HDTenEightyP/NetNews huggingface dataset:
From: deghare@daisy.uwaterloo.ca (Dave Hare)
Subject: Re: Ten equals 9.9999....
Organization: University of Waterloo
Date: Tue, 28 Jul 1992 12:06:29 GMT

In article <1992Jul27.150032.12515@phillip.edu.au> t9114145@phillip.edu.au writes:
>In article <Bry202.4BF@watdragon.waterloo.edu>, deghare@daisy.waterloo.edu (Dave Hare) writes:
>> In article <1992Jul25.145154.12498@phillip.edu.au> t9114145@phillip.edu.au writes:
>>>In article <1992Jul24.102553.17411@pasteur.Berkeley.EDU>, tjoa@cory.Berkeley.EDU (Richard Tjoa) writes:
>>>
>>>> Ok, here's something interesting...
>>>> 9.9999...  (nine point nine repeating)  =   10  (ten)
>>>
>>>There's the error. 9.9999... does not equal 10, it approaches 10.
>> 
>> No, it equals 10.  This type of question has subtleties that too often
>> are too easily dismissed.  Consider:
>
>Alright, if 9.999... = 10, then why isn't .33333... exactly 1/3?
>
>I don't see how they are equal. I know that 9.999... with infinite 9's
>is very close to 10, but that .0000...0001 is what you need to make
>exactly 10.

Um, how many 0's are there before that 1?

>>     - What exactly is a non-terminating decimal?
>
>Let's keep it simple - we're talking about recurring 9's.

Even so.  What do you think 9.999999.... means?

>>     - How will you determine that two numbers are *not* equal
>
>Why does it need a formal mathematical proof?

Everything in mathematics needs a formal mathematical proof.  Otherwise
it's not not mathematics.

>>     - If x is a non-terminating decimal, how do you define 10*x?  (Recall
>>       that for ordinary multiplication, the computation proceeds from 
>>       least significant digit to most significant digit, but x has no
>>       least significant digit.)  And how do you define the other arithmetic
>>       operations?
>
>10*x is easy. Shift the decimal place one to the right, giving
>99.9999....   which is still not equal to 10. (or 100 either.)

Prove both of those statements (that the result of multiplying by 10 is
99.999.... and that the result is not equal to 100).

>> The problem is that we are cavalierly taught in grade school that if a
>> non-terminating decimal is periodic, then we just multiply it by 10^{the
>> period}, subtract the original, divide both sides by 10^{the period}-1
>> and reduce to lowest terms, and we have that number as an exact rational.
>> 
>> These steps require justification.  HOWEVER, that justification is possible,
>> and once it is done, then statements such as 9.99999... = 10 make perfect
>> sense.

All this poster has done is to establish my point that while it is true that
9.9999.... = 10, it is not trivial to write down a careful and complete 
proof of this fact.

Dave hare
Newsgroups: rec.puzzles
From: jimhsu@athena.mit.edu (Jimmy M.D. Hsu)
Subject: To all who responded.
Organization: Massachusetts Institute of Technology
Date: Thu, 30 Jul 1992 22:01:56 GMT

There were lots of great responses to my
Killing Joke thread.  I realize that I
oversimplified the original rules.  I tried
to keep things easy, when I should have been
more explicit.  Chalk that one up to
inexperience.  Anyway, I like how many of
the respondents went ahead and made assumptions
about the rules or listed the possible options
instead of just saying they weren't sure if it
was X or Y that I meant.  Good Job to all!


----
Jimmy M.D. Hsu

The one.  The only.
THE MAN YOU TELL YOU GRANDKIDS ABOUT
The original James Amdee.  :-)
(anyone know what Amdee stands for?)
From: fxsdb@acad3.alaska.edu (Scott D. Berry)
Newsgroups: rec.puzzles
Subject: Re: 3 puzzles
Date: 27 Jul 92 23:26:54 GMT
Distribution: rec.puzzles
Organization: University of Alaska Fairbanks

In article <1992Jul27.154738.4480@cs.cornell.edu>, karr@cs.cornell.edu (David Karr) writes:
> In article <BrxA8s.Ew1@news.cso.uiuc.edu> wyg9633@uxa.cso.uiuc.edu (Wooed Sun) writes:
>>karr@cs.cornell.edu (David Karr) writes:
>>>Oh.  The problem is your question was not an English sentence.  I thought
>>>you meant, "What is the first number in alphabetical order, when all numbers
>>>are spelled out [in English --sorry for omitting this condition before]?"
>>
>>billion?
> 
> No, there is no number I would spell out as "billion."  I would write
> "one billion."

