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I think this may be a very stupid question but here goes: I have a basic understanding of complex numbers and know that you can raise a number to a complex power, etc. But it seems to me that regular integers remain in a privileged position in mathematics in the following case: If we have an equation, we talk about the number of roots. The roots might be complex, the number of roots must always be an integer from zero to n. Is there any kind of math where having a complex or even simply a fractional number of roots is possible?

This idea of the number roots of an equation being complex is just one, possibly very stupid example, but I think it illustrates what I am thinking of. I would be totally happy with someone telling me this is a stupid question not worth pursuing.

Jeff
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    I think you mean natural numbers (${0,1,2,\ldots}$) not integers — it makes no sense for an equation to have $-1$ solutions. Beyond this, the natural numbers are chosen for (and named after) the real world because we don’t experience any other numbers in our (physical) day-to-day basis. Fundamentally its historical because numbers first came about in the context of counting things — in which instance only the natural numbers were needed, and other systems (rationals, integers etc) developed alongside as and when they were needed. – aidangallagher4 Jul 27 '18 at 10:32
  • Discrete items are related to the cardinality of sets: they are "counted". Of course, you might assign a non-integer weight to the items and discuss the sum of the weights, but this does not reduce the discrete characteristic of summation. –  Jul 27 '18 at 11:03
  • Not at all sure what it might mean for $x^{\frac{1}{2}}-2$, for example, to have half a root. – gandalf61 Jul 27 '18 at 11:06
  • @aidangallagher4: yes, I understand this. and for a long time we did not accept negative integers and then i as being "real." so I am not suggesting that such mathematics be developed but was wondering if it had been. There are approximate solutions to equations, so do we say that the equation has exactly one solution or some number of solutions? If the solution is less accurate, could it be a fractional solution? If it produces results that are very wrong, could it be a negative solution? – Jeff Jul 27 '18 at 11:16
  • @Jeff I’m not sure what you mean by the last bit — are you suggesting we develop some sort of description of the accuracy of solutions to equations? In general classifying the number of solutions to an equation is not too hard, but finding the exact solutions may be – aidangallagher4 Jul 27 '18 at 11:22
  • @aidangallagher4: I am saying that if we do classify the approximate solutions to an equation by accuracy where "1" means it is an exact solution and fractions or negative values describe inexact solutions, maybe it does make some sense to speak of an equation having "half" a solution or a negative solution. Not saying this is valuable/useful. – Jeff Jul 27 '18 at 11:36
  • @Jeff: I addressed this in my comment. You can rate the roots, but they remain discrete. Root $1$ being rated $0.93$ doesn't mean that there are $0.93$ roots. –  Jul 27 '18 at 11:55
  • @YvesDaoust: if there is a single approximate solution that we say by some metric is a .93, it could be argued it has not 1 but .93 roots.( I don't know if there are equations that have only one approximate solution. I would guess there would be many approximate solutions.) – Jeff Jul 27 '18 at 12:53
  • @jeff: in the same vein, you have the fuzzy sets. Which remain perfectly orthodoxical regarding cardinality. Nothing new, nothing special. –  Jul 27 '18 at 18:41

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I think a good example of what you are searching for is Fractional Calculus. You can find more complete info for this on Wikipedia, but the gist of it is this:

Differentiation is treated as an operator, denoted $D$; i.e $D(f)$ is the first derivative of $f$. The second derivative, $D(D(f))$, yields the composed operator denoted by $D^2$; similarly the $n$-th derivative is denoted $D^n$, which covers the positive integers. We stipulate $D^0$ to be the identity function. The application of the definite integral of $f$ from $0$ to $x$ yields the operator $D^{-1}$; iterating this yields all operators $D^{-n}$.

Now for the 'W.T.F?' part: There is a general expression for $D^{-n}$ as a single integral combining the function $f$ with other terms, one of which is the factorial of $n$. This factorial is the ONLY part of the expression that actually requires $n$ to be an integer. The move here is replacement of the factorial with an equivalent call to the Gamma function! This yields an expression for $D^{-r}$ for noninteger values of $r$! One then extends this to positive fractions $r$ by applying $D^m$ to $D^{r-m}$ for some positive integer $m$.

Mind blown? Mine was!

PMar
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