Extended xor to Tetrion space

[Xorter]

Hi, everyone!

I have been writing a paper about Cayley-Dickson algebra and its higher 2^n-ion spaces. By xor extensions (from binary xoring to ternary, tetranary, pentanary,... etc.) we get 3^n-ions (Trions), 4^n-ions (Tetrions), ... . For this operation it is needed to check the existance of triads (or tetrads, pentads, hexads,... and so on to (b+1)-ads) of the (b^n)-ions. For instance the Bions (2^n-ions) exist, because there are plenty of triads, like (1;2;3). Trions are not exist, because there are no tetrads anywhere. Tetrions (4^n-ions) also exist, because there are pentads like (10;27;51). And (2^m)^n-ions exist, too.

How do I calculate it? I have extended complex bitwise (binary)  operator xor to tritwise, tetratwise, pentatwise,... n-twise (n-ary) operator xor[n]. The n-ad is a set of numbers (x;y;z), iff: (x xor[n] y = z) and (z xor[n] x = y) and (y xor[n] z = x).
 Let us suppose that a xor[n] b = (a+b) n-twise mod n.

Question 1: How to proof there is no Trions or the tetrads are empty?

Question 2: How to make a monochrome 2-D slice solution map for Tetrions? (In pari/gp if it is possible.) I would like to see the solution map of equation 10z xor[4] 27z = 51z. I know kneser.gp has a complex graphical solution, but not applied for slice map.

Thank you!
Xorter Unizo

[11Keith22]

Let us suppose that a xor[n] b = (a+b) n-twise mod n.

What does the term "n-twise" mean? The terms would seem to be montwise, ditwise, tritwise, quitwise, etc., which I don't fully understand the meaning of in relation to the mod operator, which is always binary. I am probably missing something obvious here, but oh well.

[Xorter]

What does the term "n-twise" mean? The terms would seem to be montwise, ditwise, tritwise, quitwise, etc., which I don't fully understand the meaning of in relation to the mod operator, which is always binary. I am probably missing something obvious here, but oh well.

n-twise means n-ary digit-wise. For instance:

utwise digits can be only: 1. (This is an ut.)
bitwise digits can be: 0, 1. (These are bits.)
tritwise digits can be: 0, 1, 2. (These are trits.)
tetratwise digits can be: 0, 1, 2, 3. (These are tetrats.)
pentatwise digits can be: 0, ..., 4. (These are pentats.)
and so on...
(Somewhy I prefer to the expression u- to mono-. It means one. The expression a- means zero. But I cannot find atwise digits (ats) senseful.)

So if I talk about an n-twise operator, then it means I convert its arguments to n-ary number system and then do the operator.
Let us see how it works in code pari-gp:

Code:

mod(x,y)=x-y*floor(x/y);
bttest(x,n,b)=if(x==0,0,digits(floor(x*b^50),b)[n]);
blength(x,b)=if(x==0,1,floor(real(log(x)/log(b)))+1);

h=50;
xorter(x,y)=sum(k=1,h+blength(max(x,y),3),3^(h+blength(max(x,y),3)-k)*mod(if(x==0,0,iferr(digits(floor(x*3^h),3)[k],err,0))+if(y==0,0,iferr(digits(floor(y*3^h),3)[k],err,0)),3))/3.0^h;
xorb(x,y,b)=sum(k=1,h+blength(max(x,y),b),b^(h+blength(max(x,y),b)-k)*mod(if(x==0,0,iferr(digits(floor(x*b^h),b)[k],err,0))+if(y==0,0,iferr(digits(floor(y*b^h),b)[k],err,0)),b))/b^h*1.0;

You can try it, or just plot it. E. g. ploth(x=0,10,xorter(x,3*x)) returns with a nice fractal-function.
Hope it helped to understand. Be brave to ask me.

