Even if they were, pi is not between zero and one, so would not appear in your list.

The rest of your argument is difficult to parse. If you’re interested in this kind of stuff though, I’d really recommend picking some good books on set theory and maybe number theory. Set theory in particular will cover a lot of what you’re talking about.

I am fully aware of the constructionist approach to mathematics.

But your argument goes somewhat like this:

The reals are real

In computer programs however, we can only do so much

Every computer program represents a real number

For every bounded amount of bits, there is only a countable set of computer programs

Therefore, the reals are countable.

This is not how math works. You extrapolate from a bounded set that is countable to the collection of all those bounded sets - and then claim because every one set is bounded, the collection must be bounded as well - and therefore countable.

Let me make a similar argument just to show how silly your argument is. Here we go: 1. Every set of integers with only one element is finite. Call it a trivial set, for example: {1}

2. Every finite union of such trivial sets is finite

3. Every finite union of such a finite union of trivial sets is finite

4. and so on - after many iterations, your finite set will look pretty big to the human eye - almost as if it was the set of integers.

5. Therefore, the set of integers must be finite.

Sounds pretty silly, right? Thats what you've done for a different property of a different set.

That is nonsense. This is why arguing with illiterates about math is so annoying. They think they can write non-stringent arguments and then claim they have "somewhat proof". In math, there is no such thing as "somewhat proof". You are either right or you are wrong. The reals are the reals by definition. You can not make them countable.

What you can make countable is sets of numbers that you work with. Sure. You wont ever create a computer that has access to all the reals, PRECISELY because it operates from a position of countable-ness (even quantum computers). But thats not what I care about when I talk math.

The real numbers are not accessible to in a "reality" way. They lie fully in the realm of arcane mathematics. Like many other things do. You would never claim that the hyper-exponentiation operator had anything to do with reality, and it doesn't. Neither do the reals. You're just confused about them because they appear so prominently in middle school mathematics.

Accept it and move on.

Mathematics is the realm of absolute truth. Everything we have proven is de facto true. You can make an argument that we are doing the "wrong kind of mathematics". And that may actually be true. But it doesn't change anything about our discipline. If you want to get philosophical and create a new Mathematics discipline based on different arbitrary rules (axioms), you are free to do so. But you will have to prove a lot of very mundane things and go through centuries of early day fuckups and its very likely that you will soon seek to merge with the already established mathematics.

Call it math2.0. I'll play with it. If its interesting, why not. But don't expect your new math to be more interesting than the one we've got.

Well let's count the natural numbers in binary, ..0000, ..0001, ..0010, ..0011, etc.

Now take the first digit of the first number and flip it,1 and the send digit of the second number and flip it, 1, etc. This is 2^x - 1. Is it countable because the natural numbers are countable? It's the same type of number you created on Cantor's diagonal. But it's never in the set you have counted. However, the same thing is true of x + 1.

The problem when constructing such numbers is there is not a final number in the set. So saying go to Infinity and add one or take 2 to the power of that and subtract one has no meaning.

PS: I think people don't equate them because they assume there is a difference between making a number larger and making a fraction more precise. So, 0.1, 0.11, 0.111, 0.1111 means something different than 1, 11, 111, 1111 etc.

Oh fiddlesticks, the post was deleted before I'd even finished my reply! :(

The punch-line is that there are more real numbers between 0 and 1 than there are whole numbers greater than 0. The proof basically shows that even if you imagine that you had a countably-infinitely-long list of all the real numbers between 0 and 1, you can construct a number that must be missing from the list -- which is a contradiction, and thus no such list can exist (implying that there are more numbers between 0 and 1 than numbers greater than 0).

For instance, your enumeration is missing every irrational (or even recurring decimal) number. That set of numbers is uncountably infinite.

[1]: https://en.wikipedia.org/wiki/Cantor's_diagonal_argument

You didn't prove that statement, and actually Cantor's diagonal arguments [1] proves you wrong:

Consider the set of «all real numbers which you can actually specify IN ANY FASHION AT ALL». If you can count it, you can order them in a certain fashion: n0, n1, n2, n3 … It's easy to specify a number X which is not part of this set (which is in contradiction with the definition of this set, and demonstrates ad absurdum that this set is not countable). To build such X, consider the n-th decimal of the n-th number: if it's zero, the n-th decimal of X is 1, otherwise the n-th decimal is 0. Then by construction, X is different from any number of this set, which mean X is not a part of the set. ?

[1]: https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument

It doesn't work like that. You want to show that you can't enumerate all reals, so what do you do? You start enumerating all reals. Then you build a new number, whose Nth bit is different from the Nth bit of the Nth real number. You can do that because you have enumerated them, and we have never talked about an infinite N so we are never flipping the last bit of the last number nor anything similar. We are only flipping the Nth bit of the Nth number. Number 819439243? We flip the 819439243th bit. That's all we do.

