So, disclaimer: been a while since I did analysis, and this was a random shower thought (yes, I was taking a shower lol). So, half of this is likely badly flawed in some way lol.
We will rework the terms a bit. Namely, we will establish try to establish a 1:1 between the indices and the natural numbers as follows;
#1: 1 0 0 0 0 0 0 0 0 0 0 0 0 0..... 1 in binary #2: 0 1 0 0 0 0 0 0 0 0 0 0 0 0..... 2 in binary #3" 1 1 0 0 0 0 0 0 0 0 0 0 0 0..... 3 in binary etc etc where we were left-to-right least significant to most significant digits, using binary. We generate a sequence for every natural number this way.
So now we can interpret the infinite sequences as numbers in the standard way. Now do the negation (for sequence #n, flip the nth bit), so that we get 0 0 1......
One might be tempted to think we just generated a natural number which is not a natural number - but wait, how can the natural numbers not be one-to-one with the natural numbers, that is, how can a countable set be uncountable?
Well, actually it is apparent we didn't generate a natural number even though we are flipping a countable infinite number of bits, all of which belong to natural numbers. There is clearly no successor to the sequence we generated - there's no next number. So actually, we didn't generate a natural number even though we meant to do so.
I actually stood there in the shower for like 10 minutes thinking that the generated number was a natural number loool. Hot water bad for thinking I guess o.o
P.S. Infinities are awesome, and I don't just mean the pen ones.
P.P.S. I don't suppose there is a way to do something similar with the ordinals somehow used as indices rather than powers of 2?
What you constructed is a valid binary string, but not a binary representation of a natural number. All binary representations of natural numbers e = e_1,e_2,... have some index 'n' such that for all k>n, e_k = 0, where yours was an infinite sequence of 1's after the 2nd index. ^^
On January 27 2012 10:09 ShinyGerbil wrote: What you constructed is a valid binary string, but not a binary representation of a natural number. All binary representations of natural numbers e = e_1,e_2,... have some index 'n' such that for all k>n, e_k = 0, where yours was an infinite sequence of 1's after the 2nd index. ^^
err the ... didn't signify continuing 1s. I just didnt bother to write the flipped bits of sequence #4, #5, etc. And also what you wrote about the indicies is what I meant by there is no successor so it's not actually a natural number.
On January 27 2012 10:09 ShinyGerbil wrote: What you constructed is a valid binary string, but not a binary representation of a natural number. All binary representations of natural numbers e = e_1,e_2,... have some index 'n' such that for all k>n, e_k = 0, where yours was an infinite sequence of 1's after the 2nd index. ^^
err the ... didn't signify continuing 1s. I just didnt bother to write the flipped bits of sequence #4, #5, etc. And also what you wrote about the indicies is what I meant by there is no successor so it's not actually a natural number.
You missed his point. The reason that almost everything on that list doesn't represent a natural number in binary is that there's going to be infinitely many 1s and infinitely many 0s interleaved in some way. A binary number only has finitely many 1s in its expansion.
Daydreaming impossible scenarios in the shower is real fun. I can meet any fictional characters or go skiing at any time of year. I swear I've probably had more adventures in my shower than in real life.
But you're counting with integers.... that means 4 ascends 3, and 4 is equal to 0100 (not sure why you flipped the bits backwards) regardless of how many 0's you put - they don't change the value... or maybe I am missing your point?
On January 27 2012 12:57 Ninja_Bread wrote: But you're counting with integers.... that means 4 ascends 3, and 4 is equal to 0100 (not sure why you flipped the bits backwards) regardless of how many 0's you put - they don't change the value... or maybe I am missing your point?
This seems to sum it up for me. By his logic, I could write 9, but because it's not 09 it leaves no room for ascention?
On January 27 2012 12:57 Ninja_Bread wrote: But you're counting with integers.... that means 4 ascends 3, and 4 is equal to 0100 (not sure why you flipped the bits backwards) regardless of how many 0's you put - they don't change the value... or maybe I am missing your point?
The bits were flipped backwards because otherwise for any specific number, you'd need an infinite number of 0's to the left, which is a bit weird to write and is rather nonstandard notation.
i.e. #1 = ....00000000001 instead of #1 = 100000000.....
On January 27 2012 10:09 ShinyGerbil wrote: What you constructed is a valid binary string, but not a binary representation of a natural number. All binary representations of natural numbers e = e_1,e_2,... have some index 'n' such that for all k>n, e_k = 0, where yours was an infinite sequence of 1's after the 2nd index. ^^
err the ... didn't signify continuing 1s. I just didnt bother to write the flipped bits of sequence #4, #5, etc. And also what you wrote about the indicies is what I meant by there is no successor so it's not actually a natural number.
You missed his point. The reason that almost everything on that list doesn't represent a natural number in binary is that there's going to be infinitely many 1s and infinitely many 0s interleaved in some way. A binary number only has finitely many 1s in its expansion.
Actually our statements are equivalent. The statement that there can be no successor implies there is no final 1, and vice versa.
i think what you have stumbled upon is something called a p-adic number (in this case, a 2-adic) i don't know a whole lot about them, but the number you generate ends up being 001011111111.... which is equal to -12 or something like that.