Picky, picky.  How about "a billion"?
> 
> -- David Karr (karr@cs.cornell.edu)

   -Scott (fxsdb@acad3.alaska.edu)
From: t9114145@phillip.edu.au
Newsgroups: rec.puzzles
Subject: Re: Ten equals 9.9999....
Date: 28 Jul 92 16:45:31 GMT
 <1992Jul25.145154.12498@phillip.edu.au> <Bry202.4BF@watdragon.waterloo.edu> <1992Jul27.150032.12515@phillip.edu.au>
Organization: RMIT (Bundoora Campus)

In article <1992Jul27.150032.12515@phillip.edu.au>, t9114145@phillip.edu.au writes:
> In article <Bry202.4BF@watdragon.waterloo.edu>, deghare@daisy.waterloo.edu (Dave Hare) writes:
>> In article <1992Jul25.145154.12498@phillip.edu.au> t9114145@phillip.edu.au writes:
>>>In article <1992Jul24.102553.17411@pasteur.Berkeley.EDU>, tjoa@cory.Berkeley.EDU (Richard Tjoa) writes:
>>>
>>>> Ok, here's something interesting...
>>>> 9.9999...  (nine point nine repeating)  =   10  (ten)
>>>
>>>There's the error. 9.9999... does not equal 10, it approaches 10.
>> 
>> No, it equals 10.  This type of question has subtleties that too often
>> are too easily dismissed.  Consider:
> 
> Alright, if 9.999... = 10, then why isn't .33333... exactly 1/3?

After consulting a maths lecturer I'll admit I'm wrong. It still seems
right though, after working with devices accurate to a finite number
of decimal places.
 
>> 
>> The problem is that we are cavalierly taught in grade school that if a
>> non-terminating decimal is periodic, then we just multiply it by 10^{the
>> period}, subtract the original, divide both sides by 10^{the period}-1
>> and reduce to lowest terms, and we have that number as an exact rational.
>> 

My teachers used to love saying "don't believe calculators ever! They will
tell you that things like 9.999... = 10!" :-)

>> These steps require justification.  HOWEVER, that justification is possible,
>> and once it is done, then statements such as 9.99999... = 10 make perfect
>> sense.
>> 
>> Dave Hare
> 
-- 

Adam Eberbach      		| The clever men at Oxford
t9114145@phillip.edu.au         | Know all that there is to be knowed.
Computer Science student, 	| But they none of them know one half as much
Phillip Institute of Technology.| As intelligent Mr. Toad!
Newsgroups: rec.puzzles
From: ttobler@unislc.uucp (Trent Tobler)
Subject: Re: Ten equals 9.9999....
X-Newsreader: Tin 1.1 PL4
Organization: Unisys Corporation SLC
Date: Wed, 29 Jul 1992 20:20:08 GMT

ajpierce@med.unc.edu (Andrew Pierce) writes:
: In article <1992Jul29.000030.4839@fid.morgan.com> sethb@fid.morgan.com (Seth Breidbart) writes:
: >If two numbers are different, then there's another number between
: >them.  (Actually, a lot of such numbers.)
: 
:    I agree with you but it's not that simple.  Your statement needs
: justification.  Mathematicians:  is this an axiom of number theory for
: reals?

From the 13 axioms of the real number system:  (I hope I can remember all
of them, it has been a few years...)

in the following, a, b, and c are all real.