[11Keith22]

I have been writing a paper about Cayley-Dickson algebra and its higher 2^n-ion spaces. By xor extensions (from binary xoring to ternary, tetranary, pentanary,... etc.) we get 3^n-ions (Trions), 4^n-ions (Tetrions), ... . For this operation it is needed to check the existance of triads (or tetrads, pentads, hexads,... and so on to (b+1)-ads) of the (b^n)-ions. For instance the Bions (2^n-ions) exist, because there are plenty of triads, like (1;2;3). Trions are not exist, because there are no tetrads anywhere. Tetrions (4^n-ions) also exist, because there are pentads like (10;27;51). And (2^m)^n-ions exist, too.

How do I calculate it? I have extended complex bitwise (binary)  operator xor to tritwise, tetratwise, pentatwise,... n-twise (n-ary) operator xor[n]. The n-ad is a set of numbers (x;y;z), iff: (x xor[n] y = z) and (z xor[n] x = y) and (y xor[n] z = x).
 Let us suppose that a xor[n] b = (a+b) n-twise mod n.

Thanks for answering my question. I have been busy, but now I have time to ask another question. I still don't understand why (1;2;3) is a triad or why (10;27;51) is a pentad. Let me explain my thinking so you can correct me:
(1;2;3) = (1;2;10) in base 3. 1 xor[3] 2 = 0. 0 does not equal 10.
(10;27;51) = (20;102;201) in base 5. 20 xor[5] 102 = 122. 122 does not equal 201.

Thanks,

11Keith22.

[Xorter]

Thanks for answering my question. I have been busy, but now I have time to ask another question. I still don't understand why (1;2;3) is a triad or why (10;27;51) is a pentad. Let me explain my thinking so you can correct me:
(1;2;3) = (1;2;10) in base 3. 1 xor[3] 2 = 0. 0 does not equal 10.
(10;27;51) = (20;102;201) in base 5. 20 xor[5] 102 = 122. 122 does not equal 201.

Thanks,

11Keith22.

It is not enough to convert a triad (Bion) to another number system and call it another k-ad. It does not work. You have to calculate with totally other numbers with a similar method I coded. Okay, I admit I made some mistakes in my code. Here is the good code:

Code:

mod(x,y)=x-y*floor(x/y);
blength(x,b)=if(x==0,1,floor(real(log(x)/log(b)))+1);

h=50;
xorb(x,y,b)=sum(k=1,h+blength(max(x,y),b),b^(k-1)*mod(if(x==0,0,iferr(Vecrev(digits(floor(x*b^h),b))[k],err,0))+if(y==0,0,iferr(Vecrev(digits(floor(y*b^h),b))[k],err,0)),b))/b^h*1.0;

KSearch(k,N)={s=0;for(y=1,N,for(z=y+1,N,for(w=z+1,N,if(xorb(y,z,k)==w&&xorb(z,w,k)==y&&xorb(w,y,k)==z,s+=1;print(Str(s,".: ( ",y,"; ",z,"; ",w,")"))))));return(s)};

KSearch counts the k-ads in the interval to N. And I could not see Trions, but 2^n-ions. Why?

[11Keith22]

Thanks for answering my question. I have been busy, but now I have time to ask another question. I still don't understand why (1;2;3) is a triad or why (10;27;51) is a pentad. Let me explain my thinking so you can correct me:
(1;2;3) = (1;2;10) in base 3. 1 xor[3] 2 = 0. 0 does not equal 10.
(10;27;51) = (20;102;201) in base 5. 20 xor[5] 102 = 122. 122 does not equal 201.

Thanks,

11Keith22.

It is not enough to convert a triad (Bion) to another number system and call it another k-ad. It does not work. You have to calculate with totally other numbers with a similar method I coded. Okay, I admit I made some mistakes in my code. Here is the good code:

Code:

mod(x,y)=x-y*floor(x/y);
blength(x,b)=if(x==0,1,floor(real(log(x)/log(b)))+1);

h=50;
xorb(x,y,b)=sum(k=1,h+blength(max(x,y),b),b^(k-1)*mod(if(x==0,0,iferr(Vecrev(digits(floor(x*b^h),b))[k],err,0))+if(y==0,0,iferr(Vecrev(digits(floor(y*b^h),b))[k],err,0)),b))/b^h*1.0;

KSearch(k,N)={s=0;for(y=1,N,for(z=y+1,N,for(w=z+1,N,if(xorb(y,z,k)==w&&xorb(z,w,k)==y&&xorb(w,y,k)==z,s+=1;print(Str(s,".: ( ",y,"; ",z,"; ",w,")"))))));return(s)};

KSearch counts the k-ads in the interval to N. And I could not see Trions, but 2^n-ions. Why?

