Hilbert's paradox of the Grand Hotel - An easier solution?

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Lars Krogh-Stea said:
But 010101 is a binary representation of an integer (1*1+0*2+1*4+0*8+1*16+0*32=21).
That's a finite string. They represent integers.

We are interested in infinite strings! They don't represent integers.
 
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PeroK said:
That's a finite string. They represent integers.

We are interested in infinite strings! They don't represent integers.
I understand all this.. But are you also saying that in an infinitely large set of infinitely long strings (containing only 1's and 0's) the string 100000000... wouldn't exist, and the string 1010000000... wouldn't exist?
 
Lars Krogh-Stea said:
I understand all this.. But are you also saying that in an infinitely large set of infinitely long strings (containing only 1's and 0's) the string 100000000... wouldn't exist, and the string 1010000000... wouldn't exist?
Those are infinite strings all right, but they are not integers.

I suspect the root of the problem is that you don't realize that those are not numbers.
 
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PeroK said:
That's a finite string. They represent integers.

We are interested in infinite strings! They don't represent integers.
Oh, I think I once again get the gist of your statement. I was missing the flow of your argument.
 
Lars Krogh-Stea said:
But are you also saying that in an infinitely large set of infinitely long strings (containing only 1's and 0's) the string 100000000... wouldn't exist, and the string 1010000000... wouldn't exist?
Every integer has its own place on the number line. Where would these go? The thing is, it's impossible to say, because we don't know how many zeroes are represented by the ellipses (...).
 
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valenumr said:
And really, any binary string can be interpreted as a number.
Not as an integer. Only as a real number with a decimal point somewhere.

Every integer, ##k##, has a decimal representation of the form:
$$k = k_0 + (k_1 \times 10) + (k_2 \times 10^2) + \dots + (k_n \times 10^n)$$For some finite ##n##, where the ##k_i## are the digits (binary or otherwise).

And, therefore, the integers are represented by finite strings of digits.

Every real number, ##x##, between ##0## and ##1## has a decimal representation of the form:
$$x = x_1 \times 10^{-1} + x_2 \times 10^{-2} + \dots$$Where ##x_i## form an infinite sequence of digits.

Therefore, the real numbers are represented by infinite strings of digits.

The OP's problem, I believe, is that he considers the following
$$k = k_0 + (k_1 \times 10) + (k_2 \times 10^2) + \dots$$To be an integer. Whereas, it is a divergent series and does not represent any finite number. Unless, of course, the ##k_i## are all zero after some finite value ##n##. But, then we only have the equivalent of the finite strings. We don't have every infinite sequence.
 
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Mark44 said:
Every integer has its own place on the number line. Where would these go? The thing is, it's impossible to say, because we don't know how many zeroes are represented by the ellipses (...).
I wrote them from left to right, so you would get what I mean. You say it's impossible to say how many 0's come after 1. But mirror the number and you would see that it doesn't matter. Written from right to left, they would be easy to order and map to room numbers. And thus, for every passenger on the infinitely large bus, you can find a room. One gets off the bus, his name is AABBBBBB... (or 11000000...) He will get room number 3 (the digits are flipped so it reads from right to left). If somewhere along the line of digits more 1's appear, a room number will be given accordingly. Next passengers name is ABABABBBBBB... He will get room number 13 and so on. You will need to repeat the process infinitely of course, but that applies to the other methods too. The point is that it is possible to map these names to room numbers. For every passenger that gets off the bus, you will have to read as many letters as you need to differentiate them and assign a room.
 
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Lars Krogh-Stea said:
I wrote them from left to right, so you would get what I mean. You say it's impossible to say how many 0's come after 1. But mirror the number and you would see that it doesn't matter. Written from right to left, they would be easy to order and map to room numbers. And thus, for every passenger on the infinitely large bus, you can find a room. One gets off the bus, his name is AABBBBBB... (or 11000000...) He will get room number 3 (the digits are flipped so it reads from right to left). If somewhere along the line of digits more 1's appear, a room number will be given accordingly. Next passengers name is ABABABBBBBB... He will get room number 13 and so on. You will need to repeat the process infinitely of course, but that applies to the other methods too. The point is that it is possible to map these names to room numbers. For every passenger that gets off the bus, you will have to read as many letters as you need to differentiate them and assign a room.
The digit 1 followed by infinitely many zeroes is not a (finite) number. We've been through all this so many times.

Veritasium is right. Georg Cantor was right. Every mathematician of the 20th and 21st Centuries was right. And you, I'm afraid, are wrong. No matter how many times you tell us that we cannot count, ultimately you are the one making the elementary errors.
 
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PeroK said:
The digit 1 followed by infinitely many zeroes is not a (finite) number. We've been through all this so many times.

Veritasium is right. Georg Cantor was right. Every mathematician of the 20th and 21st Centuries was right. And you, I'm afraid, are wrong. No matter how many times you tell us that we cannot count, ultimately you are the one making the elementary errors.
I'm not saying that.. I'm saying the digit 1 with infinitely many zeroes before it can be read as a binary number that you can map to room number 1. The digits 11 with infinitely many zeroes before it can be read as a binary number that you can map to room number 3.. And so on. You have already answered yes to the question, when I asked if these combinations would occur in the list. Thats why I said that if you mirror the list, so it reads from right to left, then it's possible to map the passengers to their rooms.
 
Lars Krogh-Stea said:
I'm not saying that.. I'm saying the digit 1 with infinitely many zeroes before it can be read as a binary number that you can map to room number 1. The digits 11 with infinitely many zeroes before it can be read as a binary number that you can map to room number 3.. And so on. You have already answered yes to the question, when I asked if these combinations would occur in the list. Thats why I said that if you mirror the list, so it reads from right to left, then it's possible to map the passengers to their rooms.

I think most conceptions of an infinite string of 1s and 0s does not have any such thing as infinite 0s, and then a 1. There are actual interesting ways to think about ordering in which you could imagine the set of all strings of infinitely digits and then one extra digit, but the canonical representation there is no "last digit".
 
Office_Shredder said:
I think most conceptions of an infinite string of 1s and 0s does not have any such thing as infinite 0s, and then a 1. There are actual interesting ways to think about ordering in which you could imagine the set of all strings of infinitely digits and then one extra digit, but the canonical representation there is no "last digit".
I agree, but there is a first number. That's why I say "mirror the list".

Example: 100...0 will turn into 0...001. And for this practical application of mapping people to rooms, I see that as sufficient data to get the job done.
 
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valenumr said:
You're still trying to interpret the stings of ones and zeros as binary numbers. They are not. They are instances of members of a set. There is a mathematical study of boolean logic where such thoughts apply, but it is not relevant here.
If it works, it should be relevant..
 
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Lars Krogh-Stea said:
Example: 100...0 will turn into 0...001.
Bad example. Where is 100...0 on the number line?
And why not write the latter as just plain 1?

I think we've beaten this horse to death, so I'm closing this thread.
 
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