Can base-1 represent a nonzero integer ?

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In summary: This conversation is about whether base-1 can represent a nonzero integer, and if it can be equated with tally-mark notation. The conclusion is that base-1 is a true positional system with only one position and infinite signs. However, for positive reals greater than one, the equation \left\lfloor {\frac{{\left\lfloor {\frac{a}{b}} \right\rfloor }}{c}} \right\rfloor = \left\lfloor {\frac{a}{{bc}}} \right\rfloor does not always hold. It holds for natural numbers, but not necessarily for positive reals.
  • #1
bomba923
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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.)
 
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  • #2
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
Moo Of Doom said:
.. ..why not? It's just convention that we use only digits less than the base...

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

[tex] \left\lfloor {\frac{{\left\lfloor {\frac{a}{b}} \right\rfloor }}{c}} \right\rfloor = \left\lfloor {\frac{a}{{bc}}} \right\rfloor \; {?} [/tex]
 
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  • #4
bomba923 said:
Btw, for any positive reals a,b,c, does

[tex] \left\lfloor {\frac{{\left\lfloor {\frac{a}{b}} \right\rfloor }}{c}} \right\rfloor = \left\lfloor {\frac{a}{{bc}}} \right\rfloor \; {?} [/tex]
No. Here's a counter-example:

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

[tex] \left\lfloor {\frac{{\left\lfloor {\frac{2}{3}} \right\rfloor }}{\frac{2}{3}} \right\rfloor = 0[/tex], but [tex]\left\lfloor {\frac{2}{{(3)(\frac{2}{3})}}} \right\rfloor = 1[/tex].
 
  • #5
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:
[tex] \left\lfloor {\frac{{\left\lfloor {\frac{a}{b}} \right\rfloor }}
{c}} \right\rfloor = \left\lfloor {\frac{a}{{bc}}} \right\rfloor [/tex]

Correct?
 
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  • #6
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 written 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
bomba923 said:
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:
[tex] \left\lfloor {\frac{{\left\lfloor {\frac{a}{b}} \right\rfloor }}
{c}} \right\rfloor = \left\lfloor {\frac{a}{{bc}}} \right\rfloor [/tex]

Correct?

Nope, still not :(.

Observe:

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

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...
 
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  • #8
There is just one other (just this last condition), to which I am sure there is no counterexample:

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

[tex] \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 \; {?} [/tex]
 
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  • #9
Moo Of Doom said:
No. Here's a counter-example:

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

[tex] \left\lfloor {\frac{{\left\lfloor {\frac{2}{3}} \right\rfloor }}{\frac{2}{3}} \right\rfloor = 0[/tex], but [tex]\left\lfloor {\frac{2}{{(3)(\frac{2}{3})}}} \right\rfloor = 1[/tex].
I'm sorry did no one catch this? 2/3 divided by 2/3 isn't 0.


edit: I am pretty sure I misunderstood the question, nevermind
 
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  • #10
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
Moo Of Doom said:
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?

:smile:
I take AP CompSci and just finished a basic binary integer-->decimal integer conversion program. Quite basic, no real thinking needed :rolleyes:. 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)
[tex] {\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 } } [/tex]

Now if [itex] x [/itex] is a decimal natural, using "TI-89 sequence notation" (:redface:!), the sequence
[tex] {\left\{ {a_0 ,a_1 , \ldots ,a_{\left\lfloor {\log _b x} \right\rfloor } } \right\}} [/tex]
is represented as
[tex] \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] [/tex]

|Or as I prefer, :shy:
[tex] \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 } [/tex]

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

-Thus,
[tex] \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 } [/tex]
-----------------------
And so, I came to wonder (for simplification purposes)
if this statement was true:
[tex] \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 [/tex]
:shy:
 
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  • #12
why not, you can count just by using 0

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

and so on.
 
  • #13
Anzas said:
why not, you can count just by using 0

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

and so on.

"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[itex]\ne[/itex]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
bomba923 said:
:smile:
-----------------------
And so, I came to wonder (for simplification purposes)
if this statement was true:
[tex] \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 [/tex]
:shy:

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

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

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

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

Base-1 representation is a numeral system in which each digit represents a power of 1. This means that only the digit 1 is used, and the value of each digit is determined by its position in the number.

2. Can base-1 represent all integers?

Yes, base-1 can represent all integers. This is because every integer can be expressed as a sum of powers of 1, with each power being represented by a digit in the base-1 system. For example, the number 5 can be represented as 11111 in base-1.

3. How does base-1 representation compare to other numeral systems?

Base-1 representation is unique compared to other numeral systems, as it only uses the digit 1 and has no concept of zero. This is in contrast to other systems such as base-10 (decimal) or base-2 (binary), which use multiple digits to represent numbers.

4. What is the practical application of base-1 representation?

Base-1 representation has limited practical application in mathematics or computer science. It is often used as a theoretical concept to demonstrate the fundamental principles of numeral systems.

5. Why is base-1 not commonly used in everyday life?

Base-1 is not commonly used in everyday life because it is not efficient or practical for representing large numbers. It also lacks the ability to represent fractions or decimal numbers. Other numeral systems, such as base-10, are more versatile and better suited for everyday use.

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