# Base One ?

1. Oct 2, 2005

### bomba923

Just a -very quick- clarification

Can base-1 represent a nonzero integer ?
Is there a base-1 at all?

*The digits of binary (base 2) integers contain only 0 and 1's (no 2's allowed). The digits of base-3 integers contain only 0 and 1 and 2's (no 3's allowed).

*But base-1 ??? Wouldn't it contain only zeroes ? Which would not amount to anything at all?

In some sites, I read that people equate it with "tally-mark" notation. But how can that be?? A base-one integer cannot contain "1" as a digit. Therefore, there will only be a string of zeroes, which does not amount to anything at all.

0*1 + 0*(1^2) + 0*(1^3) + .... = 0, no ?

Just for clarification, is there a base-1 ?

*Can it represent nonzero integers ?
It certaintly can't equate with tally-mark notation, can it? (i.e., tally-mark notation meaning direct representation of quantity. Therefore, ||| = 3 , |||| = 4 and so on.)

Last edited: Oct 3, 2005
2. Oct 3, 2005

### Moo Of Doom

Beyond the fact that you shouldn't have digits other than zero in base one, the entire concept of a positional system breaks down at one because 1=1^2=1^3=1^n, and thus all place values are exactly the same. Pretend that you can use any digit (why not? It's just convention that we use only digits less than the base... take 312 in base two to be 3*2^2+1*2^1+2=16(=10000 in base 2)). Then the string 241=2*1^2+4*1^1+1=7, but 421, 214, 142, 402010, .00000010000020000004, and an infinite number of other representations also equal 7. Thus the placement of the numbers is completely irrelevant, defeating the purpose of a positional system.

The tally-mark analogy is quite right, you can write any number in base one as 1111111... and just count the ones, because the number is just equal to the sum of its digits.

3. Oct 3, 2005

### bomba923

And that was my nitpick :shy: all along
------------------------
Btw, for any positive reals a,b,c, does

$$\left\lfloor {\frac{{\left\lfloor {\frac{a}{b}} \right\rfloor }}{c}} \right\rfloor = \left\lfloor {\frac{a}{{bc}}} \right\rfloor \; {?}$$

Last edited: Oct 3, 2005
4. Oct 3, 2005

### Moo Of Doom

No. Here's a counter-example:

For a=2, b=3, and c=2/3

$$\left\lfloor {\frac{{\left\lfloor {\frac{2}{3}} \right\rfloor }}{\frac{2}{3}} \right\rfloor = 0$$, but $$\left\lfloor {\frac{2}{{(3)(\frac{2}{3})}}} \right\rfloor = 1$$.

5. Oct 4, 2005

### bomba923

Hmm, but what if we restrict a,b,c to the positive reals greater than one?
(a,b,c > 1)

||Supposedly then, if one side equals to zero, the other must equal to zero as well, since no multiplier greater than one in the denominator will increase the value of the expression on the right side. (From considering relative comparisons between a,b,c)

\Thus, given positive reals a,b,c > 1, the equation will be true:
$$\left\lfloor {\frac{{\left\lfloor {\frac{a}{b}} \right\rfloor }} {c}} \right\rfloor = \left\lfloor {\frac{a}{{bc}}} \right\rfloor$$

Correct?

Last edited: Oct 4, 2005
6. Oct 4, 2005

### serge

I don't agree. there's a true base-one system.
A positional system is a convention for describing numbers with signs.
The usual convention is that 0 is the number zero, 1 is the number 1, 2 is the number two and so on, and that in base b any number is a sum of a*b^i terms with 0<= a < b

but you can say that, in base one notation, the sign '1' is the number one, '11' is the number two, '111' is the number three and so on (tally mark notation) so that any number in base one is writen with only one term a*1^1, with a being a sequence of 1s. Therefore base one is a true positional system, but with only one position and an infinite number of signs (sort of base infinite)

base one notation is often used for binary turing machines, where you use tally mark notation for numbers and 0 as a separator btw numbers.

7. Oct 4, 2005

### Moo Of Doom

Nope, still not :(.

Observe:

Let a = 10, b=9, c=11/10
$$\left\lfloor {\frac{{\left\lfloor {\frac{10}{9}} \right\rfloor }} {\frac{11}{10}}} \right\rfloor = 0$$
$$\left\lfloor {\frac{10}{{(9)(\frac{11}{10})}}} \right\rfloor = 1$$

Although it did take me longer to figure out a counter-example. Intuitively it didn't quite add up, because there can be a difference of nearly 1 between x and floor(x), and that can bump the next floor operation past an integer.

EDIT: But I have the sneaking suspicion that the difference between the two never exceeds 1, given your new constraint... I'll explore further on that...