I have no experience with some of this stuff, but this happens to be in my area of expertise. Here is my analysis why only even bases work:
First, under base b, this is a b-ion: (b/2;b^2/2;(b^2+b)/2). In digits, where c=b/2, it is (c;c0;cc). c xor[b] c0 = cc, etc. Thus, all even bases have a b-ion. In fact, replace all the digits of bions with c and you get all 2c-ions! Why?
Consider one digit from each number in (x;y;z). We have (by trivial xor properties):
x+y=z
x+z=y
We don't even need the last equation, as we can now re-arrange to form
x=z-y
x=y-z
Thus, x=-x (mod b). Add x to both sides to get
2x=0 mod b
Thus, x=0 for odd b, i.e. Trions. This argument is symmetrical and applies to both y and z, so all digits of any Trions must be 0.
Therefore, the only trion, 5-ion, etc. is (0;0;0) for non-negative integers.
Also, in even bases, b/2*2=0, thus the only digits allowed are b/2 and 0. This essentially means bions are the only interesting -ion. Also, given any two binary numbers a and b, (a;b;a xor b) is a bion, so bions (and thus other -ions) are very easy to construct.

Go ahead and make your sex-nions! (The humorous title of what would come next in base 6?)

[Xorter]

Hmm, very interesting. Somewhy I thought there are only 2^n-ions. But now I think there are only even-ions. For example according to your formula (k-ionic set of k-ads is {(x;y;z) | x xor[k] y = z and x xor[k] z = y} like (k/2; k^2/2; (k^2+k)/2)) and according to my codes I found not only 2^n-ionic but even-ionic -ads. For instance:
1st Bionic biad: (1;2;3)
1st Tetrionic tetrad: (2;8;10)
1st Sexionic sexad: (3;18;21) (or Hexionic hexad)
1st Oktionic oktad: (4;32;36)
1st Decionic decad: (5;50;55)
But how to proof there are no odd-ads, like triads? Or I ask other way: Is there any number like that could create a triad? (Complex numbers or reals?)
Anyway xor works as addition at number system base infinity.

[11Keith22]

Thanks for answering my question. I have been busy, but now I have time to ask another question. I still don't understand why (1;2;3) is a triad or why (10;27;51) is a pentad. Let me explain my thinking so you can correct me:
(1;2;3) = (1;2;10) in base 3. 1 xor[3] 2 = 0. 0 does not equal 10.
(10;27;51) = (20;102;201) in base 5. 20 xor[5] 102 = 122. 122 does not equal 201.

Thanks,

11Keith22.

It is not enough to convert a triad (Bion) to another number system and call it another k-ad. It does not work. You have to calculate with totally other numbers with a similar method I coded. Okay, I admit I made some mistakes in my code. Here is the good code:

Code:

mod(x,y)=x-y*floor(x/y);
blength(x,b)=if(x==0,1,floor(real(log(x)/log(b)))+1);

h=50;
xorb(x,y,b)=sum(k=1,h+blength(max(x,y),b),b^(k-1)*mod(if(x==0,0,iferr(Vecrev(digits(floor(x*b^h),b))[k],err,0))+if(y==0,0,iferr(Vecrev(digits(floor(y*b^h),b))[k],err,0)),b))/b^h*1.0;

KSearch(k,N)={s=0;for(y=1,N,for(z=y+1,N,for(w=z+1,N,if(xorb(y,z,k)==w&&xorb(z,w,k)==y&&xorb(w,y,k)==z,s+=1;print(Str(s,".: ( ",y,"; ",z,"; ",w,")"))))));return(s)};

KSearch counts the k-ads in the interval to N. And I could not see Trions, but 2^n-ions. Why?

