If someone came up to me and presented a proof that 1=2, l would immediately discard the proof. The OP obviously didn't present something that ridiculous but it does amount to saying that the prove Cantor gave was wrong as it proves the opposite.There isn't a third possibility here. It isn't about herd mentality here since it is mathematics.In mathematics, we stand on the shoulders of giants and it does not tolerate any weakness that we find in philosophy, religion or social sciences. I understand where you are coming from but you have to see for yourself that in this sub section, we need to be more objective and avoid beating around the bush as we normally do — Wittgenstein

I would be a bit cautious about statements about dismissing proofs like that. I can easily produce an argument using triangles and how they combine that proves 1+1 = 3. I would even go as far to say that using a very simple technique that involves very simple geometry and counting that 4=5 in some circumstances.. People get fixated with what they treat as absolutes and they will come a cropper later.

Well the theoretical phycists are welcome to have their opinion. But how can that be the case if the universe is finite in size as it would be if it were expanding from the big bang. That just sounds like alot of mumbo jumbo to me.

If someone came up to me and presented a proof that 1=2, l would immediately discard the proof. — Wittgenstein

1 dollar is worth 2 50 cent pieces.

1)First convert all numbers into binary strings. — Umonsarmon

How?

We can imagine the universe in two dimensions as a plane, which is flat and infinite (like an infinite piece of paper). ... However, with expansion, it is possible that even if the universe just has a very large volume now, it will reach infinite volume in the infinite future

That's what this should be about. Either all the mathematicians since Cantor have been idiots and retarded to not notice the fatal flaw in the proof or OP is wrong. What's the probability for each case. — Wittgenstein

Well rather than trying to deal with phantom probabilities why don't you just read the proof. The point you are forgetting is that how many people have tried to debunk Cantors proof after it was published, most people just accept it as fact. The logic is very water tight but set theory has produced paradoxes in the past and this just might be one of those cases.I only stumbled on it when i was working on a different problem.I am not closed to it being wrong it just I don't see many people trying to understand the proof. I can explain it in an even simpler way if need be.

How? — tim wood

Well the number 51.32 would be broken up into the binary for 51 and the binary for 32 and then combined into a single string.

Great! Now try π.

Great! Now try π. — tim wood

Pi would just have a never ending trail of digits but the procedure is exactly the same. That's just an infinite binary string.Cantors argument relies on numbers with infinite numbers of decimals and thats what he uses in his argument as well so if you want to cobble me for that then you have to cobble Cantor as well.Cantors argument specifically relies on having infinite strings with his slash argument. It wont work with finite strings because these can all be converted into rational fractions which we can list

Great! Now try π. — tim wood

Just to explain myself more clearly with pi you can simply covert each digit of pi into binary and do this with all the other numbers as well to generate your string.
i.e 3.141 etc = 3 in binary+1 in binary +4 in binary + 1 in binary etc etc

Well as far as I can tell any number fed into this procedure should result in a terminating rational length which will produce a set of rationals which map to the natural numbers regardless of whether it is an irrational number or not. — Umonsarmon

Yes! You can quite happily map a countable set of irrationals to any other countable set.

Now I understand the point that you could argue that the set your feeding in is uncountable but this leaves us in a strange position because the two proofs directly contradict. — Umonsarmon

1)First convert all numbers into binary strings. — Umonsarmon

Convert all real numbers to binary strings!

2)Draw a square and a line down the middle — Umonsarmon

The line is an uncountable set.

3) Starting at the middle line do the following .If the digit in your string is a 1 move half the distance to the next line to the right. If the digit is a 0 move half the distance to the next line to the left. — Umonsarmon

You have to apply this to every real number. You're mapping an interval of reals to an interval of reals with (what looks to me like) an injection.

This is entertaining, and possibly educational, At the end is a surprising insight!

https://www.youtube.com/watch?v=5TkIe60y2GI&list=PLt5AfwLFPxWKuRpivZd_ivR2EvEzKrDUu&index=3

I buy the diagonal argument - no need for infinitude to complete the demonstration. I assume you're aware that between any two rational numbers are an uncountable number of reals. That is, the ratio of the number of rational numbers to the reals approaches zero.

If you don't understand how my proof works then just say so. I am well aware of that sort of stuff, I did study maths for my degree but unlike you it seems I do question all the assumptions that I was taught and if I think something does not make sense then I will think about it. I mean you are aware that set theory has been troubled by paradoxes in the past right. This is not really anything new from that perspective

I will explain the proof again. I am aware that my presentation of it is not brilliant so here goes.

Firstly Cantor himself uses infinite decimal numbers in his diagonal argument so I don't understand why you cut him some slack but not me :)

Now first convert the real number into a binary number. We can convert all numbers to base 2 people if we need to.or we could just turn each digit into a base 2 representation

We draw a line of length 1. Point A = 0 Point B = 1/2 and point C = 1

To keep this simple I will use the binary string 101.

A B C
0 1/2 1

Now if the digit in your number is a 1 we left shift 1/2 the distance to the next point.
If the digit is 0 right shift 1/2 the distance to the next right point.

We start at B
So the first digit is 1. We left shift to the new point D

A D B C
0 1/4 1/2 1

Next digit is 0 so we right shift from D halfway to the next point which in this case is B


A D E B C
0 1/4 3/8 1/2 1


Next digit is 1 so we left shift from E halfway to the next point which in this case is D

A D F E B C
0 1/4 5/16 3/8 1/2 1

Number terminates here at 5/16 which is a rational number a/b.
Each binary number will terminate on its own unique rational a/b as will each irrational number. This can then be mapped into the natural numbers, in other words each number fed into the system comes out with a rational key regardless of how many numbers you feed into it.