Last edited: Oct 4, 2005
8. Oct 4, 2005

### bomba923

There is just one other (just this last condition), to which I am sure there is no counterexample:

Originally, the floor problem was intended $$\forall a,b,c \in \mathbb{N}$$ (but I made an unjustifiable expansion into the reals, when I originally intended this for positive integers). As a final question (although perhaps I missed a comparison),

$$\forall a,b,c \in \mathbb{N} , \; \text{Does} \; \left\lfloor {\frac{{\left\lfloor {\frac{a}{b}} \right\rfloor }}{c}} \right\rfloor = \left\lfloor {\frac{a}{{bc}}} \right\rfloor \; {?}$$

Last edited: Oct 4, 2005
9. Oct 4, 2005

### whozum

I'm sorry did no one catch this? 2/3 divided by 2/3 isn't 0.

edit: Im pretty sure I misunderstood the question, nevermind

Last edited: Oct 4, 2005
10. Oct 4, 2005

### Moo Of Doom

For natural numbers, this does seem to hold. You might want to try a proof of induction on c.

How did this come up, by the way?

11. Oct 4, 2005

### bomba923

I take AP CompSci and just finished a basic binary integer-->decimal integer conversion program. Quite basic, no real thinking needed . And then, I came upon a way (was rather bored, tried something new!) to convert decimal integers into integers of any natural base.

First, I developed this statement:
(where b=the numerical base)
$${\forall \left( {x,b} \right) \in \mathbb{N}^2 ,\;\exists \left\{ {a_0 ,a_1 , \ldots ,a_{\left\lfloor {\log _b x} \right\rfloor } } \right\}\;{\text{such that }}\forall n \in \mathbb{N} \cup \left\{ 0 \right\},\;a_n \in \left\{ {0,1, \ldots ,b - 1} \right\}\;{\text{and }}x = \sum\limits_{n = 0}^{\left\lfloor {\log _b x} \right\rfloor } {a_n b^n } }$$

Now if $x$ is a decimal natural, using "TI-89 sequence notation" (!), the sequence
$${\left\{ {a_0 ,a_1 , \ldots ,a_{\left\lfloor {\log _b x} \right\rfloor } } \right\}}$$
is represented as
$$\operatorname{seq} \left[ {\bmod \left( {\left\lfloor {\frac{x}{{b^{\left\lfloor {\log _b x} \right\rfloor - n} }}} \right\rfloor ,b} \right),n,0,\left\lfloor {\log _b x} \right\rfloor } \right]$$

|Or as I prefer, :shy:
$$\left\{ {\bmod \left( {\left\lfloor {\frac{x}{{b^{\left\lfloor {\log _b x} \right\rfloor - n} }}} \right\rfloor ,b} \right)} \right\}_{n = 0}^{\left\lfloor {\log _b x} \right\rfloor }$$

*However, as we all know $\forall (a,b) \in \mathbb{Z}^2$,
$$\bmod \left( {a,b} \right) = a - b\left\lfloor {\frac{a}{b}} \right\rfloor$$.

-Thus,
$$\left\{ {\bmod \left( {\left\lfloor {\frac{x}{{b^{\left\lfloor {\log _b x} \right\rfloor - n} }}} \right\rfloor ,b} \right)} \right\}_{n = 0}^{\left\lfloor {\log _b x} \right\rfloor } = \left\{ {x - b\left\lfloor {\frac{{\left\lfloor {\frac{x}{{b^{\left\lfloor {\log _b x} \right\rfloor - n} }}} \right\rfloor }}{b}} \right\rfloor } \right\}_{n = 0}^{\left\lfloor {\log _b x} \right\rfloor }$$
-----------------------
And so, I came to wonder (for simplification purposes)
if this statement was true:
$$\forall \left( {a,b,c} \right) \in \mathbb{N}^3 ,\;\left\lfloor {\frac{{\left\lfloor {\frac{a}{b}} \right\rfloor }}{c}} \right\rfloor = \left\lfloor {\frac{a}{{bc}}} \right\rfloor$$
:shy:

Last edited: Oct 4, 2005
12. Oct 5, 2005

### Anzas

why not, you can count just by using 0

0:0
1:00
2:000
3:0000

and so on.

13. Oct 5, 2005

### bomba923

"0" is interpreted for what it is: ZERO

0+0+0+0+0+.... = 0 :grumpy:

(unless you want your zeroes to equal 1 .... but then again, 0$\ne$1)
---------------------------------------
"Tally-mark" representation of direct quantity would require 1's:

Therefore:
1:1
2:11
3:111
4:1111
5:11111
6:111111
...and so on

14. Oct 10, 2005

### uart

Yes it is always true for natural numbers. It's fairly easy to prove directly, no need to use induction.

Just let $$a=qb + r$$, where $$r<b$$.
Then $$\lfloor \frac{a}{b}\rfloor = q$$

Now let $$q=nc + s$$, where $$s<c$$.
Then the LHS becomes, $$\left\lfloor \frac{{\left\lfloor {\frac{a}{b}} \right\rfloor }}{c}} \right\rfloor= n$$

Applying the same definitions to the RHS of the original equations results in,
$$\left\lfloor {\frac{nc + s + r/b}{{c}}} \right\rfloor$$
Clearly if the RHS is to be different to the LHS then we require $$s + \frac{r}{b} \geq c$$. But $$s \leq c-1$$ and $$r<b$$ so that is impossible.

Last edited: Oct 10, 2005
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