# On the representation of integers?

• B

## Summary:

We describe the representation of positive integers in the form of the power series ##q2^n##. This allows us to consider positive integers based on the comparison of partial sums of such power series.

## Main Question or Discussion Point

Let (a, b, c) be some arbitrary positive integers such that:
(q2^0 + q2^1+ . . . + q2^x),
(q2^0 + q2^1+ . . . + q2^y),
(q2^0 + q2^1 + . . . + q2^z),
where: q = (1, 2), (x, y, z) = (1, 2, 3, . . ., n).
In the case if and only if q = 2, we accept the following notation :
[(q-1)2^0 +(q-1)2^1 + . . . + (q-1)2^x] := \Delta{a},
[(q-1)2^0 +(q-1)2^1 + . . . + (q-1)2^y] := \Delta{b},
[(q-1)2^0 +(q-1)2^1 + . . . + [(q-1)2^z] := \Delta{c},
Indeed, it is obvious that if q = 1, then [(q-1) 2 ^ n] = 0
Polynomial Conjecture. We assume that: there exist only finitely many positive integers so that:
[(q-1)2^0 +(q-1)2^1 + . . . + [(q-1)2^z] := \Delta{c} = 0,
For example:
23=[2^0+2^1+2^2+2^3+(2^3)], where; \Delta{a}:=2^3,
40=[2^0+2^1+2^2+2^3+2^4+(2^0+2^3)], where: \Delta{b}:=2^0+2^3,
63=[2^0+2^1+2^2+2^3+2^4+2^5+( 0 )], where: \Delta{c}:=0
Is this a theorem or a conjecture ?
What does this generally mean ?
Thank you sincerely!

HallsofIvy
Homework Helper
Has it been proved? If it has been proved to be true, then it is a theorem. If it has not be proved to be true (and has not been proved to be false) then it is a conjecture.

• jedishrfu and sysprog
fresh_42
Mentor
I have difficulties to understand you. E.g. you say: Let ##a,b,c## such that ... and then they do not occur anymore? They remain unexplained. Also you should use LaTeX for a better readability. Here is how it is done:
https://www.physicsforums.com/help/latexhelp/

• sysprog
Let (a, b, c) be some arbitrary positive integers such that:
$$a:= \sum_{n=0}^x q2^n , ~~~~~ b:= \sum_{n=0}^y q2^n , ~~~~~ c:= \sum_{n=0}^z q2^n,$$
where: $$q = (1, 2), ~~~(x, y, z) = (1, 2, 3, . . ., n).$$

For example:
$$a:=23=[2^0+2^1+2^2+2^3+(2^3)],$$ where: $$\Delta{a}:=2^3,$$
$$b:=40=[2^0+2^1+2^2+2^3+2^4+(2^0+2^3)],$$ where: $$\Delta{b}:=2^0+2^3,$$
$$c:=63=[2^0+2^1+2^2+2^3+2^4+2^5+( 0 )],$$ where: $$\Delta{c}:=0.$$
One more example:
$$d:=31=[2^0+2^1+2^2+2^3+2^4],$$ where: $$\Delta{d}:=0.$$

Sincerely thank

fresh_42
Mentor
If you allow ##q \in \{\,0,1\,\}## instead, which makes no significant difference, then you simply have the binary representation of natural numbers.

##q \in \{\,1,2\,\}## and ##\Delta## only confuses this otherwise simple fact by the application of a shift operator and an artificially shortened representation by introducing ##\Delta##.