What do you do then? Well, you compare that new number you created, with the first number, the second number, and so on. And you see that this number is different to all the other numbers.

What does it mean? That it doesn't matter what your enumeration scheme is, there is at least one number that will always be out of your enumeration scheme, and so your enumeration scheme will fail (so this is an absurd also, as we were supposing that this specific enumeration scheme -that could be any enum scheme- was complete). So you don't have as many integers as reals, so QED.

Secondly, if you take the Integer set and say N = the sum of all numbers > 0 you have counted, then you will never reach N as you go to infinity.

As you see, it doens't matter, as in this proof you never need to do the sum of all numbers.

Every integer has a finite number of numerals. Proof: by definition, each natural number is either 0 or a successor to 0. 0 is finite. And if n is a finite length, then n + 1 is a finite length. So by mathematical induction, all natural numbers have finite length. A similar argument covers the negative numbers, so all integers have finite lengths.

Since the decimal expansion of certain rational numbers are infinite (e.g., 1/3 = 0.333...), the strategy of mapping each number between 0 to 1 to an integer by flipping digits over the decimal point does not work for all rationals, let alone all reals.

Each rational number has the form a / b for a pair of integers (a, b). This representation could be non-unique, e.g. 1/2 = -3/-6 are two different representations of the same rational number as fractions of integers. For this representation to be unique let's require the denominator b to be positive and a, b to have no common divisor. Let's consider (a, b) a point with integer coordinates on the lattice Z x Z, where Z is the set of integers (Z = { ..., -2, -1, 0, 1, 2, ... }).

We can count each of these points if we begin by labeling the origin (0, 0) as "0" then (1, 0) as "1" then (0, 1) as "2",(-1, 0) as "3", etc. We can order these integer lattice points first by their increasing magnitude |a| + |b| from the origin and then by the angle they make relative to the origin and count them all one by one.

Informally this shows that the set of rational numbers is not larger (in the sense of cardinality) as N_{>=0}, the set of non-negative integers. It's not smaller either as the set of non-negative integers N_{>=0} are a subset of rational numbers.

You missed the crucial property that rules out the rationals (more precisely, the rationals with their standard ordering): one way of stating it is that every sequence that has an upper bound in the set, must have a least upper bound in the set. The rationals do not satisfy this property (for example, consider the sequence of successive decimal expansions, each one to one more decimal place, of sqrt(2)), but the reals do.

The challenge for me is to understand how there can still be countable sets that also satisfy that property of the reals. (Obviously any countable set can be put into one-to-one correspondence with the rationals, but for a countable set that satisfies the least upper bound property of the reals, such a correspondence with the rationals would put an ordering on the rationals that was not the standard one.)

Countable sets are shown to be countable by providing a mapping between the natural numbers and that set. Uncountable sets are shown to be uncountable by assuming they are and creating a new element from the set of all elements and showing that it was not accounted for in the first place.

But what happens when we try to use the mapping argument on the set of the reals, and the contradiction argument in the set of the integers?

For the countability of the reals, I can setup a binary tree of infinite depth with 0 on the left and 1 on the right. I then do a depth first walk of the tree and assign an integer to each of the nodes. Then the sequence of 0s and 1s from the root of the tree to any node we care to go to represents a binary number between 0 and 1 if we treat it as a binary fraction like he does in theorem 16. So any number we care to go down to in the binary tree can be assigned a natural number, and since the binary tree is infinite it can go down to any real number. So according to this argument the reals are countable.

Now for the diagonalization-like argument for incountability of the natural numbers. Assume the positive even numbers are countable, and that we have a list of all the even numbers assigned to a natural number. I can create a new even number by adding all of the numbers in the list, since all the numbers are positive, each number is smaller than the sum of all of the numbers on the list, so the number is not on the list, and we have to conclude that our assumption was wrong, and that the even numbers are uncountable.

If the first argument is wrong, why is it wrong for the reals but not wrong for the naturals? And if the second argument is wrong, why is it wrong for the naturals but not wrong for the reals?

Eg you can try to apply Cantor's proof on the list of all integers (written in decimal form) to attempt to prove that the integers aren't countable. Or on a list of all rationals.

The proofs will fail in interesting ways.

cantors diagonalization argument is proof of that. you can't pull some silly trick to make them countable. there are many properties of R that are countable, but that doesn't make R itself countable.

I had exactly this in mind (or the proof that the set of rational numbers are countable), but mistakenly thought it was called the 'diagonal argument' because you would get the bijection to the natural numbers by counting diagonally through the table.

/me not a mathematician.

/me didn't watch the video.

Why is that?