Extended xor to Tetrion space

[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?