HDTenEightyP/NetNews huggingface dataset:
From: leag0552@uxa.cso.uiuc.edu (Luis E. Amaya)
Subject: Re: Ten equals 9.9999....
Organization: University of Illinois at Urbana
Date: Fri, 31 Jul 1992 17:41:28 GMT

In article <1992Jul31.163146.23947@dartvax.dartmouth.edu> Virginia.A.Reed@dartmouth.edu (Virginia A. Reed) writes:
>If 9.9999... + 0.00...01=10, and 0.00...01 is infinitely small(which it
>is), then it can be assumed as zero.
      ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
	 it IS zero
>therefore:
>9.9999...- 0.000..01 = 9.999...
............<some deleted lines>............
>If 9.999...=10, then
>9.9999...-0.000...01=10
>add 0.000...01 to both sides:
>9.9999...=10.000...01
>-0.000...01=0.000...01
Use what you 'assumed' at the beginning and you proved that 0=0
 
>taken to its extreme, you could prove that 7=-7, or that 2341=-2341
Incredible!!! I can not believe it!! Show me how (without multiplying by infinity on both sides)
----------
Luis
Newsgroups: rec.puzzles
From: rcutler@banach.WPI.EDU (Robert Cutler)
Subject: Re: Alien colonization
Originator: rcutler@banach.WPI.EDU
Organization: Worcester Polytechnic Institute
Date: Thu, 23 Jul 1992 07:38:31 GMT


In article <1992Jul22.153416.3564@hellgate.utah.edu>, tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
|> 
|>   The alien Ddiv lives in a universe which is three dimensional, and has
|> the planets evenly spaced throughout as in a 3D grid.
|> Ddiv is planning to have children- each child will require one full planet
|> to live on.  Ddiv is also immortal, and wishes to keep living forever.
|> He wishes to be fair to his children- so he will only have as many 
|> children as each of his children can have.  He would not have any children
|> if any single one of his offspring could not have any children, so
|> his grandkids, great grandkids, and so on must be able to have as many
|> children as he has. 
|> 
|> 
|>   How many children can Ddiv have? (Ddiv has children by dividing)



Well, this problem really reminded my of the convergence of infinite 
series I took in High School.  



	Lets say that Ddiv starts out with A_0 amount of matter.  If
Ddiv cuts himself into n pieces making n children and each one of those
children cuts itself into r pieces, then the total number of children
would be the summ of A_0 * ( r^n) with n going from 0 to infinity.
This series is equal to A_0 / 1-r.  If |r| < 1 then the series converges.
so lets say he cuts himself in 1/2 pieces (whatever that means) then 
when every singe descendent has cut himself in 1/2 pieces there will be
two full Ddiv's at the end.


	Well if you were looking for convergence of infinite series there
is one example.  


If he cut himself in 1 or 2 or any more pieces then it follows by the 
divergence test that the summ of the pieces at the end is infinite.


				Rob
From: mueller@schaefer.math.wisc.edu (Carl Mueller)
Newsgroups: rec.puzzles
Subject: 99.9... - 9.9... = 89.9... (too long)
Date: 29 Jul 92 13:20:22 GMT
Reply-To: mueller@math.wisc.edu
Organization: Univ. of Wisconsin Dept. of Mathematics


I'm sorry, but I lost the original article that dealt with this.  In it,
the author said that 99.9... - 9.9... = 89.9... and so a previous 
author's argument that 9.9... = 10 was faulty.  Well, as has been
pointed out previously, (since 9.9... does in fact equal 10 and 89.9...
does in fact equal 90, and so on) it is true that 99.9... - 9.9... = 89.9... .

Anyway, if the author believes this, and ALSO (presumably) believes that
99.9... - 10 = 89.9... (which should be pretty clear).  THEN it follows
that 9.9... = 10 = THE difference between 99.9... and 89.9... .

Also (though it's been pointed out already), the average of two numbers
lies between them (see proof in a previous article).  This is obviously
a statement about REAL numbers as the average of integers for example
need not be an integer.

Anyway, MANY people (who learned long ago that 0.999999... = 1.00000...)
are getting really tired of this thread.  Though I am one of those to whom
this fact is not new, I don't consider this a waste of bandwidth.  I think
that this fact should be taught to students reasonably early in their
mathematical education.  It's not really all that hard to understand ...
we all learned at some point that 1/2 + 1/4 + 1/8 + 1/16 + ... = 1
and this fact is of precisely the same difficulty.  Everyone that reads
this newsgroup has the ability to UNDERSTAND (not just believe) that 
0.999... = 1.000... .

