What is the significance of the structure of numbers?

[damo_clark]

I have an idea I'd like to share on this forum about the structure of numbers. I don't have a question as such but am interested to get some feedback on wheather my idea makes sense or if it's been addressed before, or maybe ways to extend on my thoughts.

I begin by defining the nth prime number as $$^{n}a$$. The first 8 prime numbers can be written as :

$$^1a = 2$$

$$^2a = 3$$

$$^3a =5$$

$$^4a =7$$

$$^5a =11$$

$$^6a =13$$

$$^7a =17$$

$$^8a =19$$

 

[damo_clark]

Now the first prime number 2, can be just expressed as "a". The superscript is ignored just as x is the same as x^1 .

2 = $$a$$

3 = $$^2a$$ but 2 = a, therefore $$3 = ^aa$$

$$4 = 2^2 = a^a$$

5 = $$^3a$$ but 3 = $$^aa$$ , therefore 5 = $$^{^aa}a$$

6 = 2 * 3 = $$a ^a \times a$$

7 = $$^4{a}$$ but 4 = $$a^{a}$$, therefore 7 =$$^{a^a}a$$

8 = $$2^3 = a^{^{a}a}$$

9 = $$3^2$$ = $$^{a}a^{a}$$

...

16 = $$2^4$$ = $$a^{a^a}$$

I could continue, but the point here, is that in the way I have defined it, the numbers have a tree like structure. Unlike a base system, every number can be represented by using a single symbol, in this case "a" . Numbers such as 3 and 4 are mirror reflections of each other, so too are 7 and 8, and 5 and 16 . The numbers 2 and 9 are mirror reflections of themselves. Another obsevation, is that the mirror reflection of 6 is 8.

My choice of notation is defining the nth prime number is rather arbitary I must admit, so it could be that that the tree like structure obseved here, doesn't represent any solid mathematical reality. That all I have to say for the time being, but might extend this idea in a later post.

 

[gendou2]

How do you represent 1, 0?

 

[jordi]

What you are using is the fact that any number can be represented as a product of primes. It always looked to me that primes are a "basis" of the natural numbers.

I have heard many times about the relationship between numbers and geometry. I have never studied it, but I guess many people did.

 

[damo_clark]

Initially I defined 1 as "a", ie.

$$^{0}a = 1$$
$$^{1}a = 2$$
$$^{2}a = 3$$

Therfore

1 = a

2 = $$2^1$$ = $$^{1}a^1$$ = $$^{a}a^{a}$$

3 = $$^{2}a^1$$ = $$^{^{a}a^{a}}a^{a}$$.

However, it became redundant. The above assumes that if there is not a exponent, or lefthand superscipt, it is 0. Normally, when we count we assume the exponent is 1, even if we don't explicitly write it, ie. when I write 2, I mean 2^1.
It makes it easier to assume the exponent and left hand superscripts are 1,if there is nothing there, as it is easier to write the numbers, in the tree like structure I explained. But strictly speaking, you can think as "a" to equal 1, as I have shown above, and if there is nothing in the exponent place then assume it is 0.

This still leaves the problem of defining zero, but for the time being I'm happy to think of zero as nothing, and leave the symbol for nothing as nothing.

 

[ModernLogic]

Its certainly a creative way to structure numbers. But as others pointed out, zero is not well-defined, not to mention negative numbers.

Also, I can't think of any interesting properties or applications to such a system.

 

[dodo]

Looks interesting, but I don't see that all naturals have a unique representation. Once you introduced the "cross" multiplication symbol, as in

$$6 = a^a \times a$$​

other numbers can be represented in multiple ways, such as

$$9 = {}^a a^a = {}^a a \times {}^a a$$​

Under what rule (on your scheme's terms) would you preferently group primes together? Of course, if the cross is not introduced, then many composite numbers have no representation.

On the other hand, if you forget the unique factorization of integers and restrict yourself to study the prime powers of primes, or (prime powers of primes) powers of primes, and so on... which numbers can be represented with a tree structure, i.e., without a cross?

 

[damo_clark]

Structure of numbers part 2

Thank-you for you points Dodo. I assume that the prime factorisation can only be expressed uniquely. For example the unique factorsation of 16 =2^4, not the other solutions shown below

eg $$2^4 = 2^3 2^1 = 2^2 2^2 = 2^1 2^1 2^2 = 2^1 2^1 2^1 2^1$$


Its probably not necesary to introdce the $$\times$$ to factorise 6, but it makes things clearer. I could instead write :

6 = $$a ^{a}a$$.

