[alok-g]

Genuinely asking:

Consider decimal numbers between 0 and 1 in binary.

Here's how I am going to synthesize this set.

Step 1: Take non-decimal binary numbers and consider them to be padded with an infinite of zeros at the left.

...0000000

...0000001

...0000010

...0000011

...0000100

...0000101

...0000110

..........

Do we agree that this will contain all the non-decimal non-negative binary numbers?

In particular, is the following number in the above set? ...111111 (all ones, not zero-padding on the left). If not, why not. And if not, this seems to be a matter of definition to me. If yes, move on.

Step 2: Place a decimal at the end of each line above and flip left and right.

0.0000000...

0.1000000...

0.0100000...

0.1100000...

0.0010000...

............

Do we agree that this contains all the decimal positive binary numbers between zero and one: [0, 1).

Let's now apply diagonalisation on this.

It says that the number 0.11111111... will not be present in the above set.

Perhaps someone can see my confusion and enlighten me. :-) Thanks!

[wayn3]

The confusion in your argument is rather simple. You construct the set of INTEGERS in a binary representation.

Then you flip them over to the other side,in an operation that you yourself do not fully understand.

The first number that you flip (0.1) turns into 0.5 decimal

The second number that you flip (0.01) turns into 0.25 decimal

The third is 0.75

0.125

and so on. You are creating a subset of the rational numbers, which is obviously countable. It is countable because you constructed it to be countable. You constructed something that was countable and then tried to prove that it is countable.

Happens all the time :)

[alok-g]

This was helpful. Thanks.

I did understand before how flipping changed the numbers into 0.5, 0.25, etc., but had missed that this process would only create rational, thereby missing the irrationals altogether. There's enough food for thought for me now. :-)

[wayn3]

What you really want to understand is "where's pi", right?

Pi looks somewhat like this: 3.1415... "and so on". Let's ditch the 3.

0.1415... and so on. flipped over equals:

...5141

Easy enough, right? But pi had infinite decimals. Infinite decimals on the left side means what? You tell me :)

Infinity is not an element in the real numbers. All the irrationals, when flipped over, get absorbed into infinity - and infinity itself has no decimal representation. For people who are not in math, saying "and so on" is fine. But you can't do math on "and so on". "And so on" is a handwaving way of saying that you lack the mathematical language to describe whats going on. Which is fine, for casuals.

[alok-g]

You have gotten to exactly what my confusion has been!

I had assumed from the middle school itself that the set {1, 2, 3, ...} includes infinity, and I still am questioning if this not being is just a matter of definition or it has to be that way. More below:

We say that the size of the set {1, 2, 3, 4} is 4, in which scenario, the number 4 happens to be an element in the set. Likewise for {1, 2, 3, 4, ..., 100000}. Now we say that the size of the set {1, 2, 3, 4, ...} is infinity, but infinity is not an element of that set.

It seems what I am missing is the formal definition of this "..." or "and so on". If these two were not allowed in any of the proofs, how would you word Cantor's Diagonalisation and other theorems in mathematics that currently involve these. (Or alternatively, what is the formal definition of "...")

PS: I do understand limits and calculus but perhaps from an engineering perspective, not for pure mathematics where I have these confusions.

Thanks! :-)

[yorwba]

It is actually quite rare that a set contains its own size as an element. E.g. {0,1,2,3} also has size 4, but does not contain 4 itself. That {1,2,3,...} does not contain infinity should not be surprising. "..." essentially means that the set is generated by adding 1 to any number already in the set.

Is infinity the result of adding 1 to a number in the set? If infinity-1 is in the set. Is infinity-1 in there? If infinity-2 is. ... This gets you a set {1,2,3,...,infinity-2, infinity-1, infinity, infinity+1, ...}

But you can do the same with foo and get the set {1,2,3,...,foo-2, foo-1, foo, foo+1, ...}. So if infinity is in {1,2,3,...}, the same should be true for foo or anything else. This obviously doesn't make sense, so {1,2,3,...} is defined to be the smallest possible set containing 1 and n+1 for each n in the set.

[wayn3]

The problem that people who don't REALLY learn mathematics is that they dont understand that EVERYTHING has to be defined in a way that makes it an absolute truth.

The definition of the number 2 is by axiom. The peano axioms state this:

1. There is a number. We call it 0

2. For every number n, there is a successor S(n). The successor of 0 is called 1

3. Let m and n be numbers. m=n is equivalent to S(n)=S(m). This means that if two numbers are the same, they have the same successor

4. For every number n, S(n) is not 0. This means that 0 is the first such number

This completely describes the natural numbers. If you want to say that 0 is not part of the natural numbers, just write down the same things and say that 1 is the first number, and construct 0 later. For some funky reason, this doesn't actually matter. Although you'd think that difference between 0 and 1 is pretty big.

