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Base system with $b \in \mathbb N$ consists of $b$ digits $d_0,d_1,d_2\dots d_{b-1}$. A number $a$ is expressed by some weighted sum of (integer) powers of $b$, where the digits $d_0,d_1,d_2\dots d_{b-1}$ are the weights.

My question is, why do we, if $b$ is a natural number greater than $1$, select the weights or the digits to be whole numbers and why do we need $b$ of them?

In the usual base $2$ system, with weights $0$ and $1$, number $15$ in base $10$, for example, would be expressed as $1111$.

However, if we say that the base two system has weights $0$ and $\frac{1}{3}$, which we name $a$, $15_{10}$ would be $a0aa0a$, ie. $\frac{1}{3} 2^5+\frac{1}{3} 2^3+\frac{1}{3} 2^2+\frac{1}{3} 2^0$.

It is a tad longer than with the usual weights. Is that the only reason to use $0$ and $1$, instead of $0$ and $1/3$?

How about base $3$ with weights $0$ and $\frac{2}{5}$, denoted by $b$? For example $15_{10}$ would be $bb0b.\overline {bbb}$ or $\frac{2}{5}3^3 + \frac{2}{5}3^2 + \sum_{n=0}^{\infty} \frac{2}{5}3^{-n}$.

Can I choose the weights or base digits in any way I want and there would always be a way to write any number in any base using them? Or do some other complications arise?

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Cause number with base $b$ has $b$ units in them and one of them being $0$. –  Inceptio Mar 26 '13 at 18:43
    
What is your goal? –  Berci Mar 26 '13 at 18:45
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@Berci Do I need a goal? –  Valtteri Mar 26 '13 at 18:47
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Using weights $0$ and $1$ in base $3$, one only gets a Cantor set of possible values, so it is impossible to represent $2$ or $5$. Likewise in your example with weights $0$ and $\tfrac25$, try to represent $2$. –  Erick Wong Mar 26 '13 at 18:48
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@Inceptio Yes, I get that now. How about the choice of those $b$ units? Or what if we use something like base $\pi$ or $3/5$? How do you select the number of units? –  Valtteri Mar 26 '13 at 19:03

3 Answers 3

up vote 2 down vote accepted

It would seem the first requirement would be to represent all naturals with just numbers in front of the fraction point. If your smallest unit is greater than $1$, this will fail with $1$. For $b=2$, as long as the weight is $\frac 1n$, you can represent all the naturals in base $b$-it is just representing $n$ times the natural you want in regular base $2$. Perhaps a bit of trouble, but it can be done. If you choose $w=\frac 23$ you can't represent any odd numbers-that is terrible.

The reason we have $b$ units in base $b$ is because that is how many it takes. Below $b^n$ you only have $n$ digits, so if you have fewer than $b$ weights you will miss some. This will apply even for units smaller than $1$ if you make $n$ large enough.

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Base of a number is the number of units. If you represent $a_1a_2\dots a_n$ in base $b$, number of possible values have to be $0 \le a_i<b$. Which means the base is upto $b-1$.

What happens if you let bases to be in fractions?

If $\dfrac{p}{q} <1$, representing $n$ is done by using digits larger than $\dfrac{p}{q}$ which is not allowed.

When $\dfrac{p}{q} >1$, representing $n$ is done by using natural numbers less than $\dfrac{p}{q}$, IF your able to represent $n$ using digits less than $\dfrac{p}{q}$, you will oserve that you won't be able to do it for $n+1$, which contradicts the definition of base.

What happens in irrational numbers?

Take an example of $\pi$, you will note that number digits that can be used are $\{0,1,2\}$. Every combination of the numbers with base $\pi$ will give an irrational except for $(1)_{\pi},(2)_{\pi}$ and $(0)_{\pi}$.

If you let rational numbers used in representation of a number $n$ in base $b$

Such that :$0 \le a_i<b$ and $a_i$ $\epsilon$ $R$, there are infinite of the rational numbers between any rational numbers. So that's not valid.

Therefore, definition of Radix(Base) is just valid to natural numbers of base $10$.

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All that is needed is that there a digit equal to each number 0 to b-1, modulo b. That is, you can add or subtract b from various d where d is in 0 to b-1.

Base 3 works with digits $0,1,2$ or $0,1,-1$ or $3,1,-1$ or $3,1,2$. Except for 3,1,-1, the bases are correctly spelt: ie sorted spelling is sorted size.

On the other hand, one can quite easily have negative bases, like $-3$, and even complex bases like $\sqrt{-3}$. Both of these rely on the digits $-1, 0, 1$ in their usual form, the latter is implemented by carrying and borrowing over two columns.

The following is a multiplication of $7+4\sqrt{3}$ by $7-4\sqrt{3}$, in base $\sqrt{3}$, which is ordinary base 3, with carry and borrow over two columns (ie 3 = 1 0 0).

     7+4 sqrt(3)     1  1  m  1  1   = 1 0 m 0 1 + 1 0 1
     7-4 sqrt(3)     1  m  m  m  1   = 1 0 m 0 1 - 1 0 1
                     -------------
                     1  1  m  1  1
                  m  m  1  m  m
              m   m  1  m  m
           m  m   1  m  m
        1  1  m   1  1
        ---------------------------
        1  0 -3   0  1  0 -3  0  1    carryless sum
        0  0  0   0  0  0  0  0  1    carry the -3 as -1 0 0.
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