0

I'm no math wiz here, but I have a question that I can't wrap my head around. In fact, I don't even know how I would even go about asking the question properly. Is there an alternative to using 0 as a placeholder?

When writing out numbers, 10 doesn't have it's own symbol. So rather we use 1 and 0(the placeholder) to represent 10. Is there a better method than using 0 or keep from going into double digits with the number 10?

Xarcell
  • 171
  • 1
  • 6
  • 2
    You could just write $X$ for 10. It was good enough for Julius Caesar... – vadim123 Mar 19 '14 at 03:06
  • What about avoiding going into double digits? It seems to me that double digits should begin with 11, not 10. Then again, I'm a math retard... – Xarcell Mar 19 '14 at 03:09
  • What about the number 0? Do you feel tempted to leave it as an empty space because it's just a placeholder for nothingness? I think this is really an issue with the number 0. – mathematician Mar 19 '14 at 03:16

2 Answers2

3

There isn't a better way, but there are ways. For example, we can use bijective base 10.

In bijective base 10, the digits are 1, 2, 3, 4, 5, 6, 7, 8, 9, A, and numbers are interpreted as

$$5A26 = 5*10^3 + 10*10^2 + 2*10^1 + 6*10^0 = 6026$$

This system is capable of uniquely representing any positive integer. However, we don't gain anything by using it, and it's more awkward to work with. It doesn't generalize as nicely to representing arbitrary real numbers, and even the natural representation of zero (the empty string) is incredibly awkward to use in an equation.

There's nothing to be gained by avoiding zero. Sure, there's no symbol for 10, but if we introduce one, then there's no symbol for 11, and 1 ends up serving a similarly placeholdy role.

user2357112
  • 2,401
1

If I'm understanding your question correctly:

It's important to remember that the number of digits a number has is only dependent on what base I'm writing the number in. We typically write things in base-$10$. That is, each "slot" can go from $0$ to $9$ before we add another "slot".

However, what if I write $10_{10}$ in base-$2$? I have $1010_2$ (count up only allowing each "slot" to get to $1$ before a new "slot" is added. You'll see where $1010_2$ comes from). In this way of counting, for any even number I'm going to be using $0$ as a "placeholder".

So, my point: you could make $11_{10}$ the fist double digit double if you wanted. Just come up with another symbol to represent $10_{10}$ and then begin writing everything in base-$11$. Whenever we use a "placeholder" or go into more digits is completely dependent on the base, so you can have the first occurrence of $0$ as a "placeholder" occur whenever you want. I hope I helped, because at least I think this is what you were asking.

foobar1209
  • 1,047
  • I tried using another symbol, but I can't wrap my head around it. For example, let's use X: 1, 2, 3, 4, 5, 6, 7, 8, 9, X, 11, 12, 13, 14, 15, 16, 17, 18, 19, ??. Can't use XX, because then you would have to use XXX for 30 and XXXXX for 50, which is cumbersome to write. – Xarcell Mar 19 '14 at 03:32
  • Alright, let's define a base-$11$ system where $10_{10}$ is represented by $X$. So, $20_{10}$ would actually be $19_{11}$. We didn't increment the second slot until one number later since we counted: $1, 2, 3, 4, 5, 6, 7, 8, 9, X, 10, 11...$ Does that make sense? – foobar1209 Mar 19 '14 at 03:50
  • but now your changing the base which breaks the base 10 system. – Xarcell Mar 19 '14 at 11:58
  • I was thinking: 1, 2, 3, 4, 5, 6, 7, 8, 9, X, 11, 12, 13, 14, 15, 16, 17, 18, 19, X2, 21, 22, 23, 24, 25, 26, 27, 28, 29, x3, 31, ... But that's still using 2 digits. – Xarcell Mar 19 '14 at 12:04
  • @Xarcell You could do that, but having every multiple of $10$ written out like that would make some arithmetic rather more unwieldy and confusing, wouldn't it? The point is that you could do it if you wanted to; there just really isn't a reason. – foobar1209 Mar 19 '14 at 16:43