The factorisation of the first 100 numbers is shown below. I have used as "Short Form" to factorise the numbers. That is, to begin by defing 2 =a, and assume that the left hand superscipts and exponents are one. If 1=a, the tree diagrams are similar, but with an "a" super scipt or exponent on any "naked a",

ie. Short Form 3 = $$^{a}a$$
Long Form 3 = $$^{^{a}a^a}a^a$$




First the first 25 primes are:


$$^{1}a^1 = 2$$

$$^{2}a^1 = 3$$

$$^{3}a^1 = 5$$

$$^{4}a^1 = 7$$

$$^{5}a^1 = 11$$

$$^{6}a^1 = 13$$

$$^{7}a^1 = 17$$

$$^{8}a^1 = 19$$

$$^{9}a^1 = 23$$

$$^{10}a^1 = 29$$

$$^{11}a^1 = 31$$

$$^{12}a^1 = 37$$

$$^{13}a^1 = 41$$

$$^{14}a^1 = 43$$

$$^{15}a^1 = 47$$

$$^{16}a^1 = 53$$

$$^{17}a^1 = 59$$

$$^{18}a^1 = 61$$

$$^{19}a^1 = 67$$

$$^{20}a^1 = 71$$

$$^{21}a^1 = 73$$

$$^{22}a^1 = 79$$

$$^{23}a^1 = 83$$

$$^{24}a^1 = 89$$

$$^{25}a^1=97$$

The first 100 numbers with their corresponding tree represnations are

2 = a

3 = $$^{2}a = ^{a}a$$

4 = $$2^2 = a^a$$

5 = $$^{3}a = ^{^{a}a}a$$

6 = $$2\cdot 3 = a \cdot ^{a}a$$

7 = $$^{4}a = ^{a^a}a$$

8 = $$2^3 = a^{^{a}a}$$

9 = $$3^2 = ^{a}a^a$$

10 = $$2\cdot 5 = a \cdot ^{^{a}a}a$$

11 = $$^{5}a = ^{^{^{a}a}a }a$$

12 = $$3 \cdot 2^2 = ^{a}a \cdot a^a$$

13 = $$^{6}a = ^{a \cdot ^{a}a}a$$

14 = $$2 \cdot 7 = a \cdot ^{a^a}a$$

15 = $$3 \cdot 5 = ^{a}a \cdot ^{^{a}a}a$$

16 = $$2^4 = a^{a^a}$$

17 = $$^{7}a = ^{^{a^a}a }a$$

18 = $$2 \cdot 3^2 = a \cdot ^{a}a^a$$

19 = $$^{8}a = ^{a^{^{a}a} }a$$

20 = $$2^2 \cdot 5 = a^a \cdot ^{^{a}a}a$$

21 = $$3 \cdot 7 = ^{a}a \cdot \: ^{a^a}a$$

22 = $$2 \cdot 11 = a \cdot ^{^{^{a}a}a }a$$

23 = $$^{9}a = ^{^{a}a^a}a$$

24 = $$2^3 \cdot 3 = a^{^{a}a} \cdot \: ^{a}a$$

25 = $$5^2 = ^{^{a}a}a^a$$

26 = $$2 \cdot 13 = a \cdot ^{a \cdot \:^{a}a}a$$

27 = $$3^3 = ^{a}a^{^{a}a}$$

28 = $$2^2 \cdot 7 = a^a \cdot ^{a^a}a$$

29 = $$^{10}a = ^{a \cdot ^{^{a}a}a}a$$

30 = $$2 \cdot 3 \cdot 5 = a \cdot ^{a}a \cdot ^{^{a}a}a$$

31 = $$^{11}a = ^{^{^{^{a}a}a }a }a$$

32 = $$2^5 = a^{^{^{a}a}a }$$

33 = $$3 \cdot 11 = ^{a}a \cdot ^{^{^{a}a}a }a$$

34 = $$2 \cdot 17 = a \cdot ^{^{a^a}a }a$$

35 = $$5 \cdot 7 = ^{^{a}a}a \cdot ^{4}a = ^{a^a}a$$


36 = $$2^2 \cdot 3^2 = a^a \cdot ^{a}a^a$$


37 = $$^{12}a = ^{^{a}a \cdot a^a}a$$


38 = $$2 \cdot 19 = a \cdot ^{a^{^{a}a} }a$$


39 = $$3 \cdot 13 = ^{a}a \cdot ^{a \cdot ^{a}a}a$$


40 = $$2^3 \cdot 5 = a^{^{a}a} \cdot ^{^{a}a}a$$


41 = $$^{13}a = ^{^{a \cdot ^{a}a}a }a$$


42 = $$2 \cdot 3 \cdot 7 = a \cdot ^{a}a \cdot ^{a^a}a$$


43 = $$^{14}a = ^{^{a^a}a \cdot a }a$$


44 = $$2^2 \cdot 11 = a^a \cdot ^{^{^{a}a}a }a$$


45 = $$3^2 \cdot 5 = \: ^{a}a^a \cdot ^{^{a}a}a$$


46 = $$2 \cdot 23 = a \cdot ^{^{a}a^a}a$$


