Posted: Apr 23, 2016 5:17 pm
by LjSpike
scott1328 wrote:
LjSpike wrote:
VazScep wrote:
scott1328 wrote:This is merely incoherent wibble.
To be fair, LJSpike admitted it was a GCSE perspective (one you typically obtain by studying maths between 14-16 years old in the UK). And maths is taught pretty shit here IMO.

Yup. Anyway I could shorten my description now, which I shall do. After using a lot of arrows on paper to try and find the most efficient description.

Integers go on infinitely, we could just say that rationals go on infinitely too, but then again, were not giving a 'concrete figure' to either of them. So if we take an increment of 1 in the integers chain, we can then express all of the rationals in this chunk as
Y/X (Y over X). X is any number from 1 to infinity. Y is any number from X * the lower end of our 'chunk' of the integer chain, to infinity. So we can say X and Y both have a range of infinite size.
So to find every combination of them we find all of the possible values of X, which we know to be infinity. We then find all the possible values of Y for all the possible values of X, as we know Y can have a range of infinite size, but can also have a range of say 1, we can do X * Y, which is Infinity * (Infinity/2) as halfway between 1, and infinity is a half of infinity, which is still infinity (as infinity is infinitely large) so we could say that its infinity * infinity or infinity^2

So for a range of 1 in integers, we have a range of infinity^2 (infinity squared) of rationals.
If we were to choose a range of 2 in integers, we can simply extrapolate the pattern found, to be 2*infinity^2 in rationals. So for a range of infinite integers, we multiply infinity^2 by infinity, giving us infinity*infinity*infinity or infinity^3

So for the total amount of integers to rationals we can say the ratio is infinity:infinity^3
Simplifying this ratio down would divide it by infinity, as that is a common factor of both of them. So we then get 1 : infinity^2 (as anything divided by itself = 1, and dividing a^x by a results in a^(x-1))

I would personally be happy to leave it at that, but as we (theoretically) can't have a number bigger than infinity, we can handle this now at the end (just like rounding in a trig question, or such, you leave it till the end to gain the best accuracy).
So we can say, infinity^2, is still infinity as nothing can be larger than infinity, giving us the answer of 1:infinity.

Read this Wiki article, it explains how the rationals can be put into a one-to-one correspondence with the natural numbers.

I looked into the puzzle further, and it isn't a one-to-one correspondence. It is infinity:infinity, but they are different infinities, thus you can't simplify them down to 1:1.
Another puzzle including different infinities (and a rather nice one I like myself) is the Hotel Paradox:
(On youtube somewhere TedEd has a brilliant video explaining it).

Again my solution, this time displayed via inequalities (compressed to a single equation without paragraphs of description)
Integers:Rationals = n:(n*x=<y=<x)/1=<x=<infinity)
n = Integer
x = unknown (denominator of rational)
y = unknown (numerator of rational)

This solution handles it without resorting both sides to infinity, presenting an at least very hard to express solution (if not impossible to express) as both infinities are different infinities, not equal infinities. (My best guess at displaying it would be)
inf1:inf2 where inf1 < inf2 for integers:rationals