Those numbers were originally spaced out alot more in my post. The terminating point is F which is 5/16

Each binary number will terminate on its own unique rational a/b as will each irrational number — Umonsarmon

The sequences don't terminate for any real number which has an infinite (non-repeating) binary expansion. EG pi/10 would never terminate. The consequence of allowing infinite sequences there means the function is simply from binary expansions to real numbers - essentially a way of encoding binary expansions. The representation of a set doesn't change any of its cardinality properties though.

It would converge on a rational number just as the sequence 1/2+1/4+1/8 converges on 1 after an infinite number of steps. Don't get me wrong, it would be quite fiendish to try and calculate where those numbers terminate but in theory they would

It would converge on a rational number just as the sequence 1/2+1/4+1/8 converges on 1 after an infinite number of steps. — Umonsarmon

I don't think this is right. Infinite convergent sequences of rationals typically converge on non-rational reals, even though all the finite sums and elements are rational. The incompleteness of the rationals demonstrates this. (Well, more precisely, convergence doesn't make much sense for infinite sequences of rationals...)

I bring this up because the incompleteness property of the rationals shows that infinite convergent sequences of rationals do not have to converge to rationals. (or more precisely that convergence breaks when you don't have ensured existence of suprema and infima)

It would converge on a rational number just as the sequence 1/2+1/4+1/8 converges on 1 after an infinite number of steps. — Umonsarmon

Really? What makes you say so?

Remember each time the point moves between the halfway point of 2 rational numbers. This in itself would be a rational number. That's just simple maths. Now I understand what your saying but even if I do that infinetley it should end up on a rational number. Now I will admit I can't be 100 certain of that so that is a bone of contention.

if I do that infinetley it should end up on a rational numbe — Umonsarmon

Try and prove it! This discussion might be helpful.

That is a tricky proof, I will have to think about that however all the distances travelled are rational numbers. A rational number + or - a rational number is a rational number. So even if I do this infinetley it will end up on a rational number. I will again say that you are cutting Cantor more slack than me :) . Would you have asked him to have written out an infinite decimal number to prove his theory even though that's impossible..

I would cut you more slack if you demonstrated more understanding. You've been thoughtful, but the rigour's lacking. This shows whenever someone spells out a mathematical idea with more precision and it turns out that the concepts (as they use them) contradict established theorems or the intuitions (conceptual understanding/imaginative background) that support them.

Specifically, your understanding of convergence needs refining. I also suspect you would discover more difficulties with your argument if you tried to define your function formally (what's its domain, what's its image, is it injective, surjective...) - one presentation of it is a way of associating reals with infinite sequences of rationals (if they do converge in the usual sense, you will successfully associate with only rationals, making the image countable, if they don't converge in the usual sense your argument doesn't work).

I think your splitting hairs personally. People can tend tend to get bogged down in details which actually mean nothing. I prefer geometric proofs which I can visualise. I will not apologise for doing that as I feel that is a much more powerful form of maths than anything I can scribble down on a piece of paper. If. If we cut through all of the banter it all boils down to whether you believe that a rational number + or - a rational number equals a rational number. No matter how many times I perform that operation it will always result in a rational number. The equation for fractions will prove that. So to throw the ball back in your court what changes when I perform that operation ad infinitum, well the honest answer is that nothing does.

If I'm understanding, you have an "instruction" set of zeros and ones. You start on a point on the number line that is a rational number. Then you move through your instruction set and as you hit a zero or a one, you move to left or right by a rational distance. the idea being that at all times the point you land on is a rational number. That sounds ok to me, but so what? What does that have to do with the number of zeros and ones in your instruction set? What does it have to do with the convergence of anything?

If we cut through all of the banter it all boils down to whether you believe that a rational number + or - a rational number equals a rational number. — Umonsarmon

Finite sequences don't automatically have the same properties as infinite sequences. It doesn't boil down to that at all.

No matter how many times I perform that operation it will always result in a rational number. — Umonsarmon

$\sum_{i=0}^{N}(-1)^{i}\frac{(2i-3)!!}{2i!!}$

where $x!!$ is all the even numbers up to $x$ multiplied together. Is always rational for any $N$.

$\sum_{i=0}^{\infty}(-1)^{i}\frac{(2i-3)!!}{2k!!}$

Is $\sqrt{2}$

Ok the termination point of the number is a rational number, thats what I'm arguing. This rational number is a key for the binary number so to speak. I put the rational number in a set of other rational key numbers which as we all know can then be listed by the naturals. The convergence is to do with whether an infinite binary number would still converge on a rational key. You can think of the key as the address of the binary number.. The point is that each binary number will have a unique key that it terminates on regardless of its size , this is critical, this is what allows them to be listed.

I would just use a -a/b value and then list that next to its a/b twin — Umonsarmon

But then zero would be mapped to -(1/2) and -1 too would be mapped to -(1/2) which would result in a failure of the necessary bijection. There would be two real numbers (0 and -1) mapped to only one point -(1/2) in your scheme.

Now that is an interesting equation. I would have to see the break down of all the calculations for that. If it is correct then I will accept your point