47 = $$^{15}a = ^{^{^{a}a}a \cdot ^{a}a }a$$


48 = $$2^4 \cdot 3 = a^{a^a} \cdot ^{a}a$$


49 = $$7^2 =\: ^{a^a}a^a$$


50 = $$2 \cdot 5^2 = a \cdot ^{^{a}a}a^a$$


51 = $$3 \cdot 17 = \: ^{a}a \cdot ^{^{a^a}a }a$$


52 = $$2^2 \cdot 13 = a^a \cdot ^{a \cdot ^{a}a}a$$


53 = $$^{16}a = ^{a^{a^a} }a$$


54 = $$2 \cdot 3^3 = a \cdot \: ^{a}a^a$$


55 = $$5 \cdot 11 = ^{^{a}a}a \cdot ^{^{^{a}a}a }a$$


56 = $$2^3 \cdot 7 = a^{^{a}a} \cdot ^{a^a}a$$


57 = $$3 \cdot 19 = \: ^{a}a \cdot ^{a^{^{a}a} }a$$


58 =$$\: 2 \cdot 29 = a \cdot ^{a \cdot ^{^{a}a}a}a$$


59 = $$^{17}a = ^{^{^{a^a}a }a}a$$


60 = $$2^2 \cdot 3 \cdot 5 = a^a \cdot ^{a}a \cdot ^{^{a}a}a$$


61 = $$^{18}a = \:^{a \cdot ^{a}a^a}a$$


62 = $$2 \cdot 31 = a \cdot ^{^{^{^{a}a}a }a }a$$


63 = $$3^2 \cdot 7 = \: ^{a}a^a \cdot ^{a^a}a$$



64 =$$2^6 = a^{a \cdot ^{a}a }$$


65 = $$5 \cdot 13 = ^{^{a}a}a \cdot ^{a \cdot ^{a}a}a$$


66 = $$2 \cdot 3 \cdot 11 = a \cdot ^{a}a \cdot ^{^{^{a}a}a }a$$


67 = $$^{19}a = ^{^{a^{^{a}a} }a }a$$


68 = $$2^2 \cdot 17 = a^a \cdot ^{^{a^a}a }a$$


69 = $$3 \cdot 23 = ^{a}a \cdot ^{^{a}a^a}a$$


70 = $$2 \cdot 5 \cdot 7 = a \cdot ^{^{a}a}a \cdot ^{a^a}a$$


71 = $$^{20}a = ^{a^a \cdot ^{^{a}a}a}a$$


72 = $$2^3 \cdot 3^2 = a^{^{a}a} \cdot ^{a}a^a$$


73 = $$^{21}a = ^{^{a}a \cdot \: ^{a^a}a}a$$


74 = $$2 \cdot 37 = a \cdot ^{^{a}a \cdot a^a}a$$


75 = $$3 \cdot 5^2 = \:^{a}a \cdot ^{^{a}a}a^a$$


76 = $$2^2 \cdot 19 = a^a \cdot ^{a^{^{a}a} }a$$


77 = $$7 \cdot 11 = \:^{a^a}a \cdot ^{^{^{a}a}a }a$$


78 = $$2 \cdot 3 \cdot 13 = a \cdot ^{a}a \cdot ^{a \cdot ^{a}a}a$$


79 = $$^{22}a =\: ^{a \cdot ^{^{^{a}a}a }a}a$$


80 = $$2^4 \cdot 5 = a^{a^a} \cdot ^{^{a}a}a$$


81 = $$3^4 = \:^{a}a^{a^a}$$


82 = $$2 \cdot 41 = a \cdot ^{^{a \cdot ^{a}a}a }a$$


83 = $$^{23}a = ^{^{^{a}a^a}a }a$$


84 = $$2^2 \cdot 3 \cdot 7 = a^a \cdot ^{a}a \cdot ^{a^a}a$$


85 = $$5 \cdot 17 = ^{^{a}a}a \cdot ^{^{a^a}a }a$$


86 = $$2 \cdot 43 = a \cdot ^{^{a^a}a \cdot a }a$$


87 = $$3 \cdot 29 = \:^{a}a \cdot ^{a \cdot ^{^{a}a}a}a$$


88 = $$2^3 \cdot 11 = a^{^{a}a} \cdot ^{^{^{a}a}a }a$$


89 = $$^{24}a = ^{a^{^{a}a} \cdot \: ^{a}a }a$$


90 = $$2 \cdot 3^2 \cdot 5 = a \cdot \:^{a}a^a \cdot ^{^{a}a}a$$


91 = $$7 \cdot 13 = ^{a^a}a \cdot ^{a \cdot ^{a}a}a$$


92 = $$2^2 \cdot 23 = a^a \cdot ^{^{a}a^a}a$$


93 = $$3 \cdot 31 = ^{a}a \cdot ^{^{^{^{a}a}a }a }a$$


94 = $$2 \cdot 47 = a \cdot ^{^{^{a}a}a \cdot ^{a}a }a$$


95 = $$5 \cdot 19 = ^{^{a}a}a \cdot ^{a^{^{a}a} }a$$


96 = $$3 \cdot 2^5 = \:^{a}a \cdot a^{^{^{a}a}a }$$


97 = $$^{25}a = ^{^{^{a}a}a^a }a$$


99 = $$2 \cdot 7^2 = a \cdot \: ^{a^a}a^a$$


100 = $$2^2 \cdot 5^2 = a^a \cdot ^{^{a}a}a^a$$


