This my answer seems to be absolutely relevant to the question:


This looks like some personal attack.

The answer is below. If it is undeleted I would like to expand it considerably to include other examples.

Example of 3D numbers.

Take $\mathbb{R}^3$ with Hadamard product. In other words, triplets of numbers with element-wise multiplication.

Now assign $(1,1,1)=1,(-1,1,1)=j, (1,1,-1)=k$.

A number would be written in the form $a+bj+ck$. Algebraically it will be a commutative ring with zero divisors (hence, not a field, but that's OK). For instance $(j-1)(k-1)=0$.

Here is a Mathematica code to experiment with:

Unprotect[Power]; Power[0, 0] = 1; Protect[Power];
$Pre = (# /. {j -> {-1, 1, 1}, k -> {1, 1, -1}}) /. {x_, y_, z_} -> 
     x/2 + z/2 + (j (y - x))/2 + (k (y - z))/2 &;

Using this code you can see that






$\log (j+k+1)=\frac{1}{2} j \log (3)+\frac{1}{2} k \log (3)$


A division formula would be: $\frac{a_1+b_1 j+c_1 k}{a_2+b_2 j+c_2 k}=\frac{j}{2} \left(\frac{a_1+b_1+c_1}{a_2+b_2+c_2}-\frac{a_1-b_1+c_1}{a_2-b_2+c_2}\right)+\frac{k}{2} \left(\frac{a_1+b_1+c_1}{a_2+b_2+c_2}-\frac{a_1+b_1-c_1}{a_2+b_2-c_2}\right)+\frac{a_1+b_1-c_1}{2 \left(a_2+b_2-c_2\right)}+\frac{a_1-b_1+c_1}{2 \left(a_2-b_2+c_2\right)}$

If you add a complex unity $i$, you will get a 6-dimensional number system.

Particularly, you will see that


$j^i=\frac{1}{2} \left(1-(-1)^i\right) j+\frac{(-1)^i}{2}+\frac{1}{2}$


$\log (j k)=i \pi-\frac{i \pi j}{2}-\frac{i \pi k}{2}$


$\sqrt{j}=\left(\frac{1}{2}-\frac{i}{2}\right) j+\left(\frac{1}{2}+\frac{i}{2}\right)$

  • 9
    $\begingroup$ Do you mean math.stackexchange.com/questions/200776/… maybe? Because if you can copy-paste the same answer on two separate questions, both of which are kinda old, and you actually do that, then you're using the site wrong. $\endgroup$
    – Asaf Karagila Mod
    Aug 5 at 10:24
  • 6
    $\begingroup$ This is besides the point, but, your answer leaves me feeling very unsatisfied. I can believe on a casual read that it is a comm ring with zero divisors perhaps. But then you talk of powers like $j^j$? Why is this defined? Why is it relevant? And also a division formula? What was that about zero divisors then..? And then the examples IMO unnecessarily uses a lot of vertical space. If you are writing for a reader other than yourself, perhaps try to be more cohesive $\endgroup$ Aug 5 at 10:31
  • 1
    $\begingroup$ Well, what are $j^j$ and $k^k$? $\endgroup$ Aug 5 at 12:15
  • $\begingroup$ @ArcticChar $j^j=j$, $k^k=k$ (this is in the post as you can see). $\endgroup$
    – Anixx
    Aug 5 at 12:21
  • $\begingroup$ @CalvinKhor yes, it as zero divisors. But this is beyond the point, as I wanted to expanded the answer wit 2 more examples, but found it was deleted 3 months ago. $\endgroup$
    – Anixx
    Aug 5 at 12:25
  • $\begingroup$ So you define $j^j$ to be $j$? $\endgroup$ Aug 5 at 12:26
  • $\begingroup$ @ArcticChar this uses hadamard product, so this is a consequence. $\endgroup$
    – Anixx
    Aug 5 at 12:28
  • $\begingroup$ @ArcticChar all functions act element-wise on the triplets. $\endgroup$
    – Anixx
    Aug 5 at 12:29
  • $\begingroup$ @ArcticChar $j^j=(-1,1,1)^{(-1,1,1)}=(-1^{-1},1^1,1^1)=(-1,1,1)=j$ $\endgroup$
    – Anixx
    Aug 5 at 12:31
  • 9
    $\begingroup$ "This looks like some personal attack." -- As a moderator of other places, I always roll my eyes at posts that assert this with no real reasoning; it comes off as assuming bad-faith from pretty much any authority, without a real reason to do so. The least you could do is ask about why the deletion occurred and wait for a response, before actually asserting some sort of abuse going on. $\endgroup$ Aug 5 at 22:32
  • 2
    $\begingroup$ it seems OP has ‘found a home’ for the deleted answer and the two new examples here. $\endgroup$ Aug 6 at 7:20

1 Answer 1


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