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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
2 months, 1 week ago

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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?

Patrick Corn - 3 weeks, 2 days ago

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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.

Halvor Bratland - 3 weeks, 2 days ago

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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.

Patrick Corn - 3 weeks, 2 days ago

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@Patrick Corn 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.

Halvor Bratland - 3 weeks, 2 days ago

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