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

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 $\sum\limits_{i=0}^{n-1}d_i\cdot10^i$.

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.

Thank you.

  • 2
    $\begingroup$ If the $10$ makes you uncomfortable, replace it by the letter $t$, or $B$. $\endgroup$ – André Nicolas Jan 31 '15 at 18:36
  • $\begingroup$ I think this answer answers your question, you don't need to define $10$ separately. $\endgroup$ – Git Gud Jan 31 '15 at 18:37
  • $\begingroup$ @AndréNicolas Then you need the symbols $\{0,1,2,...,9,t\}$. Just joking. $\endgroup$ – Pp.. Jan 31 '15 at 18:42
  • $\begingroup$ @AndréNicolas: Thanks (it's an honor BTW :)... Anyway, that what I've basically asked towards the end of the question - "it seems that an additional symbol has to be added". Yet, it doesn't quite make sense to add a symbol just for the sake of one element. $\endgroup$ – barak manos Jan 31 '15 at 18:45
  • $\begingroup$ @GitGud: Thanks. $\endgroup$ – barak manos Jan 31 '15 at 19:01

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.

  • $\begingroup$ So are you saying that I've confused between the representation of a number and the calculation of the quantity which the number represents? $\endgroup$ – barak manos Jan 31 '15 at 18:47
  • $\begingroup$ Yes. You need some way to describe the calculation that defines the decimal numerals, but that description doesn't have to use the decimal notation itself. $\endgroup$ – MJD Jan 31 '15 at 18:58
  • $\begingroup$ OK, I sort of understand why I've mixed those two things (though I need to think about it for a minute. Thank you very much! $\endgroup$ – barak manos Jan 31 '15 at 18:59

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.