The attachment of the tree diagrams for the first 50 numbers is attached. It expresses for numbers in Long Form.

What I am interested in doing is finding the mirror reflections of the numbers. I think this may be an interesting property of these numbers. If I define a function $$\theta (n)$$, to be a function that returns the mirror reflection of a number, n.

eg
$$\theta (3) = 4$$
$$\theta (4) = 3$$
$$\theta (7) = 8$$
$$\theta (8) = 7$$

The inverse function

$$\theta ^{-1} (3) = 4$$
$$\theta ^{-1} (4) = 3$$

The property that I think may or may not be interesting is that the inverse of the mirror function can have more than one solution.

$$\theta ^{-1} (8) = 6 , 7$$

This is because $$\theta (6) = \theta ( ^{a}a \cdot a) = a^a \cdot a = 4 \cdot 2 = 8$$

Also $$\theta^{-1} (2^8) = \hspace{5}^7a,\hspace{20} ^{6}a, \hspace{20} ^{8}a \cdot a, \hspace{20} ^{16}a \cdot ^{4}a, \hspace{20} ^{1}a \cdot ^{2}a \cdot ^{16}a, \hspace{20} ^{1}a \cdot ^{3}a \cdot ^{4}a$$


note that $$\theta(7) = 8$$
$$\theta(6) = 8$$
$$\theta(8) +\theta(1) =8$$
$$\theta(16) + \theta(4) = 8$$
$$\theta(1) +\theta(2) + \theta(16) = 8$$
$$\theta (1) + \theta(3) +\theta(4) = 8$$



In this system if I was asked to find the google'th prime it would be easy. I would write

the googlth prime = $$^{10^{100}}a\hspace{20} = \hspace{20} ^{a^{\hspace{2}a^{a}\hspace{2}\cdot\hspace{2} ^{^{a}a}a^{a} } \hspace{5}\cdot\hspace{5} ^{^{a}a}a^{\hspace{2}a^{a} \hspace{2}\cdot\hspace{2} ^{^{a}a}a^{a} }}a$$

However that it not a very useful result, it is like saying the googlth prime is "the googlth prime" . What would be more interesting is to find the googlth prime as a sum of smaller numbers.

If you could find a couple of solutions to $$\theta^{-1} (2^{googlth\:prime})$$ , say $$^{N}a \hspace{5}\cdot \hspace{5} ^{M}a$$ ,then $$\theta(N) +\theta(M)$$ = googlth prime. If N and M could also be expressed as decimal numbers, then you would have a decimal representation of the googlth prime. The problem would be to find the inverse solutions, a problem which could be harder than finding the google prime by brute force.

I can establish some propeties of $$\theta(n)$$. For example

$$\theta(A \cdot B) = \theta(A) \cdot \theta(B)$$

if gcd(A,B)=1

Which bring me to a question, (sorry if my posting has been so long and mostly explanatory up until this point)

Knowing this property of $$\theta(n)$$ does this give me any information about it inverse function?

 

[DeaconJohn]

I have an idea I'd like to share on this forum about the structure of numbers. I don't have a question as such but am interested to get some feedback on wheather my idea makes sense or if it's been addressed before, or maybe ways to extend on my thoughts.