Carl Mueller (mueller@math.wisc.edu)

P.S. (Sorry I rambled a bit.)
From: tskelly@ccvax.ucd.ie
Newsgroups: rec.puzzles
Subject: Re: vacuuously false?
Date: 30 Jul 92 14:45:05 GMT
Organization: University College Dublin

In article <1992Jul29.125234@IASTATE.EDU>, macster@IASTATE.EDU (Michael C Mccarty) writes:
> In article <Bs5pw4.MEp@rahul.net>, dhesi@rahul.net (Rahul Dhesi) writes:
>> Normally an assertion
>> 
>>      if A then B
>> 
>> is considered to be vacuously true if A is known to be false.  But
>> consider this assertion:
>> 
>>      if today is Monday then tomorrow is Tuesday
>> 
>> Is this vacuously true on Friday?
>> --
>> Rahul Dhesi <dhesi@rahul.net>
>> also:  dhesi@cirrus.com
> 
> Ahhh....  Good question.   Well, at least, I think it is, anyway.  I am not
> familiar with the term 'vacuously true', but I think I have derived a fairly
> good idea of its meaning (but please enlighten me to its meaning if I am
> completely off-track).  
> 	Logically, the first statement (often denoted, A -> B) above has a very rigid
> definition.  This statement is considered 'valid' (valid is what I am assuming
> is roughly what is meant by 'vacuously true') iff (iff = if and only if) when A
> is true, B is true, but NOT vice versa.  This allows B to be true even if A is
> not.  What causes an implication to become 'invalid' is when A is true but B is
> false, thereby denying the meaning of the statement. Any other state of
> condition has no effect upon the validity of the implication.
> 	What must be realized is that the 'natural' (ie English) interpretation of the
> above statement ("if today is Monday...") includes an implicit meaning that its
> logical counter-part does not.  Symbolically, it would be tempting to write the
> sentence as:
> 
> 	(today is Monday) -> (tomorrow is Tuesday).
> 
> 	This allows (tomorrow is Tuesday) to be true even when (today is Monday) is
> false.  This, of course, is wrong.  The correct and definitive way to state such
> an assertion is: 'Tomorrow is Tuesday if and only if today is Monday.'
> (represented by A <-> B)
> 
> 					Mike
> 
> P.S.  Ugh.  Can I babble or what?!?!?   :-)



Sorry Mike,
	that's not quite the right interpretation.
Logic states that the assertion A -> B is TRUE if A is FALSE. This means that
the *assertion itself* is TRUE if A is FALSE but doesn't say anything about B
in this case - i.e. B could be TRUE or FALSE. The *assertion itself* would be
FALSE if for any A and B, A is TRUE and B is FALSE.
What this means is that the assertion 
	"If today is Monday tomorrow is Tuesday"
is TRUE because when today IS Monday tomorrow IS Tuesday but when today IS NOT
Monday it doesn't matter what tommorow is.
The IFF ( <-> ) operator is a different matter altogether and while the
assertion "Today is Monday IFF tomorrow is Tuesday" is also correct there are
many cases in normal language usage where an "IF" assertion is TRUE while the
corresponding "IFF" assertion is FALSE.

	Hope that's reasonably clear,
	sorry if I misinterpreted your argument,

	TSKELLY.

Newsgroups: rec.puzzles
From: carl@SOL1.GPS.CALTECH.EDU (Carl J Lydick)
Subject: Re: Proof of -1 = 1
Reply-To: carl@SOL1.GPS.CALTECH.EDU
Organization: HST Wide Field/Planetary Camera
Date: Fri, 24 Jul 1992 05:21:07 GMT

In article <chan.711948973@s1.elec.uq.oz.au>, chan@elec.uq.oz.au (Chuen Wen Chan) writes:
>
>I hope this problem hasn't been posted before, but here it is anyway:
>
>	sqrt(-1) = sqrt(-1)
>
>	sqrt( -1 / 1 ) = sqrt( 1 / -1 )
>
>	sqrt(-1)    sqrt(1)
>	-------- = ---------
>	sqrt(1)     sqrt(-1)
>
>	now cross multiply and we get:
>
>	-1 = 1  QED
>
>ie:  Everything we learnt in classical maths is wrong
>   => no more maths subjects in future
>   => all mathematicians lose their jobs. Yippe!
>
>PS:  If no one gets the solution by Monday, it will be posted.

