+1}+1)0...00&=a_na_{n-1}...a_{k+1}99...9 + 1\\&=a_n\cdot 10^n+...+a_{k+1}\cdot10^{k+1}+9\cdot10^{k}+...+9+1\\&=a_n\cdot 10^n+...+a_{k+1}\cdot10^{k+1}+9\cdot10^{k}+...+(9+1)\\&=a_n\cdot 10^n+...+a_{k+1}\cdot10^{k+1}+9\cdot10^{k}+...+9\cdot10^1+10\\&=a_n\cdot 10^n+...+a_{k+1}\cdot10^{k+1}+9\cdot10^{k}+...+10\cdot(9+1)\\&=a_n\cdot 10^n+...+a_{k+1}\cdot10^{k+1}+9\cdot10^{k}+...+10^2\\&\ldots\\&=a_n\cdot10^n+...+(a_{k+1}+1)\cdot10^{k+1}\end{align}.$$

You could also do things the other way around. Have the symbols $\{0,1,...,9,t\}$, define how sums of powers of $t$ ought to behave. In particular $9+1=t$. And then define $a_n...a_0$, with $a_i\neq t$ as the short-hand notation for $$a_n\cdot t^n+...+a_0.$$

If I remember correctly Zeilberger advocated in one of his Opinions to do something like this other way.