# 1 is 0.9999999999999............

Discussion in 'Alternative Theories' started by chinglu, Oct 27, 2013.

1. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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I'll get back to you on that. It is the middle of the night for me in Brazil. (I woke up knowing my first post had some errors. but am sleepy again now that I think I have it done.) Try to answer your question with my general room assignment calculation formula while I sleep.

3. ### Aqueous Idflat Earth skepticValued Senior Member

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I doesn't sound like that would fill up the even numbered rooms.

5. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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A personal Cartesian coordinates point (b , p), unique for Passenger number, P#, on Bus number, B#, has been specified.
Welcome (1,37) to the Billy T infinite hotel, more modern (twice faster elevators*), than the old Hilbert Hotel. The empty room numbers are: 2{( S'-1)^2 + O'}. Please go directly to your room: 2{36^2 + 1} = 2,594. That plus your credit card number, extending it as a decimal fraction, will be your internet pass word and will be used for billing.
Towels and sheets are changed every week, and as Mr. Hilton always requested: Please make sure the shower curtain is INSIDE the tub.

Recall, S' is the greatest coordinate of your personal point in the XY plain so your S' is 37. The "square corner guests" (who have equal X & Y point coordinates) have O' = S', but your "Order number", O, is 1 (equal to your X coordinate) as your personnel point was on Y coordinate = S' line and you are not a "corner guest." If your point had been on the line X = S' but not the corner point, then your O would have been (your Y coordinate + S') This is different from the old Hilbert Hotel as we are "positive people" and don't force you to ever subtract numbers larger than 1.

Just to be perfectly clear, (or skip this paragraph if you already are) note the personal point (S' , {S' - 1}) has O = S' +1 or in general personal point (S' , {S' - n}) has "Order number", O = S' +n and of course {S' - n} is never zero or negative as ALL personal points are INSIDE the "first quadrant."

* Twice as many, with half stopping only at even floor numbers, half at odd floors, means only 25% of the "elevator delay" at the old Hilbert hotel so we do charge 5% more per night, but here there is a FREE internet connection in every room! Don't stay in an old, run-down, hotel without internet connection in every room.
Stay at the modern Billy T infinite hotel with immediate room assignments by simple formulae.

Last edited by a moderator: Nov 15, 2013

7. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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That is correct. For example, room number 6 always remains empty. So not only is that assignment method very slow (finding next large prime) it is economically inefficient (leaving room(s?)* empty always.) An infinite number of his "potential guests" will gladly pay 5% more to stay in the Billy T hotel with IMMEDATELY calculated room assignment numbers.

* I'm nearly sure that more than half of the first trillion even number rooms never are used and the "utilization factor" continues to decrease for blocks of larger even number rooms.

Last I heard was that the Monimonika hotel was closed by the court as in bankruptcy proceedings and unable to even find a buyer to re-open it, as it was such a money loser.
(Mortgage cost on ALL rooms but nearly half were always unoccupied).

8. ### MonimonikaRegistered Member

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59
Actually, room #6 would be filled by Passenger 1 of Bus 3. (B3+1)th prime = 4th prime = 7, so 7^P1-1 = 7^1-1 = 7-1 = 6.

But you are correct that there will be an infinite ton of empty rooms such as room 86, since my method does not cover numbers (room# + 1) that have more than one prime when factored. So I cannot fill room 86, since 87 has both prime factors 3 and 29. The only thing my method guarantees is that no room will be double-booked.
Hellooooo, bankruptcy!!

I haven't had time yet to thoroughly go thru your method and revise it down to less confusing wording (passengers really shouldn't have to have a grid in front of them to figure out their room number), and to straighten out what values go to what variables. I will do this tonight when I have more time. Thank you!

9. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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They don't. They just tell the hotel receiving agent* their bus number and passenger number and he (or she) almost immediate tells them that their "personal point" coordinates in the XY plain is (b, p).

We have an infinite number of them so every one gets immediate assistance as they get off the bus. It is not well known, but they all live almost for free in the infinite adjoining annex; but get no salary for helping arrivals. (Their low rent is still an infinite income.) The old Hilbert hotel did not have an annex so the average wait to get help was infinitely long.

Thanks for correct my "room 6 is always free" error. I'm sorry about the bankruptcy, but if you need a job or a place to almost for free live, you can stay in room 0 of the annex and help arriving guest. (Once you understand the room assignment method, that is.) Assuming that works out well, I plan to make you the "chief agent" whose only job is to manage the others. This because you must have great "people skills" to have not gone under years ago. BTW, if you can bring a few dozen of your best cooks with you, please do so. Hilbert has only French cooks and they are good, better than mine, so I must charge slightly less, now to compete with his kitchen.