I have no experience with some of this stuff, but this happens to be in my area of expertise. Here is my analysis why only even bases work:
First, under base b, this is a b-ion: (b/2;b^2/2;(b^2+b)/2). In digits, where c=b/2, it is (c;c0;cc). c xor\[b\] c0 = cc, etc. Thus, all even bases have a b-ion. In fact, replace all the digits of bions with c and you get all 2c-ions! Why?
Consider one digit from each number in (x;y;z). We have (by trivial xor properties):
x+y=z
x+z=y
We don't even need the last equation, as we can now re-arrange to form
x=z-y
x=y-z
Thus, x=-x (mod b). Add x to both sides to get
2x=0 mod b
Thus, x=0 for odd b, i.e. Trions. This argument is symmetrical and applies to both y and z, so all digits of any Trions must be 0.
Therefore, the only trion, 5-ion, etc. is (0;0;0) for non-negative integers.
Also, in even bases, b/2*2=0, thus the only digits allowed are b/2 and 0. This essentially means bions are the only interesting -ion. Also, given any two binary numbers a and b, (a;b;a xor b) is a bion, so bions (and thus other -ions) are very easy to construct. (1)

Go ahead and make your sex-nions! (The humorous title of what would come next in base 6?)[/b]

Hmm, very interesting. Somewhy I thought there are only 2^n-ions. But now I think there are only even-ions. For example according to your formula (k-ionic set of k-ads is {(x;y;z) | x xor[k] y = z and x xor[k] z = y} like (k/2; k^2/2; (k^2+k)/2)) and according to my codes I found not only 2^n-ionic but even-ionic -ads. For instance:
1st Bionic biad: (1;2;3)
1st Tetrionic tetrad: (2;8;10)
1st Sexionic sexad: (3;18;21) (or Hexionic hexad)
1st Oktionic oktad: (4;32;36)
1st Decionic decad: (5;50;55)
But how to proof there are no odd-ads, like triads? Or I ask other way: Is there any number like that could create a triad? (Complex numbers or reals?)
Anyway xor works as addition at number system base infinity.

Therefore, the only trion, 5-ion, etc. is (0;0;0) for non-negative integers. (Above, maybe you were skimming or something.)
However, this does not address negative integers nor non-integers. I assume negative integers are not defined bitwise, however. Taking the 2's complement, however, does define negative integers. In binary, 1 xor -1 = 2**. Thus, for example, a negative triad would be (-1;1;2**) (1).

The 2's complement, by the way, is a way to define bitwise negatives. (If you already know this, then skip this paragraph.) Take the number -5, for example:
dec = bin
5 = 101
Not 5 = ...11010
(Not 5) + 1 = ...11011 = -5
Here is the intuition behind this:
0 = 0
Not 0 = ...111
(Not 0) + 1 = ...111 + 1 = 0 (wrap-around) = -0
Basically, 2's complement assumes ...111 + 1 = 0, which works if you assume 2^infinity = 0, which works in some contexts like this.

For non-integers, it is important to define what our current version of an integer is. Right now we are dealing with bitwise integers, so really, it might be more useful to think of individual digits as our fundamental number system rather than the numbers themselves. For example, (please don't assume I know group theory, I am just using it as a reference here), a cyclic group of order 2 (bits under xoring) has the property that x=-x is true for all (both) of its members. A cyclic group of order 2k+1 has it that x=-x iff x=0 and a cyclic group of order 2k has it that x=-x iff x=0 or x=k. All of these properties explain why, as stated previously, only even-ads exist (besides (0;0;0)) and they are all of only two digits in their respective bases. Only if a digit is its own inverse can it be in a 2k-ad. If you want to explore other number systems, I suggest considering the above.

What are you using this for, anyway? What spooky number systems can you create with sexads?

*EDIT: Rings are probabaly better for describing digits, not groups.