1) a + b = b + a
2) a * b = b * a
3) a + ( b + c ) = ( a + b ) + c
4) a * ( b + c ) = ( a * b ) + ( a * c )
5) a + 0 = a
6) a * 1 = a

7) for every a and b, there exists c s.t. a + c = b  :==  c = b - a
8) for every a not zero and b, there exists c s.t. a * c = b  :==  c = b / a

9) for a and b, exactly one of the following holds:
	a = b
	a < b
	b < a

10) a < b and b < c implies a < c
11) a + c < b + c
12) a * c < b * c for 0 < c

13) let S be any set of real numbers, and T be the set of all numbers
     x with x <= y for all y in S.  If T is non empty, then there exists
     one element p in T, such that p <= q for every q in T.   (or,
     if you prefer, use one of the equivilant theorems)

(If you have not taken a number theory course, or read any number theory
 books, the first 12 define two number systems, rationals and reals..
 #13 restricts it to the reals.  Well, actually, we may have to have one
 more, and that is that 0 < 1)


I assume you already can prove the common arithmetic properties...

Suppose we have two numbers, a and b, with a < b.  Prove there exists
some number c such that a < c < b.

    a < b
    a + a < b + a                   and   a + b < b + b
    ( a + a ) / 2 < ( b + a ) / 2   and   ( a + b ) / 2 < ( b + b ) / 2
    a < ( a + b ) / 2               and   ( a + b ) / 2 < b
    let c = ( a + b ) / 2.
    a < c                           and  c < b
    a < c < b
    Therefore, for any two distinct numbers a and b, there exists a number
      c such that a < c < b.

Is that formal enough for you? ;)

Note, this applies to both the system of rationals, and reals.. we don't
invoke axiom #13 at all.

--
  Trent Tobler - ttobler@unislc.slc.unisys.com
Newsgroups: rec.puzzles
From: radcliff@csd4.csd.uwm.edu (David G Radcliffe)
Subject: Cyclic combinations
Organization: UW-Milwaukee Math Department
Date: Thu, 30 Jul 1992 03:30:33 GMT

This problem was inspired by James Amdee's "Nine spots" puzzle.

You have a circle with n spots on the perimeter, equally spaced.
In how many ways can you cover the spots with r red buttons and
b blue buttons, where r + b = n ?  As in the original problem,
two patterns are considered identical if they differ by a rotation.

Please do not interpret this problem so as to be trivial.  If I
had stated this problem in a strictly formal manner, it would be
unintelligible to large numbers of readers.  I am aware that you
can cover all of the spots with one sufficiently large button, 
for example.

--
David Radcliffe                     
radcliff@csd4.csd.uwm.edu          
Newsgroups: rec.puzzles
From: andsol@is.rice.edu (Andrew J. Solberg)
Subject: Re: Spacewarp wonders 
Organization: Rice University
Date: Thu, 23 Jul 1992 20:07:27 GMT

In article <1992Jul23.133333.4885@hellgate.utah.edu>, tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:

Marvelous puzzles.

|>   Your spaceship has been sucked through a spacewarp, and then crashed
|> onto a place.  The place is foggy, so one can not see more than a few
|> kilometers clearly.  There is always a fairly bright light here.
|> Also, you have access to whatever science gear you might like, as it
|> was aboard your craft.  There are lots of life forms, all of them huge.
|> 
|> 1) How would you be able to tell if the planet was a sphere, or a nearly flat
|>    plane extending in all directions?

Set up a radar beacon at the crash site.  Bring an odometer with me and walk
to a point exactly 1 km from the beacon.  Time the radar send-and-return time
to the beacon.  Now walk in a circle around the beacon, using the radar signal
to stay at the same radius.  Measure the distance travelled.

On a spherical planet, the circumference of the circle as measured by the 
odometer will be less than 2*pi km.  Assuming a high degree of accuracy and
a small enough radius of curvature, we can use spherical geometry to find
the radius of the planet.

|> 
|> 2) How would you be able to tell if you had travelled forwards or backwards
|>    through time 5 billion years or more?

If the universe's background radiation can penetrate the fog (gamma rays can
get through just about anything) we can measure the intensity of the night
sky's 'heat'.  This will tell us how far from the beginning of the universe
we were, assuming we are still in the same universe.

|> 
|> 3) How would you be able to tell if you and your space ship were somehow
|>    shrunken in passage?

Assuming that the laws of physics do not change, the Bohr radius of an
atom would be immutable.  There are a number of experiments we can devise
to find the Bohr radius accurately.  We then compare the 'found'
radius to the known value to find whether we have changed in size.