Nowhere in this does infinity appear. Infinity does, in fact, not belong to the natural numbers. Infinity as a NUMBER is only defined much later in mathematical literature. As a concept, it just means "big". When we talk about infinity in terms of the reals it just means "if we construct a series of numbers that get succeedingly bigger and they are not bounded, then they 'tend toward infinity'". But infinity is not a thing. Its just a name for "this shit gets big brutha".

The problem is that I can't give you a formal definition for "and so on", because it can mean many things.

I will give you a formal definition for a number instead.

Pretend again that we look at the number .111111111...

and now we flip it around. We look at the number ...111111.0

What this means is the following. This number is the sum over all the exponents of 10. The number is obviously identical to 1 + 10 + 100 + ... = 10^0+10^1+10^2+...

We still get the and so on. Now we need to find language that seeks to eliminate the notion of "and so on". Keep in mind that this number is not part of the real numbers, because it would be identical to infinity.

We prove this in the following way:

Our number is obviously bigger than 1, because it is 1+10+...

It is obviously bigger than 10, because it is 1+10+...

Now, we perform something devious. An argument by induction. This happens in the first week of math education and it is precisely when we leave the realm of "all the things we can write down".

We prove that for every n, the number ...1111.0 is bigger than 10^n.

1. We already know this for n=1, because 1+10+... is bigger than 10^1 = 10.

2. For n>1, 10^(n-1)+10^(n) is clearly bigger than just 10^(n), and our number contains all the exponents of 10.

(^Up there is a tautological proof. You already know this, I just write it down "semi formally". Proving a fact about natural numbers by induction is always somewhat tautological because the mechanism that makes induction possible IS the natural numbers themselves.)

Now we know that for every exponent of ten, our number is bigger than that. This means that for every natural number that we can possibly think of, our number is bigger than that. Therefore, the number we are looking at is not part of the natural numbers and by extension, not part of the real numbers.

This is because every natural number has a successor and by definition, the successor is bigger than the number itself. But our number is bigger than all the numbers already. In order for it to be a natural number, it would have to have a successor, which would be bigger than itself. But our number is bigger than all the numbers. This sounds a bit clunky. I could write it down mathematically, but that would just confuse you, probably.

Now I'm just being wordy. The proof was over long ago.

edit: Writing this down mathematically goes like this: Let b=...1111 and n be an arbitrary natural number. b>n => n is not successor of b => b is not in N. (You see that I had to reduce the argument to something I already knew - the successor axiom - in order to wrap it up).

I've shown that your number is bigger than all the possible number on the real line and if its not on the real line, it is not part of the real numbers.

Now you would ordinarily say that it is infinite. And that's kind of true, but you can only say that once you understand that infinity is just a concept, not a number in the way you understand numbers.

tl;dr: Infinity is not part of any of the ordinary sets of numbers (natural numbers N, integers Z, rationals Q, reals R, complex C, ... - yes, there are more of these)

[alok-g]

Very helpful. Thanks! I can now see why infinity cannot be a natural number, under Peano's axioms. I follow mathematical induction arguments also. Thanks a lot! :-)

[BuuQu9hu]

The number ...1 in your first notation is actually -1, using 2-adic interpretation. The number 0.1... is actually 1, which lies outside your expected range.

Maybe another base will help. In base 3:

0.0... 0.10... 0.20... 0.010...

Once again, a naive diagonalization yields 0.1... but this time at least we have not extended past 1. Is there a way to avoid that convergence? Sure, we can change the 0 digit to either 1 or 2 randomly, or based on some pattern or enumeration, instead of just 1. So now we can take many diagonals, and they all look like:

0.121212... 0.122122... 0.121121... 0.22212221... ...

Augh! What happened? Our diagonalization appears to have revealed an infinity of missed reals! This is similar to the construction of the Cantor set (https://en.wikipedia.org/wiki/Cantor_set) and hopefully illustrates the problem with your enumeration of the reals.

[alok-g]

Thanks. This was helpful. I am not sure that I fully get it, but I think I do. :-)

[powera]

Step 1 fails because there is no such number that is an infinite number of 1s. Only infinity has infinitely many digits.

Step 2 fails because you can't create a 1-1 correspondence between the real numbers between zero and one, and the integers. That is what you are trying to prove. I'm too tired to figure out if your proof is valid or not, but your conclusion is correct.

[pwdisswordfish]

> decimal positive binary numbers

Make up your mind.

[alok-g]

Clarifications:

decimal: Not base 10, but fractional, those with a decimal point binary: Not just 0 or 1, but base 2.

0.01 is a binary positive decimal number which is 0.25 in base 10.