Ten Equals 9.9999.... (new twist!?)

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re...@vax5.cit.cornell.edu

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Aug 4, 1992, 7:57:58 PM8/4/92
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Greetings,

I just found this newsgroup recently and I have been having a great
time reading it. Undoubtedly one of the hottest topics is
10 = 9.9999...., and I kind of hate to bring it up.

First of all, I'm convinced, 10 = 9.9999.... as far as I'm concerned.

But then I remember a long time ago a puzzle that asked,
"If you started at one end of a room and walked half the
distance to the other side of the room, then walked half
the remaining distance, then half the remaining distance
again, ad infinitum, would you ever get to the other
side of the room?"

Well, this sounds like a similar puzzle to 10 = 9.9999....
But for the longest time I believed that you would NEVER get
to the other end of the room; it just seemed obvious.

So, I cracked open my calc textbook and started playing with
geometric series and figured out the problem stated above is
a geometric series a + ar + ar^2 + ... where a = 1/2 and r = 1/2,
which means the series converges to a/(1-r) = (1/2) / (1 - 1/2)
= 1. Now, it's true that the series CONVERGES to 1, so the total
distance walked converges to 1. But does that mean the total
distance EQUALS 1? I really am horrible at series, and I
really never understood convergence and divergence. I don't
even know if that's where my problems lie.

I'm going to stop and ask someone to pick up here, and explain
where my logic is headed, and point out any flaws in my thinking.

I hope this is an interesting twist to this problem, and not a
blatant oversight...?

David S. Rowell
RE...@Vax5.cit.cornell.edu
.sig available upon request

Carl Mueller

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Aug 4, 1992, 9:02:49 PM8/4/92
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In article <1992Aug4.1...@vax5.cit.cornell.edu> re...@vax5.cit.cornell.edu writes:
>Greetings,

>
>I remember a long time ago a puzzle that asked,
>"If you started at one end of a room and walked half the
>distance to the other side of the room, then walked half
>the remaining distance, then half the remaining distance
>again, ad infinitum, would you ever get to the other
>side of the room?"
>
>I cracked open my calc textbook and started playing with
>geometric series and figured out the problem stated above is
>a geometric series a + ar + ar^2 + ... where a = 1/2 and r = 1/2,
>which means the series converges to a/(1-r) = (1/2) / (1 - 1/2)
>= 1. Now, it's true that the series CONVERGES to 1, so the total
>distance walked converges to 1. But does that mean the total
>distance EQUALS 1? I really am horrible at series, and I
>really never understood convergence and divergence. I don't
>even know if that's where my problems lie.
>
>I'm going to stop and ask someone to pick up here, and explain
>where my logic is headed, and point out any flaws in my thinking.
>
>David S. Rowell
>RE...@Vax5.cit.cornell.edu

(1/2) + (1/4) + (1/8) + (1/16) + (1/32) + ... = 1.

The formula you used shows this just fine. However, if you start walking
across the room as in your original problem, moving half way to the far
wall with each step, it will take you forever to get there (assuming that
there is a minimum time required for you to take a step). This does not
mean that you can't cross a room (or that motion itself is impossible),
it simply means that as we move across a room, we cover vanishingly
small fractions of the width of the room in vanishingly small units of time.

Carl Mueller (mue...@math.wisc.edu)

Mike Levis

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Aug 4, 1992, 9:02:12 PM8/4/92
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>First of all, I'm convinced, 10 = 9.9999.... as far as I'm concerned.

A very, very easy way to see that 10 = 9.999... is as follows:
10 = 10/3 + 10/3 + 10/3
= 3.333... + 3.333... + 3.333...
= 9.999...

likewise:
1 = 1/3 + 1/3 + 1/3
= 0.333... + 0.333... + 0.333...
= 0.999...