• sysprog
Then maybe like this:
$$[a]:= \sum_{n=0}^x (q-1) 2^n :=2^3$$
Although I still need to think.
Sincerely thank

Let (x, y) be some arbitrary positive integers such that:
$$x:=\sum_{n=0}^kq2^n, ~~~ y:=\sum_{n=0}^{k+1}q2^n, ~~~\text{where: q = (1, 2), k = (0, 1, 2, 3, . . ., n).}$$
For example:
$$x:=[2^0+2^1+2^2+2^3+(2^3)]= 23, ~~~\text{where:} ~~~\sum_{n=0}^x(q-1)2^n = 2^3,$$
$$y:=[2^0+2^1+2^2+2^3+2^4+(2^0+2^3)]=40, ~~~\text{where:}~~~ \sum_{n=0}^y(q-1)2^n =2^0+2^3.$$
In the case if and only if q = 2, we accept the following notation:
$$[x]:=\sum_{n=0}^k(q-1)2^n,~~~[y]:=\sum_{n=0}^{k+1}(q-1)2^n$$
(As a disjunctive proposition.) We say that for any arbitrarily large positive integer X, there exists a positive integer Y greater than X whenever for both the following condition is fulfilled so that:
$$[x]:=\sum_{n=0}^k(q-1)2^n=0, ~~~[y]:=\sum_{n=0}^{k+1}(q-1)2^n=0.$$
That is:
$$\forall x\in\mathbb{N}\setminus{0}\,\exists \,y>x\,|\,y\in\mathbb{N}\setminus{0}\,:\,y=\sum_{n=0}^{k+1}(q-1)2^n = 0.$$
For example:
$$x:=[2^0+2^1+2^2+2^3+2^4]=31, ~~~\text{where:} \sum_{n=0}^k(q-1)2^n =0,$$
$$y:=[2^0+2^1+2^2+2^3+2^4+2^5]=63, ~~~\text{where:} \sum_{n=0}^{k+1}(q-1)2^n = 0.$$

Mark44
Mentor
Let (x, y) be some arbitrary positive integers such that:
$$x:=\sum_{n=0}^kq2^n, ~~~ y:=\sum_{n=0}^{k+1}q2^n, ~~~\text{where: q = (1, 2), k = (0, 1, 2, 3, . . ., n).}$$
This seems unnecessarily complicated. The series for y is the same as the series for x, but with one additional term, ##k2^{n + 1}##.
velikh said:
For example:
$$x:=[2^0+2^1+2^2+2^3+(2^3)]= 23, ~~~\text{where:} ~~~\sum_{n=0}^x(q-1)2^n = 2^3,$$
$$y:=[2^0+2^1+2^2+2^3+2^4+(2^0+2^3)]=40, ~~~\text{where:}~~~ \sum_{n=0}^y(q-1)2^n =2^0+2^3.$$
I'm not following your examples at all.
Based on your definition for x, if q = 1 and k = 3, then ##x = \sum_{n=0}^kq2^n = 2^0 + 2^1 + 2^2 + 2^3 = 15##.
Using your definition for y, again with q = 1 and k = 3, then ##y = \sum_{n=0}^{k + 1}q2^n = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 = 31##. Clearly we get the same numbers plus the last term ##2^4##.
velikh said:
In the case if and only if q = 2, we accept the following notation:
$$[x]:=\sum_{n=0}^k(q-1)2^n,~~~[y]:=\sum_{n=0}^{k+1}(q-1)2^n$$
Why? What is the significance of brackets around x and y?
velikh said:
(As a disjunctive proposition.) We say that for any arbitrarily large positive integer X, there exists a positive integer Y greater than X whenever for both the following condition is fulfilled so that:
$$[x]:=\sum_{n=0}^k(q-1)2^n=0, ~~~[y]:=\sum_{n=0}^{k+1}(q-1)2^n=0.$$
This makes no sense to me. If q = 1, then q - 1 = 0, and you have two finite sums all of whose terms are 0. Of course both sums are zero. If q = 2, then you have two finite sums whose terms are nonnegative powers of 2, so neither sum can be zero.

