# Why this my upvoted answer has been deleted by moderator with no explanation?

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 $$\sqrt{j}=\left(\frac{1}{2}-\frac{i}{2}\right) j+\left(\frac{1}{2}+\frac{i}{2}\right)$$ • 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. – Asaf Karagila Mod Aug 5 at 10:24 • 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 Aug 5 at 10:31 • Well, what are$j^j$and$k^k$? Aug 5 at 12:15 • @ArcticChar$j^j=j$,$k^k=k$(this is in the post as you can see). Aug 5 at 12:21 • @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. Aug 5 at 12:25 • So you define$j^j$to be$j$? Aug 5 at 12:26 • @ArcticChar this uses hadamard product, so this is a consequence. Aug 5 at 12:28 • @ArcticChar all functions act element-wise on the triplets. Aug 5 at 12:29 • @ArcticChar$j^j=(-1,1,1)^{(-1,1,1)}=(-1^{-1},1^1,1^1)=(-1,1,1)=j\$ Aug 5 at 12:31
• "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. Aug 5 at 22:32
• it seems OP has ‘found a home’ for the deleted answer and the two new examples here. Aug 6 at 7:20