>David S. Rowell
>RE...@Vax5.cit.cornell.edu
>.sig available upon request


--
====== Mike Levis mle...@ringer.cs.utsa.edu ======
"I believed what I was told, I thought it was a good .--.
life, I thought I was happy. Then I found something (OS/2)
that changed it all..." -- Anonymous, 2112 ~--~

Bill Taylor

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Aug 5, 1992, 1:50:51 AM8/5/92
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>But then I remember a long time ago a puzzle that asked,
>"If you started at one end of a room and walked half the
>distance to the other side of the room, then walked half
>the remaining distance, then half the remaining distance
>again, ad infinitum, would you ever get to the other
>side of the room?"

This reminds me of a former calculus lecturer who used to introduce limits,
by saying..

"We'll get all the girls here to line up along one wall, and all the boys to
line up along the opposite wall. Now after 5 seconds, we'll halve the distance
between them; and after another 5 seconds, we'll halve the remaining distance;
and another 5 and halve again; and so on, and so on.
Now clearly, the two lines of bodies will never quite meet; but after a
certain time, they'll be close enough for all practical purposes........"

David Karr

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Aug 5, 1992, 11:53:24 AM8/5/92
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>"If you started at one end of a room and walked half the
>distance to the other side of the room, then walked half
>the remaining distance, then half the remaining distance
>again, ad infinitum, would you ever get to the other
>side of the room?"

That depends on how long it takes you to halve the distance each time.
If you take 1 second for each halving, then in fact you never reach
the other side, as long as you adhere to the mathematical model you
described. (I believe when you get down to some subatomic scale,
quantum effects take over and you are no longer able to walk
precisely half the distance each time :-) .)

On the other hand, suppose the room is 1 meter long (for simplicity)
and you walk at a rate of exactly 1 meter per second. Then in 1/2
second you have walked halfway, the next halving takes only 1/4 second,
the next takes 1/8 second, and so forth. It takes you exactly one
second to traverse the room at this rate. (Of course, if you
believe a certain man named Zeno, then you still never get there :-) .)

>So, I cracked open my calc textbook and started playing with
>geometric series and figured out the problem stated above is
>a geometric series a + ar + ar^2 + ... where a = 1/2 and r = 1/2,
>which means the series converges to a/(1-r) = (1/2) / (1 - 1/2)
>= 1. Now, it's true that the series CONVERGES to 1, so the total
>distance walked converges to 1. But does that mean the total
>distance EQUALS 1? I really am horrible at series, and I
>really never understood convergence and divergence. I don't
>even know if that's where my problems lie.

I believe that the sum of the infinite series is DEFINED to be the
limit of the partial sums, if they converge to a limit (or it is
defined in some other equivalent way). You cannot define the sum in
terms of repeated applications of binary (i.e. two-operand, not base
2) addition, because any number of applications of binary addition
produce only a finite series.

>David S. Rowell
>RE...@Vax5.cit.cornell.edu
>.sig available upon request

-- David Karr (ka...@cs.cornell.edu)


Mark Christensen

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Aug 6, 1992, 10:21:47 AM8/6/92
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Okay ladies and gentlemen, I've decided to come out and play
again, despite my last encounter with this newsgroup. Mostly
positive, some negative, but lots of followup puzzles.
Please forgive me if this next puzzle has either been submitted
before or is old to you. And please don't get all wrapped
up in my wording. I'm not an English major.

Here goes:

An alien comes down to earth with the intention of bringing
back all the information contained in the Encyclopedia
Britannica. He has only a metal rod (no special technology,
just your simple metal rod and it is short) and a method
of inscribing on it. He is allowed to make ONLY ONE (1...EIN)
mark on this rod. He has no special technology so please don't
come up with any bizarre scenarios. Is it THEORETICALLY possible
for him to make only one mark on this metal rod to transcribe
all the data in the encyclopedia? Is it PHYSICALLY possible?
Why or why not? Let's assume that the mark must be no longer
than the circumference of the metal rod (which by the way is round).
That way I don't get a alien writing cursive in very tiny letters.