Last edited by a moderator: Nov 15, 2013
10. ### TrippyALEA IACTA ESTStaff Member

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Assiging the passengers from the first bus is easy - you just assign them to room 2p where p is the passenger number.

If we were dealing with an infinite number of busses with finite numbers of passengers on them, I would say that the room number is 2(n(b-1) +p) - at least I think that's the right formula, I've used it on a spreadsheet to generate a unique identifier for a set of tables that were generated by considering all combinations of a pair of variables (aquifer type, vadose zone type). B is the bus number, n is the number of passengers on the bus, and p is the passenger number. This method, of course, has an obvious drawback when you have an infinite number of passengers, although it will work for all finite numbers.

It seems to me that the easiest way to answer the problem is not with a formula or an algorithm, but to use a booking table and I would lay it out thusly:

Where one of the axes is the bus number and the other is the passenger number, and the numbers in the table body are the room number. Doing it this way will map all of the passengers on all of the busses to all of the available even numbered rooms with a 1:1 correspondence, but, I have no idea is there is a function to describe the mapping, or if there might be an alternative way of doing the mapping that lends itself to a function.

But the problem didn't specifically ask for a function or algorithm, it asked for a method to allocate room numbers, and a booking table is a valid method.

11. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Trippy: see post 763 for simple formulae valid (I think) for infinite number of infinite capacity busses, all full, and arriving simultaneously. If correct there is no need for your infinitely heavy book of tables or buses to have only finite capacity.* Post 763 is a condensed version of post 760, with a minor change to make the formulae more elegant,** but 760 does motivate / explain why it works and leaves no empty rooms. Look at 760 too, if these things are not immediately obvious to you, but be warned here may be a few math errors and typos still in post 760. (I awakened in the middle of the night to correct an error but was still sleepy and knew a better / more compact/ way to tell the solution to Pete's room assignment problem, so did not check 760 more.)

* Neglecting technical problems with overloaded tires, etc. of course, when busses with finite number of tires carry an infinite load.
** Multiplication and addition only required except for one subtraction of 1, once for each passenger.

12. ### TrippyALEA IACTA ESTStaff Member

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Heaven forbid anyone should ever come up with a viable alternative to your answer billy.

Look, honestly. We're dealing with an infinitely large hotel with an infinite number of rooms, and an infinitely large car park filled with an infinite number of busses each of which contains an infinite number of passengers and you're quibbling over the weight of the booking table?

If you're going to try and correct me at least have the grace to provide some constructive criticism like "why don't you try changing the filling order".

13. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Your solution is not general, but limited to only a finite number of passengers on each of the infinite number of buses If I understand your post. You said in post 767:
"If we were dealing with an infinite number of busses with finite numbers of passengers on them. ... B is the bus number, n is the number of passengers on the bus, and p is the passenger number. This method, of course, has an obvious drawback when you have an infinite number of passengers, although it will work for all finite numbers."

Also does it leave any rooms empty? (I think all will fill, but ask with slight concern that the column widths in any real table may need to be finite.) That is why Monimonika's infinite hotel went bankrupt - infinite number of rooms with mortgage cost but no rent.

There is nothing of hostile tone in my post 768. I was only directing your attention to what I think is a simple GENERAL solution to Pete's puzzle. Why are you being so hostile?

Last edited by a moderator: Nov 15, 2013
14. ### TrippyALEA IACTA ESTStaff Member

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Clearly then you don't understand my post.

Look again at the table.

Do you see the ellipses?

What do ellipses mean in this context Billy? They mean 'ad infinitum'.

No Billy. It fills every room, that's the point, it's a 1:1 mapping of passengers to even numbered rooms. What's more, I have an improved solution, which I shall write up momentarily.

15. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Good. I look forward to seeing it. I hope it does not require table column widths approaching infinitely wide columns. I. e. hope you produce a formulae version.

I had not noticed the ellipses, but now that I do, and look more closely at your "looping path" it seems to be a specific version of Someguy1's post 755. If so, you "snaking thru the even numbers" has a clearly indicated path, and his did not, but still I'm concerned about how practical a tabular approach is with infinitely many columns needing extreme width.

16. ### arfa branecall me arfValued Senior Member

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7,832

The table has fairly obvious rules for its construction. The rule(s) means all rows are strictly increasing, and all the columns are too.
It seems the table can grow without limit, and that's without bothering to formalise the method with some formula.