Alternatively
	1^2 = 1  so sqrt(1) = 1
	(-1)^2 = 1 so sqrt(1) = -1
	-1 = 1  QED
--------------------------------------------------------------------------------
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.
Newsgroups: rec.puzzles
From: tsmaster@media.mit.edu (David LeCompte)
Subject: Re: carrying the torch
Organization: MIT Media Laboratory
Distribution: rec
Date: Mon, 27 Jul 1992 20:17:49 GMT

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).

Which makes the original question meaningless.

		Dave "Gotta love standardized tests" LeCompte

-- 
---------------------
ObSig
From: dik@cwi.nl (Dik T. Winter)
Newsgroups: rec.puzzles
Subject: Re: Easy prob. question
Date: 26 Jul 92 00:17:49 GMT
Organization: CWI, Amsterdam

In article <1992Jul24.225745.19641@EE.Stanford.EDU> mack@isl.Stanford.EDU (mack) writes:
 > This looks to be an easy one:
It is not because it is ill defined.
 > 
 > Q: Two cards are picked at random from a complete deck of cards. What
 >    is the probability that they are of the same suit?
 > 
What is a complete deck?  I have several.  The most complete is the Tarot
deck as it is currently used in France to play the game of Tarot (yes, a
game, no future-telling or somesuch).  78 cards.  14 cards to each of 4 suits
plus 22 trumps.  Another complete deck is an Australian deck of 62 cards
for the game of 500.  Added: 11's, 12's and red 13's.  And also my Spanish
deck of 40 cards, 10 cards to each suit (no 8's, 9's or 10's).  Etc.
 > Have fun.   :-)
 > 
Do I have fun now?
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland
home: bovenover 215, 1025 jn  amsterdam, nederland
dik@cwi.nl
Newsgroups: rec.puzzles
From: tsmaster@media.mit.edu (David LeCompte)
Subject: Re: Ten equals 9.9999....
Keywords: nine ten repeating digits
Organization: MIT Media Laboratory
Date: Wed, 29 Jul 1992 17:50:22 GMT

In article <5688@osc.COM> Joe Keane <jgk@osc.com> writes:
>I'm afraid you made a mistake in your subtraction.  The problem is that
>99.999... minus 9.9999... is 89.999..., not 90 like you have.

How do you come up with this? 
   _           _
99.9 - 10 = 89.9
                                                    _
I'll grant that as obvious. But if you claim that 9.9 != 10, then I
don't see your claim at all.
                      _
No, really, really. 9.9 = 10. It does. 
  _
9.9 = 9+.9+.09+.009+... forever.
  _
 .9=.9+.09+.009+... forever.

you REALLY can term by term subtract ALL the 9s after the decimal
points and get all 0s after the decimal point.

Dave LeCompte
-- 
---------------------
ObSig
Newsgroups: rec.puzzles
From: cdl0915@ritvax.isc.rit.edu (Christian Dylan Losinger)
Subject: permutations
Reply-To: cdl0915@ritvax.isc.rit.edu
Organization: Rochester Institute of Technology
Date: Thu, 23 Jul 1992 13:18:16 GMT


	I need a good algortithm for producing all permutations of a sequence
of items .
	The one I'm triing now (and finding is not working is) :
	
	swap item 1 with 2
	swap item 1 with 3
	swap item 1 with 4
	....
	all the way to the number of items, then repeat

	this does produce many permutations, but has some repetition.

	i'm stumped as to a better one. 
	anyone?

	
        --------------------------------------------------------------
        	c  h  r  i  s   l  o  s  i  n  g  e  r   r  i  t 
Newsgroups: rec.puzzles
From: jimhsu@athena.mit.edu (Jimmy M.D. Hsu)
Newsgroups: rec.puzzles
Subject: Killing King (variation)
Date: 29 Jul 92 02:09:39 GMT
Organization: Massachusetts Institute of Technology

Very good, Seth!  Try this variation.

"The Killing Joke"
Part 2

        Brian, upon further thought, realizes
that he reasoned incorrectly.  He needed to take
into account the games he loses.  He decides the
best strategy is to take only one card at a time
and to discard immediately.  Brian computes an
advantage of about 3.5% over Sam's strategy.
        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!
(He said that the four kings are evil, but it
takes two to conspire together to have enough
strength for a loss.)  In fact, this could be
even more difficult to solve if you use the
fifth spade, or any of the numerous possible
variations.
        Sam thinks a bit longer and realizes
a fundamental flaw in the computation of valid
strategies.  Sam is very perturbed at this.