On January 27 2012 12:57 Ninja_Bread wrote: But you're counting with integers.... that means 4 ascends 3, and 4 is equal to 0100 (not sure why you flipped the bits backwards) regardless of how many 0's you put - they don't change the value... or maybe I am missing your point?
This seems to sum it up for me. By his logic, I could write 9, but because it's not 09 it leaves no room for ascention?
So one would be inclined to think that we have enumerated every possible infinite binary string - thus the total number of possible binary strings of undetermined length is infinite, but countably infinite. By countably infinite I mean we can put them into a 1-1 correspondence with the natural numbers - pair them up exactly one for one by some clever scheme.
However, this isn't the case. Above, we have clearly assigned every natural number an infinite binary string. However, we will now create a binary string which clearly isn't any of the strings above. We do this by flipping bits. For sequence #x, we flip the xth bit. Thus sequence #x is not the same as sequence #1, or #2, #3, etc no matter what sequence #n we use since whatever sequence we compare, one of the bits has been flipped.
So there is no one-to-one correspondence between the infinite binary strings and the natural numbers.
On January 27 2012 15:14 infinitestory wrote: i think what you have stumbled upon is something called a p-adic number (in this case, a 2-adic) i don't know a whole lot about them, but the number you generate ends up being 001011111111.... which is equal to -12 or something like that.
It's unclear what you're trying to show. Could you tell us?
It feels like you're trying to show that the set of infinitely long binary strings is uncountable? If so, you can just use Cantor's diagonalization argument instead of the awkward procedure you used in your first post.
On January 27 2012 16:10 supbros wrote: It's unclear what you're trying to show. Could you tell us?
It feels like you're trying to show that the set of infinitely long binary strings is uncountable? If so, you can just use Cantor's diagonalization argument instead of the awkward procedure you used in your first post.
I wasn't trying to show anything, just laying out the confusion I had while in the shower wherein I thought I created a natural number during diagonalization which is clearly absurd.
On January 27 2012 12:57 Ninja_Bread wrote: But you're counting with integers.... that means 4 ascends 3, and 4 is equal to 0100 (not sure why you flipped the bits backwards) regardless of how many 0's you put - they don't change the value... or maybe I am missing your point?
The bits were flipped backwards because otherwise for any specific number, you'd need an infinite number of 0's to the left, which is a bit weird to write and is rather nonstandard notation.
i.e. #1 = ....00000000001 instead of #1 = 100000000.....
So shouldn't you write #1 as 1000000000000 instead of 0000000001? The only thing you're changing is the base, the same principles still apply....
I feel like you're thinking of something completely different than what you're demonstrating
On January 27 2012 09:07 EtherealDeath wrote: One might be tempted to think we just generated a natural number which is not a natural number - but wait, how can the natural numbers not be one-to-one with the natural numbers, that is, how can a countable set be uncountable?
So would you please try to explain yourself clearer....?
From what I can gather, it's flipping one of the digits value, which just offsets which number is the resultant. It's still 1:1 if I'm understanding it right though....
010 is 2, but if you flip the right 0 you end up with 011 (three) If you had 011 and flip the right digit, it becomes 010. Essentially, odd/even numbers would flip values, but it's 1:1
if you flip a digit of higher value, the difference between the paired numbers increases (001 [1] <-> 011 [3]) But it's still 1:1
On January 27 2012 15:14 infinitestory wrote: i think what you have stumbled upon is something called a p-adic number (in this case, a 2-adic) i don't know a whole lot about them, but the number you generate ends up being 001011111111.... which is equal to -12 or something like that.
Why do you ALWAYS beat me to the punch in all the math blogs . Basically that though, natural numbers will have 'tails' of 0s, whereas you're generating one with 1s.
On January 27 2012 12:57 Ninja_Bread wrote: But you're counting with integers.... that means 4 ascends 3, and 4 is equal to 0100 (not sure why you flipped the bits backwards) regardless of how many 0's you put - they don't change the value... or maybe I am missing your point?
The bits were flipped backwards because otherwise for any specific number, you'd need an infinite number of 0's to the left, which is a bit weird to write and is rather nonstandard notation.
i.e. #1 = ....00000000001 instead of #1 = 100000000.....
So shouldn't you write #1 as 1000000000000 instead of 0000000001? The only thing you're changing is the base, the same principles still apply....
I feel like you're thinking of something completely different than what you're demonstrating
On January 27 2012 09:07 EtherealDeath wrote: One might be tempted to think we just generated a natural number which is not a natural number - but wait, how can the natural numbers not be one-to-one with the natural numbers, that is, how can a countable set be uncountable?
So would you please try to explain yourself clearer....?
That is how I wrote it. Was just responding to a guy asking why I didn't write it the other way, and who clearly didn't get that it was an countably infinite sequence of sequences, and that the bit flipping is to show there is no bijection.
As for the second part, I was just saying how initially I somehow confused myself into thinking I had a new natural number, since previously I was using a natural number interpretation of the sequences. That leads to various contradictions, which is what the rest of that sentence is. But of course, the sequence that consists of the bit flips is not actually a natural number so no problem there.
EPT
