# Operations (not a puzzle)

Acolyte
Posers and Puzzles 04 Jun '04 08:46
1. Acolyte
04 Jun '04 08:462 edits
I just thought of a set of operations corresponding to the integers which go as follows:

(i) a&lt;0&gt;b = a+b
(ii) a&lt;n+1&gt;b = exp(log(a)&lt;n&gt;log(b))

If you 'reverse' (ii) you get
(iii) a&lt;n-1&gt;b = log(e^a&lt;n&gt;e^b), allowing one to define &lt;-1&gt; etc.

&lt;0&gt; and &lt;1&gt; correspond in some sense to + and x, and &lt;n&gt; looks like it's associative and commutative. Unfortunately, I can't think of a nice set on which I can make &lt;n&gt; well-defined and closed for n other than 0. Does anyone have any ideas?

Edit: I think I can see how to make &lt;2&gt; well-defined over R\{0,1}, so maybe this idea's going somewhere after all. Can anyone tell me whether &lt;n+1&gt; distributes over &lt;n&gt; in general? At the moment I'm getting an identity element for &lt;n&gt; if n is at least 0, but inverses are hard to come by. Can anyone see how I could get inverses (and an identity for n&lt;0), and make &lt;n&gt; a group operation?
2. psychopath42
Green Slime
20 Jun '04 18:22
Originally posted by Acolyte

(i) a<0>b = a+b
(ii) a<n+1>b = exp(log(a)<n>log(b))
(iii) a<n-1>b = log(e^a<n>e^b)
let's let c_n = a&lt;n&gt;b, that is, c_n is the &quot;sum&quot; applied to a and b.

for convenience, let log^n (.) = log log ... log (.)
and let exp^(n) (.) = exp exp ... exp (.)
and log^0 (.) = (.)

the operation &lt;n&gt; is equivalent to
log^n c = log^n a + log^n b

(or for &lt;-n&gt;: exp^n c = exp^n a + exp^n b)

and hence we can deduce that the zero element is exp^n (0) for non-negative n. for &lt;-1&gt; i think the &quot;zero element&quot; is negative infinity, but for anything less than that, no zero element exists, methinks..
3. Acolyte
22 Jun '04 09:46
Originally posted by psychopath42
let's let c_n = a<n>b, that is, c_n is the "sum" applied to a and b.

for convenience, let log^n (.) = log log ... log (.)
and let exp^(n) (.) = exp exp ... exp (.)
and log^0 (.) = (.)

the operation <n> is equivalent to
log^n c = log^n a + log^n b

(or for <-n>: exp^n c = exp^n a + exp^n b)

and hence we can deduce that the zero element ...[text shortened]... nt" is negative infinity, but for anything less than that, no zero element exists, methinks..
That sounds about right, unless there's some trick involving complex logs for n &lt; -1. The next question is, can I turn &lt;n&gt; into a group operation for n at least -1?

As for &lt;n+1&gt; distributing over &lt;n&gt;:

a&lt;n+1&gt;(b&lt;n&gt;c) = exp(log(a)&lt;n&gt;log(b&lt;n&gt;c)) = exp(log(a)&lt;n&gt;(log(b)&lt;n-1&gt;log(c)))
= exp((log(a)&lt;n&gt;log(b))&lt;n-1&gt;(log(a)&lt;n&gt;log(c))) provided &lt;n&gt; distributes over &lt;n-1&gt;
= exp((log(a)&lt;n&gt;log(b))&lt;n&gt;exp(log(a)&lt;n&gt;log(c))
= (a&lt;n+1&gt;b)&lt;n&gt;(a&lt;n+1&gt;c)

&lt;1&gt; distributes over &lt;0&gt;, so &lt;n+1&gt; distributes over &lt;n&gt; for non-negative n. I think a similar argument applies for negative n.

So if I can get &lt;n&gt; and &lt;n+1&gt; to be group operations over the same set, we'll have a sequence of fields!
4. Acolyte
22 Jun '04 10:10
The inverse of a under &lt;n&gt;, n non-negative, if it exists, is b = exp^n(-1)&lt;n+1&gt;a:

b = exp^n(-1)&lt;n+1&gt;a = exp^n(-1&lt;1&gt;log^n(a)) = exp^n(-log^n(a))
=&gt; log^n(a) + log^n(b) = 0

and similarly (I think) for negative n. That just leaves closure and uniqueness of inverses.

exp and log are closed on the extended complex plane. But are the inverses unique there?
5. psychopath42
Green Slime
23 Jun '04 06:13
i don't think any inverse exists if n is negative: if there is no zero element, there definitely can't be an inverse element.

i think ultimately the whole procedure depends on the domains; if we restrict ourselves to the reals, then the problem hinges on finding the domain for a given &lt;n&gt;. But the exponential function maps R -&gt; R++, and the logarithm function maps R++ -&gt; R.

Since the calculation involved in &lt;n&gt; implies multiple application of logs and exps in succession, we have to make sure that we don't fall out of the domain.

For &lt;0&gt;, we know that the elements can be in R.

For &lt;1&gt;, even though a&lt;1&gt;b simplifies to ab, by the construction of &lt;1&gt; on &lt;0&gt;, the elements can only be in R++ (since we are taking logs of a and b).

For &lt;2&gt;, we'd require a and b to be numbers strictly greater than 1. Then log^2 (a) and log^2 (b) can be defined. I don't quite think the domain can be R\{0,1}, as you said...?

For &lt;3&gt;, we'd need a and b to be numbers strictly greater than exp(1), and so on.

Interestingly enough, for the negative n's, a and b can be elements of R, and a&lt;n&gt;b will be well defined.

at this point (if anyone's read this far), i'd like to point out that if we are dealing with real numbers, exp is a one-to-one function, and hence has an inverse, which is defined as log.

exp is not a one-to-one function if we are dealing with complex numbers, for example, exp(i*2pi) = 1, and exp(0) = 1, assuming exp is to the base e. the inverse logarithm &quot;function&quot; hence isn't a function, since it has multiple values... however, it does have a principal value... it can be assigned a value in a subset of the complex plane. (i *think* it's the infinite band that contains complex numbers of the form x+iy where x is in R and -pi &lt; y &lt;= pi).

i have no idea where i'm going in this post, so i think i'll end it here. ðŸ˜€