        Now, what is the best strategy?  What
is a generic formula for computing probability?
I welcome comments on my puzzles.  Else, I
might stop, if nobody reads and nobody cares.
From: mmass@casbah.acns.nwu.edu (Michael Massing)
Newsgroups: rec.puzzles
Subject: Re: vacuuously false?
Summary: Logic discussion
Keywords: Logic, assertions, vacuously true
Date: 29 Jul 92 16:13:45 GMT
Organization: Northwestern University, Evanston Illinois.


dhesi@rahul.net (Rahul Dhesi) says in <Bs5pw4.MEp@rahul.net>:

> Normally an assertion

>     if A then B

> is considered to be vacuously true if A is known to be false.  But
> consider this assertion:

>      if today is Monday then tomorrow is Tuesday

> Is this vacuously true on Friday?

Of course.  Since it's Friday, and not Monday, tomorrow can be anything it
wants.  Only on Monday must tomorrow be Tuesday.
--
Mike Massing                              mmass@casbah.acns.nwu.edu
"In America, having a nice lawn is considered a major cultural acheivement,
like owning a hardcover book or watching 'Meet The Press.'  Americans would
rather live next to a pervert heroin addict Communist pornographer than a
person with an unkempt lawn."   -- Dave Barry
------------------------------------------------------------------------------
Disclaimer: All opinions expressed here are solely my own.
Newsgroups: rec.puzzles
From: Dave Dodson <dodson@convex.COM>
Subject: Re: Plane Division
Originator: dodson@bach.convex.com
Reply-To: dodson@convex.COM (Dave Dodson)
Organization: Engineering, CONVEX Computer Corp., Richardson, Tx., USA
Date: Wed, 22 Jul 1992 18:57:09 GMT
X-Disclaimer: This message was written by a user at CONVEX Computer
              Corp. The opinions expressed are those of the user and
              not necessarily those of CONVEX.

In article <1992Jul22.154053.3498@alf.uib.no> s290@brems.ii.uib.no ( Per Ivar Roervik) writes:
>  This is adopted from an oldie given by Gardener.
>  
>  In a closed and bounded area of any 2-D shape, 1 million random
>  points are chosen.
>  Is it always possible to bisect the area AND the points into two
>  eqal parts with a straight line?
>
>  Proof or disproof needed.  

If I get to choose the shape and the points, the answer is no.

Let the shape be a circle.  Then any line that divides the shape into equal
areas must be a diameter.  Put 1 point at the center.  If the rules require
the dividing line to miss all the points, you already can't find a dividing
line.

If a line can pass through a point, we have to decide on the rule telling
how the point is "divided" by the line.  I adopt the convention that a
line through a point divides the point in half, whatever that means.  Then
I put 1 point halfway between the center and 12:00 o'clock and put the rest
of the points on the segment between the center and 3:00 o'clock.  Now try
to find a diameter that separates the points into halves.

----------------------------------------------------------------------

Dave Dodson		                             dodson@convex.COM
Convex Computer Corporation      Richardson, Texas      (214) 497-4234
Newsgroups: rec.puzzles
From: dwr2560@zeus.tamu.edu (RING, DAVID WAYNE)
Subject: Re: carrying the torch
News-Software: VAX/VMS VNEWS 1.41    
Organization: Texas A&M University, Academic Computing Services
Distribution: rec
Date: Sun, 26 Jul 1992 05:09:00 GMT

>>   >3 6 9 27 ?

I guess...
>>   3^3^3, whatever that means.

ph@vortex.ama.caltech.edu (Paul Hardy) writes...
>No.  What's the rule for the progression?

If you don't know the rule, how can you say it's wrong? Consider this one
to be another puzzle. What's the sixth term? Hint: the first term '3' is a
little ambiguous, but is reasonable.

Dave Ring
dwr2560@zeus.tamu.edu
Newsgroups: rec.puzzles
From: hhassler@iaik.tu-graz.ac.at (Hannes Hassler)
Subject: Re: Gummy Drop-bears
Organization: Institut fuer Angewandte Informationsverarbeitung, TU Graz
Date: Thu, 30 Jul 92 13:37:08 GMT

In article <Jul.29.03.30.41.1992.18050@remus.rutgers.edu> Chris Long writes:
>
>Real gummy drop-bears mass 10 grams, while imitation gummy drop-bears mass 9 
>grams.  Spike has 7 cartons of gummy drop-bears, 4 of which contain real gummy
>drop-bears, the others imitation.  Using a scale only once and the minimum 
>number of drop-bears, how can Spike determine which cartons contain real gummy
>drop-bears?
>
I can do it with 53 bears: Take 0, 1, 2, 4, 7, 14, and 25 bears, respectively.
No two distinct triples over these numbers add up to the same sum.  Hence, you
can identify all imitations.  I think that 53 bears is best possible; it is
easy to see that 0 must occour in the optimum solution.