I don't get what you're trying to do here.
velikh said:
That is:
$$\forall x\in\mathbb{N}\setminus{0}\,\exists \,y>x\,|\,y\in\mathbb{N}\setminus{0}\,:\,y=\sum_{n=0}^{k+1}(q-1)2^n = 0.$$
For example:
$$x:=[2^0+2^1+2^2+2^3+2^4]=31, ~~~\text{where:} \sum_{n=0}^k(q-1)2^n =0,$$
$$y:=[2^0+2^1+2^2+2^3+2^4+2^5]=63, ~~~\text{where:} \sum_{n=0}^{k+1}(q-1)2^n = 0.$$
[/quote]

• sysprog
Thank you very much for your participation and the questions asked. I will try, as far as possible, to explain some perhaps controversial definitions.

The fact is that ##x## and ##y## are not power series. However, ##x## and ##y## are positive integers that can be considered as partial sums of the same power series.

By the way, some of these partial amounts include recurring members (in this case ##q = 1, 2##). At the same time, some of them include each member of the sequence ##q2 ^ n## that corresponds to a certain value of n only once (in this case ##q = 1##).

Additional terms of the power series for which ##q = 2## are called a "number growth" and are denoted by ##[x]## or ##[y]##.

So, such partial sums that contain ##2 ^ n## only once (for a certain value of ##n##) are rarely meet. The essence of the question is how long can this last? In other words, how often such ##x## or ##y## for which ##q = 1## occur in relation to others for which ##q = (1, 2)## ?

"Whether there exist a limit of a power series ?" should read as "Whether there exist a limit on the number of partial sums of power series (for which ##q = 1, 2##) ?"

fresh_42
Mentor
The point is, that you can easily convert your system from ##q_i\in \{\,1,2\,\}## to ##\tilde{q}_i\in \{\,0,1\,\}## which would result in a binary representation of integers. Neither of the above is a power series, they are all finite sums. If you converted the representation into binary, you can still do whatever you want with the highest powers. What you did is simply ##x=b_0\ldots b_n = 111\ldots 1 + R## where the ##b_i## are the binaries of ##x## and then talk about ##R##.

• sysprog and velikh
Before that I did not work with binary system of calculus. Nevertheless, if I understand correctly, namely:
$$63 = 111111$$
$$43 = 101011 = 31 + 12 = 11111 +  := N + [x]$$
$$31 = 11111$$
$$20 = 10100 = 15 + 5 = 1111 +  := N + [x]$$
However, I do not quite understand how it is possible to describe in binary representation the condition that:
$$\sum_{n=0}^k(q-1)2^n = \sum_{n=0}^k(1-1)2^n + \sum_{n=0}^k(2-1)2^n = 0 + \sum_{n=0}^k2^n~~~ \text{whenever :}~~~ q = (1, 2).$$
I suppose, however, I doubt that
$$\text{if}~~~\tilde{q_i}={0\wedge 1}, \text{then}~~~ q_i = 0,$$
$$\text{else if} ~~~\tilde{q_i}={0 \vee 1}, \text{then}~~~ q_i = 1.$$

BvU
Homework Helper
2019 Award
What does the subscript i for the q stand for ?

$$x=b_0b_1\dots b_{i-1}b_i=1110101\dots 01$$

Mark44
Mentor
Before that I did not work with binary system of calculus. Nevertheless, if I understand correctly, namely:
$$63 = 111111$$
$$43 = 101011 = 31 + 12 = 11111 +  := N + [x]$$
$$31 = 11111$$
$$20 = 10100 = 15 + 5 = 1111 +  := N + [x]$$
Most of the above is correct, but it would be better to identify binary numbers using a subscript so that they aren't misinterpreted as decimal (base-10) number.