I begin by defining the nth prime number as $$^{n}a$$. The first 8 prime numbers can be written as :

$$^1a = 2$$

$$^2a = 3$$

$$^3a =5$$...

Nope, never seen it before. But, it brings up and interesting question. Namely, exactly what is it that makes a mathematical result interesting?

The usual answer from an operational viewpoint is the demonstration of a relationship between the result in question and previous results.

Of course, that's not a complete answer. I suppose that "all mathematicians worth their salt" would agree that the prime numbers are inherently interesting simply because they arise so naturally from one of the most the basic structure of mathematics, the natural numbers.

However, this is definitely a potental difficulty with appreciation of the results of your study by the mathematical community at large.

A mathmatical result (like 3x7x11 = 231) that is not interesting to the mathmatical community will not be remembered for long. At least, it will not be remembered for its mathmatical value.

On the other hand, this is certainly a gray area.

Consider the "rep units," for example. Many mathematicians would not consider that they are inteeresting in themselves because their very definition is based on the decimal system, and our use of the decimal system seems to be more an accident of history than anything else.

On the other hand, all "rep units" with a prime number of ones are prime.

For example, 1111111, that has seven ones, is prime. (Please correct me if I'm wrong on this.)

This follows from the fact that the second cyclotomic polynomial (p2(x)) obtained factoring x^p-1 where p is prime is irreducible (over the rationals, Q). (x^p-1 = (x-1)p2(x).)

p2(x) = x^p + x^(p-1) + ... + 1.

Now suppose that the first time somebody figured out that p2(x) is irreducible they figured it out from the fact that the rep units (of prime length) are prime.

Then the "rep units" would become very interesting.

(This example is somewhat of a "cheat" because rep units of prime length are prime in any base.)

DJ

 

[gel]

For example, 1111111, that has seven ones, is prime. (Please correct me if I'm wrong on this.)

1111111 = 239 × 4649

well, you did ask me to correct you :)

 

[DeaconJohn]

1111111 = 239 × 4649

well, you did ask me to correct you :)

Gel,

Thank you so much. I was seriously confused on that point. Once again you came to my rescue. (The first time was a few days ago when I was stymied by the definition of ramification in algebraic number theory.)

In case anybody else is interested: Wikipedia reports that the cyclotomic polynomial argument (that I had mis-understood in my post above this one) tells us that the only prime rep units are the ones with a prime number of digits. It does not say (and, as Gel points out, it is not true) that the rep units with a prime number of digits are prime.

http://en.wikipedia.org/wiki/Repunit#Repunit_primes

Then converse to this fact fails very badly. The number of digits in the first five rep unit primes are 2, 19, 23, 317, and 1031, and the rep unit primes spread thinner and thinner as "n" gets bigger and bigger. It is conjectured that there are an infinite number of rep unit primes.

DJ

 

[dodo]

Getting back to trees, perhaps it would be easier to switch to the following notation:
1: 1
2: a
3: (a, 1)
4: (1, a)
5: ((a, 1), 1)
6: a (a, 1)
7: ((1, a), 1)
8: (1, (a, 1))
9: (a, a)
10: a ((a, 1), 1)

Note that, whenever you have branches, the trunk is always "a", so it is not mentioned here.

Also, such 'numbers' should be treated as sets (sets of trees, or 'forests'), since the order of the trees is irrelevant: in composite numbers like 6, a (a, 1) = (a, 1) a. in other words, the above was shorthand for
1: {1}
2: {a}
3: { (a, 1) }
...
6: { a, (a, 1) }
...
and the juxtaposition was shorthand for set union.

Then, rather than saying that trees (or forests) are numbers (which is ambiguous and leads to functions having more than one result), perhaps you should define the value of a tree, as a function v(T), maybe something like: v(1) = 1, v(a) = 2, v( (x, y) ) = (prime sub v(x)) ^ v(y), and v ( x y ) = v(x) . v(y). Thus several trees can have the same value.

The mirror function can also be easily expressed, as theta(1) = 1, theta(a) = a, theta( (x, y) ) = (y, x), and theta( x y ) = theta(x) theta(y). Now the inverse function is the mirror function itself; what you would study is how the values are changed by the mirroring.

 

[DeaconJohn]

Nope, never seen it before. But, it brings up and interesting question. Namely, exactly what is it that makes a mathematical result interesting?

The usual answer from an operational viewpoint is the demonstration of a relationship between the result in question and previous results.

Damo_Clark,

Of course, trees! Why didn't I see it before? Or, forests. There's your relation to previous work. My impression is that computer scientists have done a lot of work on them. There IS a significant connection to previous work. :).

DJ