-- 
Andrew Solberg          |"If I were your wife I'd poison your tea!" 
Undying University Mooch|              Anon. Outraged British Woman
andsol@owlnet.rice.edu  |"And if I were your husband I'd drink it!"
Phone:713-529-8627      |              Winston Churchill

			
Newsgroups: rec.puzzles
From: radcliff@csd4.csd.uwm.edu (David G Radcliffe)
Subject: Re: Something to look at.
Organization: University of Wisconsin - Milwaukee
Date: Wed, 29 Jul 1992 06:42:38 GMT

jimhsu@athena.mit.edu (Jimmy M.D. Hsu) writes:
 
: Suppose you had a pattern like this:
: 
:         O   O
:      O         O
:           O
:      O         O
:         O   O
: 
: With nine points, one in the middle.
: How many different ways are there
: of placing four red buttons, four
: blue buttons, and one white button
: on the nine spots.  If you can rotate
: the pattern to get a new one, it does
: not count as "different."  Is there
: a generic formula for x red, y blue,
: and z white with x+y+z=9?  How about
: for more points in the circle?

Case 1:  Blue button in center.

  Place the white button anywhere on the outside.  Moving clockwise
  from the white button, we place 4 red and 3 blue buttons.  This
  gives C(7,3) = 35 distinct patterns.

Case 2: Red button in center.

   Likewise, 35 distinct patterns.

Case 3: White button in center.

   There are C(8,4) = 70 ways of placing 4 red and 4 blue buttons
   on the outside.  C(4,2) = 6 of them are symmetric with respect to
   rotation by 180 degrees.  These 6 configurations correspond
   to 2 distinct patterns: rrBBrrBB and rBrBrBrB.

   There are 64 nonsymmetric configurations, which are generated
   by 64/8 = 8 distinct patterns.

Thus, there are 35 + 35 + 2 + 8 = 80 distinct patterns.

--
David Radcliffe                                
radcliff@csd4.csd.uwm.edu            
Newsgroups: rec.puzzles
From: dwr2560@zeus.tamu.edu (RING, DAVID WAYNE)
Subject: Re: Killing King (variation)
News-Software: VAX/VMS VNEWS 1.41    
Organization: Texas A&M University, Academic Computing Services
Date: Wed, 29 Jul 1992 03:38:00 GMT

jimhsu@athena.mit.edu (Jimmy M.D. Hsu) writes...
>        Brian, being very clever, decides to
>create a variation that is more difficult to
>compute probabilities with.  He decides to make
>the King the death card, not the Joker.  And,
>the first King does not kill, only the second!

>        Now, what is the best strategy?  What
>is a generic formula for computing probability?

I assume the second king that _either_ of them gets is the death card.
Then taking a fraction f of the deck has a probability 6f^2 of instant
death (for two kings). This is a crude approximation, good for small f.
The purpose of having more cards than your opponent is to force him
to draw once the first king shows up. Since you can force him to draw
f*N cards, you have a probability of 3f of forcing him to draw the second
king. (same crude approximation). If neither occurs, then the game is even,
so you want to maximize 3f-6f^2 which occurs at f=1/4. Note that the 
approximations are then rather poor.

Dave Ring
dwr2560@zeus.tamu.edu
From: markc@eomer.hw.stratus.com (Mark Christensen)
Newsgroups: rec.puzzles
Subject: Re: 3 puzzles
Date: 29 Jul 92 16:22:26 GMT
Organization: Stratus Computer, Inc.

In article <1992Jul27.145517.7957@cbfsb.cb.att.com> mmokhtar@attmail.com writes:
>In article <1992Jul26.134105.13917@ens.fr> courtes@trimaran.ens.fr (Francois Courtes) writes:
>>>>>	What is the first number, when spelled out (e.g. O-N-E) is
>>>>>in alphabetical order?
>>>>40  (F-O-R-T-Y)
>>In French the first one is 2 (D-E-U-X).
>>Is there a language in which 1 is so ?
>
>  Eins in German.
>
>Mazen Mokhtar

Okay already!  A guy tries to be nice and pose a few simple
puzzles for the enjoyment of all, and all I get is grief about
sentence structure and technicalities about what language.
Man o man.  Is this newgroup read by a bunch of eggheads or
something.  Everyone tries to find a way around giving the
real answer.  

