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

https://math.stackexchange.com/a/4436362/2513

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

$j^2=k^2=1$

$jk=j+k-1$

$j^j=j^k=j$

$k^k=k^j=k$

$\sqrt{j+k}=\frac{j}{\sqrt{2}}+\frac{k}{\sqrt{2}}$

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

$0^{j+k}=1-\frac{j}{2}-\frac{k}{2}$

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

$i^j=ij$

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

$i^{j+k}=1-j-k$

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

and

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

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ "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
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    $\begingroup$ 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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$\begingroup$

The answer was a duplicate of this answer. The general policy across SE is to delete duplicate answers.

DO NOT post the same answer to multiple questions. If one answer adequately addresses more than one question, answer only one of those questions, and nominate the other(s) for closure as duplicates. The goal here is to consolidate questions and answers, not to spread them around in a large number of disparate places.

Duplicate answers can—and often will—be deleted without further comment.

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