Bart Goddard

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Aug 6, 1992, 11:12:26 AM8/6/92
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Distribution: rec

In article <53...@transfer.stratus.com> ma...@eomer.hw.stratus.com (Mark
Christensen) writes:
> [...]


>
> An alien comes down to earth with the intention of bringing
> back all the information contained in the Encyclopedia
> Britannica. He has only a metal rod (no special technology,
> just your simple metal rod and it is short) and a method
> of inscribing on it. He is allowed to make ONLY ONE (1...EIN)
> mark on this rod. He has no special technology so please don't
> come up with any bizarre scenarios. Is it THEORETICALLY possible
> for him to make only one mark on this metal rod to transcribe
> all the data in the encyclopedia? Is it PHYSICALLY possible?
> Why or why not? Let's assume that the mark must be no longer
> than the circumference of the metal rod (which by the way is round).
> That way I don't get a alien writing cursive in very tiny letters.

Using base 26 (52? higher? how ever many characters one needs), treat
the data in the encyclopedia a (rather long) string of digits. Thus it
represents a (rather large) number, N. Let
x=arctan(N)*(circumference). Make a mark x long around the rod (an arc
of a circle in a plane perpendicular to the axis of the rod.) He can
use the distance d of the mark from the end of the rod to record
another string of information (such as the base used or other decoding
instructions) in the same manner.

So, I guess it's theoretically possible, but it's going to be hard to
measure the last part of the scratch on that last electron while it's
still spinning......

Bart

David Karr

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Aug 6, 1992, 11:44:33 AM8/6/92
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In article <1992Aug6.1...@cs.rose-hulman.edu> god...@NeXTwork.Rose-Hulman.Edu (Bart Goddard) writes:
>Using base 26 (52? higher? how ever many characters one needs), treat
>the data in the encyclopedia a (rather long) string of digits. Thus it
>represents a (rather large) number, N. Let
>x=arctan(N)*(circumference).

An even simpler method: Place a decimal point to the left of your number.
Now it represents a base B fraction between 0 and 1. (In fact it is a
rational number.) Let that number be x, and supposing the rod is 1 meter
long, make a mark x meters from the end. If you can't tell one end from
the other, make a mark x/2 meters from the end. To read the data back
you measure the distance to the mark and expand it to a base B fraction.

Of course making a mark at a spot measured to arbitrary precision requires
not only "special technology" but suspension of the known laws of physics...
;-)

In practical terms, I suspect the answer is "can't be done." Not if you're
limited to a simple mark on a reasonably small bar and an electron microscope
under which to view it.

-- David Karr (ka...@cs.cornell.edu)

Mark Phaedrus

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Aug 6, 1992, 1:22:55 PM8/6/92
to
In article <53...@transfer.stratus.com> ma...@eomer.hw.stratus.com (Mark Christensen) writes:
>An alien comes down to earth with the intention of bringing
>back all the information contained in the Encyclopedia
>Britannica. He has only a metal rod (no special technology,
>just your simple metal rod and it is short) and a method
>of inscribing on it. He is allowed to make ONLY ONE (1...EIN)
>mark on this rod. He has no special technology so please don't
>come up with any bizarre scenarios. Is it THEORETICALLY possible
>for him to make only one mark on this metal rod to transcribe
>all the data in the encyclopedia? Is it PHYSICALLY possible?
>Why or why not? Let's assume that the mark must be no longer
>than the circumference of the metal rod (which by the way is round).
>That way I don't get a alien writing cursive in very tiny letters.