I'll think twice before offering you guys any entertainment.
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Newsgroups: rec.puzzles
Subject: Re: Easy prob. question
Date: 25 Jul 92 00:28:07 GMT
Reply-To: carl@SOL1.GPS.CALTECH.EDU
Organization: HST Wide Field/Planetary Camera

In article <1992Jul24.170441.20263@hellgate.utah.edu>, tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
>>This looks to be an easy one:
>>
>>Q: Two cards are picked at random from a complete deck of cards. What
>>   is the probability that they are of the same suit?
>
>You pull out one card.  There are 12 left of that suit, so it is 12/51 or
>12/53 depending on what type of deck you have.

For a deck without jokers, the 12/51 looks right.  But for the deck with
jokers, how do you treat the jokers?  If you treat them as a suit in
themselves, then you've got a 2/54 chance of picking a joker as the first card
and a 52/54 chance of picking a card from a suit.  If you picked a joker, then
the chances of picking the other joker as the second card are 1/53; if you
picked a card from a suit, then the odds of picking another card from that suit
are 12/53.  So the odds of two cards from the same suit are
	2/54 * 1/53 + 52/54 * 12/53 = 626/2862 = 313/1432
If jokers match anything, then if you picked a joker as the first card, you are
guaranteed two cards of the same suit;  if you picked a card from a suit, then
the odds of picking another card of the same suit or a joker are 14/53, so the
chances of two cards from the suit in this interpretation would be
	2/54 * 53/53 + 52/54 * 14/53 = 834/2862 = 417/1431
--------------------------------------------------------------------------------
Carl J Lydick | INTERnet: CARL@SOL1.GPS.CALTECH.EDU | NSI/HEPnet: SOL1::CARL

Disclaimer:  Hey, I understand VAXen and VMS.  That's what I get paid for.  My
understanding of astronomy is purely at the amateur level (or below).  So
unless what I'm saying is directly related to VAX/VMS, don't hold me or my
organization responsible for it.  If it IS related to VAX/VMS, you can try to
hold me responsible for it, but my organization had nothing to do with it.
Xref: sparky rec.puzzles:5254 sci.math:9569
From: hoey@AIC.NRL.Navy.Mil (Dan Hoey)
Newsgroups: rec.puzzles,sci.math
Subject: Digital Invariants (was Number Theory)
Date: 25 Jul 92 22:13:33 GMT
Followup-To: rec.puzzles
Organization: Naval Research Laboratory, Washington, DC

uunet!eliot!andyc (Andy Panda Collins) writes:

> I don't know much about this news-group, but I thought
> that I might post this list of number-theoretic numbers.

>   Let N be an n-digit number which is separated into it's
>   decimal digits, raise each digit to the nth power and sum
>   them. Let N be a unique number if the sum is equal to N.

Well, this topic is more the province of recreational math than of
number theory, so I've redirected it to rec.puzzles such topics are
more customarily treated.

According to David Wells's book _The_Penguin_Dictionary_of_Curious_
_and_Interesting_Numbers_ these numbers are called digital invariants.
Wells notes that

    When G. H. Hardy wished, in his book _A_Mathematician's_Apology_,
    to give examples of mathematical theorems that were not `serious',
    he chose two examples, `almost at random', from Rouse Ball's
    _Mathematical_Recreations_.

    The first was the fact that 8712 and 9801 are the only 4-digit
    numbers that are multiples of their reversals.

    The second was the fact that, apart from 1, there are just 4
    numbers that are the sums of the cubes of their digits.

I found all the numbers that are multiples of their reversals last
year (and coined the term `palintiples' to refer to them) so I suppose
it is not so surprising that I should be interested in this problem as
well.