I.e., like this: ##63 = 111111_2##
The parts I don't understand are ##N + [x]## that you used in two of the equations above. What is this supposed to mean?
velikh said:
However, I do not quite understand how it is possible to describe in binary representation the condition that:
$$\sum_{n=0}^k(q-1)2^n = \sum_{n=0}^k(1-1)2^n + \sum_{n=0}^k(2-1)2^n = 0 + \sum_{n=0}^k2^n~~~ \text{whenever :}~~~ q = (1, 2).$$
I suppose, however, I doubt that
$$\text{if}~~~\tilde{q_i}={0\wedge 1}, \text{then}~~~ q_i = 0,$$
$$\text{else if} ~~~\tilde{q_i}={0 \vee 1}, \text{then}~~~ q_i = 1.$$
Again, this seems like an unnecessarily complicated way to write something that is fairly simple.
If q = 2, then ##\sum_{n=0}^k(q-1)2^n = \sum_{n=0}^k2^n##
If q = 1, then ##\sum_{n=0}^k(q-1)2^n = 0##

It is not true to say that
##\sum_{n=0}^k(q-1)2^n = \sum_{n=0}^k(1-1)2^n + \sum_{n=0}^k(2-1)2^n = 0 + \sum_{n=0}^k2^n~~~
\text{whenever :}~~~ q = (1, 2)##

We are currently at 14 posts, and it is still not clear to me what you are trying to say or trying to do.

• sysprog
Mark44
Mentor
Before that I did not work with binary system of calculus.
This really has nothing to do with calculus (meaning differential calculus or integral calculus).

The binary representation of integers and even floating point numbers is very well-known in computer science. Any positive integer can be represented in binary as a finite sum of powers of 2; i.e., as $$\sum_{k = 0}^n a_k2^k$$, where ##a_k## is either 0 or 1.
For example, the binary representation of 23 is ##1\cdot 2^4 + 0 \cdot 2^3 + 1 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0##, or more compactly, as ##10111_2##.

Negative numbers can also be represented in binary form, using what is called twos-complement notation.

Floating point numbers can also be represented, with the two types most commonly used being 32-bit single-precision (float) or 64-bit double-precision (double).

@velikh, instead of trying to reinvent the wheel by coming up with your own representations for integers, do a search to see what is already there. A search using "binary representation of integers" brings up lots of hits.

Last edited:
As it is not difficult to notice, the first of these two sums ##N = b_ob_1b_2b_3. . . b_i## increases very slowly. While the second of these two sums ##[x] = R## alternately changes either increases or decreases.
Further, patterns of the change in the first sum ##N## are obvious and do not cause difficulties for understanding. However, on the contrary, changes in the second sum ##[x] = R## do not show any laws and seem arbitrary; in order to verify this, it suffices to trace the behavior of these two sums for the first ##100## positive integers.
But since this should not be so, then this is of interest. Whether there exist a pattern of the change in the second sum ##[x]## and what are these changes?
As a complement, there exists another way to describe positive integers in binary.
Let ##x## be some arbitrary positive integer such that:
$$x:=\sum_{n=0}^k\tilde{q_i}2^n,~~~\text{where:}~~\tilde{q_i}\in\{0, 1\},~~~(k, i)=(0, 1, 2, 3, \dots , n).$$
For example:
$$x=40=2^0+2^1+2^2+2^3+2^4+[2^0+2^3]= b_0b_1b_2b_3b_4b_5+[b_0b_1b_2b_3],$$
$$x:=\sum_{n=0}^k\tilde{q_i}2^n | \tilde{q_i}\in\{1\} + \sum_{n=0}^k\tilde{q_i}2^n | \tilde{q_i}\in\{0, 1\}:=N+R,$$
$$\textit{where:}~~~(k, i) = (0, 1, 2, 3, . . ., n).$$
$$x=11111_2 + 1001_2=9_{10}$$
We assume that do not exist two equal sums such that: $$R= \sum_{n=0}^k\tilde{q_i}2^n | \tilde{q_i}\in\{0, 1\}.$$

Mark44
Mentor
Since you haven't made clear what it is you're trying to do by this point, I am closing this thread. Possibly this is due to inexperience in communicating in English.