Well, the solution I've always heard is this: Just come up with some
scheme for converting the contents of the Britannica to n-digit integers.
(For the text you could ASCII codes, for instance; for illustrations, use GIF
or something.) Now, concatenate all these numbers together, in the order they
appear in the text, to produce one very long decimal number. (For example,
the start of this paragraph would be .009087101108108044032116... and so on.)
TAB W e l l , SPC t
Now, once he's reduced Britannica to a single decimal number, he gets
an extremely accurate ruler and an extremely accurate saw, measures precisely
that far from one end of the bar (in terms of a percentage of the total length
of the bar; so for the number above, he'd measure out precisely .9087101...% of
the total length of the bar), and make a mark at precisely that point. (Of
course, since the alien can't make an extra mark to tell which end of the bar
was measured from, he had better keep careful track.) Then, when he gets home,
he just gets out another extremely accurate ruler, measures the precise
distance of the mark down the bar again, reconstructs the original decimal
number, and reconstructs the Britannica from that. (Since the measurement is
in terms of a percentage of the length of the bar, temperature-indiced
expansion or contraction of the bar won't affect the result, as long as the
whole length of the bar expands or contracts equally.)
In theory, this is a workable scheme. In practice, even with a bar the
length of the planet, no matter how precise the marking and measuring tools,
I would wager that quantum effects would prevent the mark from being made or
measured precisely enough to accurately record and decode a number with enough
digits to store the entire Britannica.
--
\ Mark Phaedrus, Computer Science Major, Univ. of Washington, Seattle, WA \*\
*\ Work: phae...@cs.washington.edu Play: phae...@u.washington.edu \*\
\*\ Cute .sig quote still under alpha-testing; planned for release in 1993. \*\
- Hi! I'm an anti-virus utility! Install me in your .signature right away! -

Kenneth Tolman

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Aug 6, 1992, 7:28:29 PM8/6/92
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>Of course making a mark at a spot measured to arbitrary precision requires
>not only "special technology" but suspension of the known laws of physics...
>In practical terms, I suspect the answer is "can't be done." Not if you're
>limited to a simple mark on a reasonably small bar and an electron microscope
>under which to view it.

It is even worse than this... a number of problems follow:

1) The Hiesenberg uncertainty principle will limit us to Plancks length,wich
will allow for a great amount of information but it is limited.

2) Just recording an encyclopedia is no good- how will the alien understand
it? The alien must presumably learn the language, all the customs and
background information too- does this inforamtion not count? If he can
learn all that, why not put the whole encyclopedia in his head too? Then
he can throw the rod out. This is a major problem- and cannot be quickly
dismissed, it involves "assumed information" which is very important and is
not accounted for here. Otherwise, he could say that the number 157384
represented the code for the whole encyclopedia set and that then he could
decode it when he got home. Clearly, that is ridiculous.

m...@here.now

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Aug 7, 1992, 2:47:21 AM8/7/92
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In article <1992Aug6.1...@hellgate.utah.edu>, tolman%asylum.cs...@cs.utah.edu (Kenneth Tolman) writes:
|> >Of course making a mark at a spot measured to arbitrary precision requires
|> >not only "special technology" but suspension of the known laws of physics...
|> >In practical terms, I suspect the answer is "can't be done." Not if you're
|> >limited to a simple mark on a reasonably small bar and an electron microscope
|> >under which to view it.
|>
|> It is even worse than this... a number of problems follow:
|>
|> 1) The Hiesenberg uncertainty principle will limit us to Plancks length,wich
|> will allow for a great amount of information but it is limited.

[Second part deleted]

Is it possible to form a series that generates the decimal, sum the series,
and then use that result?

__
i.e. : 0.27272727

Can be put into a series and summed, and you get 3/11.

Right?

Nazman
<nas...@rpi.edu>

Mark Phaedrus

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Aug 7, 1992, 4:43:23 AM8/7/92
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In article <339...@rpi.edu> nas...@rpi.edu writes:
[concerning the alien-rod problem; encoding the encyclopedia by making an
extremely long decimal number, and marking that far down the bar; technical
impossibilities in implenting this]

>Is it possible to form a series that generates the decimal, sum the series,
>and then use that result?
>
> __
>i.e. : 0.27272727
>
>Can be put into a series and summed, and you get 3/11.
>
>Right?

