Just an idea, please explain if this could happen.

[mardlamock]

Hello everyone, I was just thinking about the possible solutions to a certain function with or without complex numbers and I just wanted to know if there could be any way to express a function that between any two values has as many solutions as impossible solutions. And that if for example i was to take another two random values within the original ones i would also have equal number of solutions and impossible solutions (i really dont know what is the opossite of solution in math). What do you think, is it possible to have something like that?

[bivouac]

This sounds eerily similar to the P vs NP problem.

[lajoswinkler]

Depends on the domain of a function. If the domain of the function are real numbers, â„Â, then there is an infinite amount of impossible solutions because real numbers are dense, uncountable.

If you define a proper domain, you can make a function with equal number of possible and impossible solutions.

[mardlamock]

I dont mean a function that has an equal amount of solutions and impossible solutions, that is easy. Ie sqrt x. its got infinite solutions on the positive side and infinite impossible solutions on the negative side (disregarding imaginary numbers). I kind of tried to explain it here http://imgur.com/L8NeyQr

[Rhidian]

What exactly do you mean by an impossible solution? Do you just mean imaginary numbers being in the solution would qualify it as impossible, or is there some other meaning?

[K^2]

I'm still trying to figure out what he means by solution to a function...

[mardlamock]

By solutions i mean points where f is undefined

[K^2]

Ok, so you need a function that has an equal number of defined an undefined points on any arbitrary continuous interval? Hm. It's definitely possible if we only consider rational numbers. Let me think for a bit if there are any contradictions with requiring this is also true for reals.

Edit: Here is basically what it boils down to. If you require that on any interval the number of points are equal, it automatically demands that both sets are dense with respect to each other. So we need to split real numbers into two sets which are dense in each other, and both of these sets must be uncountable. That will satisfy conditions, but I'm still not sure if this is possible. All of the methods of picking a dense subset of reals I'm coming up with give me a countable subset, which doesn't satisfy the "equal number" requirement.

Edit 2: So apparently, the answer is yes, but I'm having trouble following it. You might encounter the same problem.

Edited January 31, 2014 by K^2

[mardlamock]

I didint expect this to be so darn complicated, I somewhat understand it. It was just a question that came up to my mind when my math s teacher was explaining something about planes intersecting lol. A guy on IRC managed to explain it quite well to me. Thanks for the link man!