Andy lists all the digital invariants up to 14 digits, and I wrote a
program that confirms his list.  The next few are:

15 digits: none		
16 digits: 4338281769391371
           4338281769391370 
17 digits: 35875699062250035
           35641594208964132
           21897142587612075
18 digits: none
19 digits: 4929273885928088826
           4498128791164624869
           3289582984443187032
           1517841543307505039
20 digits: 63105425988599693916
21 digits: 449177399146038697307
           128468643043731391252
22 digits: none
23 digits: 35452590104031691935943
           28361281321319229463398
           27907865009977052567814
           27879694893054074471405
           21887696841122916288858
24 digits: 239313664430041569350093
           188451485447897896036875
           174088005938065293023722 
25 digits: 4422095118095899619457938
           3706907995955475988644381
           3706907995955475988644380
           1553242162893771850669378
           1550475334214501539088894.

I have also verified that there are no digital invariants of 46 or
more digits.

David Wells lists one more digital invariant, though, of 39 digits:

    115132219018763992565095597973971522401

and he says it is the largest one known.  However, I am not sure but
what this may have been proven to be the largest one that can exist by
whoever discovered it.  I do wish Wells were a little more
conscientious with his attributions.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From: ph@tesla.ama.caltech.edu (Paul Hardy)
Newsgroups: rec.puzzles
Subject: Re: carrying the torch (MAJOR HINT)
Date: 29 Jul 92 01:27:11 GMT
	<25JUL199200043304@zeus.tamu.edu>
	<PH.92Jul25131320@vortex.ama.caltech.edu>
	<1992Jul27.201749.21434@news.media.mit.edu>
Distribution: rec
Organization: California Institute of Technology

In article <1992Jul27.201749.21434@news.media.mit.edu> tsmaster@media.mit.edu (David LeCompte) writes:

>   In article <PH.92Jul25131320@vortex.ama.caltech.edu> ph@vortex.ama.caltech.edu (Paul Hardy) writes:
>   >In article <25JUL199200043304@zeus.tamu.edu> dwr2560@zeus.tamu.edu (RING, DAVID WAYNE) writes:
>   >
>   >>   >3 6 9 27 ?
>   >>
>   >>   3^3^3, whatever that means.
>   >
>   >No.  What's the rule for the progression?
>
>   hey, with a little bit of work, one can work out a 5th degree polynomial
>   that will make 3^3^3 be the value of P(5).

SPOILER...















Sure, but usually you're given a few terms that follow from an easily
observed rule!  That series might have been too tricky, so here's a clue (and
a similar series).  I thought of giving this originally instead of the one
I posted, but decided it would probably be too easy.  It follows the same
rule as the series 3, 6, 9, 27, and has the same fifth term.  If this isn't
enough to figure out the series, I don't know what is!

			      4 6 9 27 ?

				--Paul
--
This is my address:         ph@ama.caltech.edu
This is UUCP:               ...!{decwrl,uunet}!
This is my address on UUCP: ...!{decwrl,uunet}!caltech.edu!ama!ph
Any questions?

"Does Emacs have the Buddha nature?"  --Paul Hardy   "Yow!" --Zippy
From: sharkey@robots.oxford.ac.uk (Paul Sharkey)
Newsgroups: rec.puzzles
Subject: Re: Tiling a plane
Date: 28 Jul 92 08:41:11 GMT
Organization: Robotics Research Group, Engineering Science Dept, Oxford, UK.

In article <711892126.F00001@csource.oz.au> joe@csource.oz.au (Joe Slater) writes:

>A plane can be tiled with three regular geometric shapes: squares,
>triangles and hexagons. It can also be tiled with pairs of shapes:
>hexagons and triangles, or octagons and squares.

>Are there any other pairs that will tile a plane? What about
>triplets? Answers requiring fractal-like series to fill up gaps aren't
>counted.

If you are familiar with the work of Esher you will find many examples
of a plane being tiled with birds, fish, frogs and so on --- no gaps.

A simple triplet would be rectangles chopped into 3 pieces. That is
trivial but easily extended to general patterns as Esher showed.

Paul.

-------------------------------------------------------------------
Paul M Sharkey                      Phone:  +44 (0)865 273 150
Robotics Research Group             Fax:    +44 (0)865 273 908
Dept Engineering Science            
Parks Road, Oxford OX1 3PJ          Email:  sharkey@uk.ac.ox.robots