**Should be -2, I made a mistake. Flipped a sign. Classic, eh?

[Xorter]

Therefore, the only trion, 5-ion, etc. is (0;0;0) for non-negative integers. (Above, maybe you were skimming or something.)
However, this does not address negative integers nor non-integers. I assume negative integers are not defined bitwise, however. Taking the 2's complement, however, does define negative integers. In binary, 1 xor -1 = 2. Thus, for example, a negative triad would be (-1;1;2) (1).

The 2's complement, by the way, is a way to define bitwise negatives. (If you already know this, then skip this paragraph.) Take the number -5, for example:
dec = bin
5 = 101
Not 5 = ...11010
(Not 5) + 1 = ...11011 = -5
Here is the intuition behind this:
0 = 0
Not 0 = ...111
(Not 0) + 1 = ...111 + 1 = 0 (wrap-around) = -0
Basically, 2's complement assumes ...111 + 1 = 0, which works if you assume 2^infinity = 0, which works in some contexts like this.

For non-integers, it is important to define what our current version of an integer is. Right now we are dealing with bitwise integers, so really, it might be more useful to think of individual digits as our fundamental number system rather than the numbers themselves. For example, (please don't assume I know group theory, I am just using it as a reference here), a cyclic group of order 2 (bits under xoring) has the property that x=-x is true for all (both) of its members. A cyclic group of order 2k+1 has it that x=-x iff x=0 and a cyclic group of order 2k has it that x=-x iff x=0 or x=k. All of these properties explain why, as stated previously, only even-ads exist (besides (0;0;0)) and they are all of only two digits in their respective bases. Only if a digit is its own inverse can it be in a 2k-ad. If you want to explore other number systems, I suggest considering the above.

What are you using this for, anyway? What spooky number systems can you create with sexads?

*EDIT: Rings are probabaly better for describing digits, not groups.

Well, according to pari gp binary operator xor, 1 xor -1 = -2, however according to my primitive xor extension by simple addition and mod it returns 1 xor -1 = 0. But I guess this last result is wrong, because the original xored function does not look like extended xored function to other number systems at negatives. I suppose x xor y should be extended with more precisely than a simple (x+y) mod 2. We should complete this extension because of the negatives.
You mentioned a logical thesis at cyclic groups in where you had - I think - a preconception: not not x = x. I sign the operator negation with not. I guess you think of it or something like this at that passage. So, what if we construct numbers, or operated combination of numbers whose triple negation are itselves. So not not not x = x while not not x isn't x as usual, and this x will make a trionic triad whose 1st is: (x/2, x^2/2,(x+x^2)/2) or something like this. But I am sure that we should investigate higher dimensional numbers.

We know Bions, the set of Bions is {reals, complexes, quaternions, octonions, sedenions, and so on}. These are bionic because the number system base is binary.
What if the base is not binary? If it is tetranary, we get the Tetrions.
If it is hexanary, we get the Hexions or Sexions.
If it is octanary, we get the Oktions/Octions.
... etc.
These are infinite sets of higher dimensional number sets. The composition of imaginary units and the type of the multiplication make the difference. I must investigate them, because this is the key for the understanding of the Multiverse and its patterns. I have been writing a paper about it for weeks. And I need others' help like yours. So thank you for replying.

[Xorter]

Okay, now it works because of the complements:

Code:

xorb(x,y,b)={x=if(x<0,b^h+x,x);y=if(y<0,b^h+y,y);z=sum(k=1,h+blength(max(x,y),b),b^(k-1)*mod(if(x==0,0,iferr(Vecrev(digits(floor(x*b^h),b))[k],err,0))+if(y==0,0,iferr(Vecrev(digits(floor(y*b^h),b))[k],err,0)),b))/b^h*1.0;if(z>b^(h-1),return(z-b^h),return(z));};

I have tested on the complex interval [[-3;3];[-3i;3i]] but triads are not found.
But what about using natural logarithm of quaternionic base units, like i, j, k? Well, the logarithm makes multiplication from the addition, like this way:
log i + log j = log ij = log k
So, can (log i; log j; log k) be a triad?

(Let us notice that log i[x] not equals to i[x](pi/2 + 2npi) in this context.)