******************************************************************************
 Hannes Hassler
******************************************************************************
Newsgroups: rec.puzzles
From: ajpierce@med.unc.edu (Andrew Pierce)
Subject: Re: Ten equals 9.9999....
Keywords: nine ten repeating digits
Organization: UNC-CH School of Medicine
Date: Wed, 29 Jul 1992 04:13:20 GMT

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?
     -Andy
ajpierce@med.unc.edu
Newsgroups: rec.puzzles
From: tsmaster@media.mit.edu (David LeCompte)
Subject: Re: Ten equals 9.9999....
Keywords: nine ten repeating digits
Organization: MIT Media Laboratory
Date: Wed, 29 Jul 1992 18:03:14 GMT

In article <1992Jul28.232532.3577@hellgate.utah.edu> tolman%asylum.cs.utah.edu@cs.utah.edu (Kenneth Tolman) writes:
>>If two numbers are different, then there's another number between
>>them.  (Actually, a lot of such numbers.)
[deleted]
>
>SAY WHAT?  This is the silliest bit of nonsense ever.  "If two numbers are
>different, then there's another number between them".  Let's consider the
                                                        ^^^^^^^^^^^^^^^^^^
>integers, and I will give you 2 and 3.  Now, if they are different, there
 ^^^^^^^^

Oh, yeah. THAT sounds like a valid simplification of the problem. How
about just the two integers 5 and 12. This is as silly a subset as the
integers. I'll easily grant that there are no numbers in the set
{5,12} between 5 and 12, but I have no problem with the claim that Ken
disputes. 
>must be an integer between them.  Well, what is it?
            ^^^^^^^
No, a number. Try 3.5. You're oversimplifying.

>
>Now, if you assert that the above is true for REAL's then you better have
>some reason, for I can just as easily assert the opposite.  As a matter of
>fact, I can show you two numbers which are different, which do not have any
>number between them...
>  41.999999.... and 42
       _
No, 41.9 = 42.

-- 
---------------------
ObSig
From: shapiraa@pleiades.cs.rpi.edu (Andrew Shapira)
Newsgroups: rec.puzzles
Subject: Re: 3 puzzles
Date: 23 Jul 92 17:40:39 GMT
Article-I.D.: rpi.sgtxr0
Distribution: rec.puzzles

markc@eomer.hw.stratus.com (Mark Christensen) writes:

>Here are three puzzles:

>[1]
>	What is the first number, when spelled out (e.g. O-N-E) is
>in alphabetical order?

e, i, or a small constant from physics.
From: t9114145@phillip.edu.au
Newsgroups: rec.puzzles
Subject: Re: Ten equals 9.9999....
Date: 25 Jul 92 19:51:54 GMT
Organization: RMIT (Bundoora Campus)

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.

>               _
>             9.9  =  x  (yes, I mean "repeating" with the bar over it.)
>               _
>            99.9  =  10x  (multiply by 10)
>               _
>          -  9.9  = -  x  (subtract that first thing)
>           --------------
>            90    =   9x  (oooh, no more decimal)
>            10    =    x  (divide by 9)
>               _
>             9.9  =   10  (voila!  [QED])
> 
> 
> -Richard
> (My personal favorite)
> How about 1 = 2 later?
-- 


Adam Eberbach      t9114145@phillip.edu.au          fnord
Computer Science student, Phillip Institute of Technology.
From: ags@seaman.cc.purdue.edu (Dave Seaman)
Newsgroups: rec.puzzles
Subject: Re: Plane Division
Date: 22 Jul 92 18:51:14 GMT

In article <1992Jul22.114343.12817@hemlock.cray.com> de@cray.com (Duane  
Eitzen) writes:
> _________________________
> |                       |
> |           X           |
> ----------------------| |
>                       | |
> ----------------------- |
> |                       |
> |                       |
> -------------------------
> 
> Where "x" is the roughly the area where the 1 million points are. In  
this
> case the straight line will either trisect the area (not bisect), or
> will not cut the shape into two equal parts. 

There are infinitely many lines (one for each direction) that cut the  
shape into two equal parts. Given any line, just translate it until the  
areas on the two sides are equal.

Some of the lines thus constructed (the horizontal one, for example), will  
completely miss the area where the points are. However, if you consider  
what happens to the family of bisecting lines as the direction changes,  
you will notice that you start out facing east with all the points on your  
left, and after making a half rotation you are facing west with all the  
points on your right. At some stage during the turning process, you must  
have had an equal number of points on each side of the line.

--
Dave Seaman
ags@seaman.cc.purdue.edu