I'm not sure where you're going with this, but I don't believe it solves
the problem. The encoding of the encyclopedia, at least by the method we
described, is not going to produce an endlessly-repeating series; since the
encyclopedia has a finite number of characters, and each character generates
a finite number of digits, the resulting decimal number will have a finite
length. Yes, you can then convert that decimal number to a fraction; any
finite decimal can be converted to a fraction. But having it in fractional
form instead of decimal form doesn't make it any easier to encode it on the
bar. You still have only one mark available, so you still have to mark off
exactly that fraction of the bar, just as you had to mark off exactly that
decimal portion of the bar before. So you face the same problems of
imprecision and quantum uncertainty that you did before. Even if, through some
freak of chance, the decimal converts to some "easy" fraction like 3/8, that
still doesn't help; it's as difficult to exactly encode 3/8 on the bar with a
single mark as it is to exactly encode some fraction with a much more
complicated numerator and denominator. (The alien doing the measuring at the
other end has no way of knowing that it's supposed to be an "easy" fraction
like 3/8, and not some much more complicated fraction that happens to be
extremely close to 3/8; if the aliens are allowed to remember additional facts
about the fraction like the fact that it's simple, the bar becomes irrelevant,
since they may as well just remember the fraction in the first place, or even
the encyclopedia itself.)
Consider this: Let's say that we have a long enough bar, and precise
enough marking and measuring tools, that we can place our mark in any of a
decillion (10^33) uniquely identifiable positions. This is probably quite
a few orders of magnitude better than current technology and quantum
uncertainty would let us get away with, but let's say we could anyway. Now
consider that this level of precision still won't even let us store as
much information as 16 bytes of computer memory, which can store over 10^38
unique values. This means that if you can come up with a way to compress
an arbitrary encyclopedia onto this rod, you'll have come up with a way to
compress an arbitrary encyclopedia-sized file into 16 bytes of storage.
You will be very, very, *very* rich if you can slap a patent onto this scheme.
But I wouldn't hold my breath in the meantime. :)

chris kiick

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Aug 12, 1992, 9:23:48 PM8/12/92
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ka...@cs.cornell.edu (David Karr) writes:
>In article <1992Aug6.1...@cs.rose-hulman.edu> god...@NeXTwork.Rose-Hulman.Edu (Bart Goddard) writes:
>>Using base 26 (52? higher? how ever many characters one needs), treat
>>the data in the encyclopedia a (rather long) string of digits. Thus it
>>represents a (rather large) number, N. Let
>>x=arctan(N)*(circumference).
>Of course making a mark at a spot measured to arbitrary precision requires
>not only "special technology" but suspension of the known laws of physics...
>;-)
>In practical terms, I suspect the answer is "can't be done." Not if you're
>limited to a simple mark on a reasonably small bar and an electron microscope
>under which to view it.

Given that we can represent the enclyclopedia as a large number or fraction,
what if we used a REALLY compact form of numeric notation?

Something like
(b!+c) b
a^ +d Or a^ ![c]+d

(note: x![n] is meant to represent x! n times. ie x![4] = x!!!!)

Which with four carefully chosen numbers could represent a HUGE number.
Then we encode a as the distance from the end of the rod, b as the
lenght of the line, c as the angle of the line from the perpendicular,
and d as the width of the line. We can re-arrange these to suit.
We could even squeeze a couple of other numbers out by using a curved
line and noting the arc-radius, or measuring the depth of the mark.
The numbers themselves can be re-interpreted as convenient ratios.

This just might make the physical measurments possible.

