# A More Symmetric Exponentiation

Exponentiation is distributive over multiplication, but it isn't commutative or associative like addition and multiplication are.

Is there a binary operation that is distributive over multiplication, and also commutative and/or associative?

In order to find one such operation, I assumed that there is an identity element. An easier question than the one above is: Is the identity element 0, 1, or neither?

If you find an operation that works, can you then find a commutative and/or associative operation that is distributive over that?

Note by Halvor Bratland
10 months, 2 weeks ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Well, here's one: $a*b = \begin{cases} 1&\text{if } ab \ne 0 \\ 0&\text{if } ab = 0 \end{cases}.$ Is that the kind of thing you had in mind?

- 8 months, 4 weeks ago

Yeah, that definitely works. And it’s pretty easy to find a similar operation that’s distributive over that one. The one I had in mind was a^(log_sqrt(2)(b)), which gives a wider range of outputs, but rarely integer ones. Yours also works better with negative numbers.

- 8 months, 4 weeks ago

Ah ok. If I change your sqrt(2) to an e, I get something symmetric-looking like $$e^{\ln(a)\ln(b)},$$ which is pretty nice. But it doesn't work on negative numbers. I guess maybe if you put absolute values on the $$a$$ and $$b$$? Does that work? And you can probably even fill it in at 0 by setting $$0 * b = 0.$$ I haven't checked all the details.

- 8 months, 4 weeks ago

I chose the square root of two because it continues the pattern of 2+2 = (2)(2) = 4, and the tangents at that point increasing by a factor of 2.

- 8 months, 4 weeks ago