17. ### DinosaurRational SkepticValued Senior Member

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4,885
Quantum Quack: You must be careful of typo’s. Check the following carefully, counting the number of nines in each term.
Code:
10*0.99999 = 9.99990
Minus  0.99999 = 8.99991
Similarly for more nines.
Using the notions of ordinary arithmetic, multiplying a string of nines by ten can be accomplished by shifting the decimal point & appending a zero. Subtracting the original string of nines results in 8.9999 . . . . 1

Others:The following proof is invalid because it uses the notions of ordinary arithmetic, which requires appending a zero when multipying by 10
Code:
x =  0.99999 . . . .
10*x = 9.99999 . . . .
10*x - x = 9
9*x = 9
Ergo: x = 1
Of course the conclusion is at least reasonable, if not valid.

Note that the proof is not valid for a google of nines or a googleplex of nines. How can it be asserted that it is valid for some magic number of nines called an infinite number?

Note also that a reasonable and/or valid conclusion does not imply that a proof is valid.

You must deal with these problems using methods more advanced than the notions of ordinary arithmetic.

Using the limit of a geometric series, it is easy to prove that 0.999999 . . . . . has a least upper bound of one,
implying that asserting it to be equal to one is valid due to resulting in no paradox.

18. ### TrippyALEA IACTA ESTStaff Member

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So. Booking table v2.0

Each passenger has two numbers (p,q) p is the bus number, q is the passenger number.

Here's the new booking table:

Let us consider the diagonal lines on the table, as that is how I have filled the table. We're only considering the '45 degree' diagonals - if that makes sense, and the elements are room numbers. So Diagonal 1 (D1) has the elements {2}. D2 has the elements {4,6}. D3 has the elements {8, 10, 12} and so on and so forth ad infinitum, as I have attempted to indicate with strategicly placed elipses.

The difference between this booking table and the last booking table is that the last booking table reversed the direction of filling with each diagonal, but with this table, the diagonals all fill in the same direction.

Now. Every unique passenger (p,q) lies on one of these diagonals.

The first passenger on bus p is on diagonal p. So the first passenger of bus 7 is on diagonal 7.
Passenger q is member q of the p+(q-1) diagonal so, Passenger 1 bus 7 is the first member of diagonal 7, passenger 2 bus 7 is the 2nd member of the 8th diagonal, passenger 3 of bus 7 is the 3rd member of the 9th diagonal, but, passenger 1 of bus 8 is the first member of diagonal 8, passenger 2 is the second member of the 9th diagonal and so on and so on, ad infinitum.

This brings us to the next step.

The starting number of each diagonal is as follows:

D1 = 2
D2 = 4 (2 + 2)
D3 = 8 (2 + 2 + 4)
D4 = 14 (2 + 2 + 4 + 6)
D5 = 22 (2 + 2 + 4 + 6 + 8)

So the starting number of the k[sup]th[/sup] diagonal, D[sub]k[/sub]is $2+\sum^{n=1}_k{2(n-1)}$

Bringing it all together, for any unique passenger {p,q}

Their unique room number, n, is given by $q + 2 + \sum^{n=1}_{p+(q-1)}{(2(n-1))}$

Assuming i've done everything correctly.

No need to mess about with cartesian planes or infinite tables of infinite weight.

19. ### TrippyALEA IACTA ESTStaff Member

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Thanks. I was stumbling over the rules for formalizing the filling of the table, but I noticed that if I changed the order the table was filled in it great simplified things for me of course Trippy = over tired + hungry + a shade hungover + two whiney children in the background, so, concentration isn't my forte at the moment.

And yeah, that was the idea, I was trying to construct a table that was able to grow without limit in a stirctly increasing way, and counting along rows or columns wasn't working. Then it occure to me to use the diagonals.

20. ### rpennerFully WiredValued Senior Member

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4,833
Simply because any finite string of 9's has an end while no infinite string of 9's has an end.
Thus all infinite strings of 9's have a 9 in every possible place, so
$9.999... - 0.999... = 9$

21. ### Quantum QuackLife's a tease...Valued Senior Member

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23,328
ahh I see your point... and yes I agree..
It has been mentioned earlier that 0.999... = 1 is derived by definition and not calculation again to avoid the issue of the paradox as you mention.
IMO The "paradox" btw exists and is managed in so many ways but can never actually resolve it.
Finitely infinite?

Infinitely finite?

'tis just the rational man's inability to deal or cope with that which is infinite.
To cope with the infinite takes Faith and funnily enough 0.999...=1 has to be accepted on faith that the defining process [placing pseudo finite limits] is adequate.

22. ### TrippyALEA IACTA ESTStaff Member

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Bah! I should have been more careful naming my variables. Oh well. Rooom number r rather than n.

23. ### Quantum QuackLife's a tease...Valued Senior Member

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23,328
but rpenner
if 0.999...= 1
then 9.999... - 0.999... = 8.999...

the problem is using the proof to prove the proof I think...
at this point in the proof formulation you are relying on 0.999... = 0.999... and not 1. yes?
yet you rely on this to prove that 0.999... = 1 and that's a confusing thing IMO