I am assuming that since we have the computational capability to transcribe
an entire encyclopedia into one number in a reasonable time, that we
can also determine the four numbers needed to fit the equation.
--
***********************************************************
Chris J. Kiick | work phone: (205) 730-6171
Programmer at Large | Email: ingr!b11!flinx!kiick
#include <disclaimer.h> | New!!: ki...@flinx.b11.ingr.com
Ever try to boot an Indigo?
***********************************************************

David Karr

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Aug 13, 1992, 1:48:32 PM8/13/92
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In article <1992Aug13.0...@b11.b11.ingr.com> ki...@b11.b11.ingr.com (chris kiick) writes:
>Given that we can represent the enclyclopedia as a large number or fraction,
>what if we used a REALLY compact form of numeric notation?
>
>Something like
> (b!+c) b
> a^ +d Or a^ ![c]+d
>
>(note: x![n] is meant to represent x! n times. ie x![4] = x!!!!)
>
>Which with four carefully chosen numbers could represent a HUGE number.

I take it you mean four carefully choses SMALL numbers. So each of
your four numbers lies in a relatively small range, say 1..100 for
example. (So to represent these numbers accurately on the bar with
one mark, you need to be able to measure length along the bar to
1/(100^4), i.e. 1 part in 100 million, or 1/100th of a micron if the
bar is one meter long.)

Then with these four variables you can represent a total of 100 million
different numbers. Suppose the encyclopedia could be coded in a string
of just one million digits. (That's a very small encyclopedia--just
500 pages of 2000 characters each if you encode one character per digit.)
Now there are 10 to the 1,000,000th power different possible encodings
that the encyclopedia might have. The alien has a formula that can
generate a short encoding for 10 to the 8th power of these. So there
is only a 1 in 10 to the 999,992nd power chance that the encoding of the
encyclopedia is one of the few that the alien can represent.

You suggest that there are many different formulas that you could use,
but if you allow the alien to use any of the 10 to the whatever power
(at least 10 to the 999,992nd power) different formulas that are
needed to "hit" all the large number values, he or she also needs to
encode which formula was used. Now just recording a number from 1 to
10 to the 999,992nd power to specify which formula is used is about as
hard as recording the encyclopedia in the first place. And any other
scheme for specifying the formula, e.g. an encoding of its symbols, is
bound to be even worse.

Essentially what you are trying to do is to use data compression to
reduce an arbitrary volume of text the size of an encyclopedia down to
just a few characters. There are texts for which this might be done,
for example the fictitious volume whose text is, "See Figure 1 for the
first 10,000,000 decimal digits of pi," the rest of the book being
occupied by Figure 1. Now you see you could (in principle) reproduce
the entire text of this 10,000,000-plus character book from the fewer
than 200 characters it took me to describe it. You will have a hard
time convincing me that any encyclopedia is such a text.

-- David Karr (ka...@cs.cornell.edu)


Abreu

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Aug 17, 1992, 5:53:12 PM8/17/92
to
In <53...@transfer.stratus.com> ma...@eomer.hw.stratus.com (Mark Christensen)
writes:
> An alien comes down to earth with the intention of bringing
> back all the information contained in the Encyclopedia
> Britannica. He has only a metal rod (no special technology,
> just your simple metal rod and it is short) and a method
> of inscribing on it. He is allowed to make ONLY ONE (1...EIN)
> mark on this rod. He has no special technology so please don't
> come up with any bizarre scenarios. Is it THEORETICALLY possible
> for him to make only one mark on this metal rod to transcribe
> all the data in the encyclopedia? Is it PHYSICALLY possible?
> Why or why not? Let's assume that the mark must be no longer
> than the circumference of the metal rod (which by the way is round).
> That way I don't get a alien writing cursive in very tiny letters.

The whole contents of the Encyclopaedia Britannica can be encoded by one
unique number. The number is arrived at by devising a sort of Godel
numbering for the alphabet plus all other characters printed, then
encoding the contents to finally arrive at the encyclopaedia's Godel
number.

Variations on this could be used to encode different kinds of
information; for example, we could extend the coding strategy to include
graphics.

It turns out that this number would be BIG (real BIG). Given that the
rod is finite, and a short one at that, the engraving could be done as
a fraction of some unity: a centimeter, say. The accuracy necessary
to make the engraving would be far beyond the physical diameter of an
atom.

So, although theoretically possible to do it, it is impractical.
Hamilton

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