1.9...is it 2?
Steve Whittet
>
>In article <kwqp4.212$Rh4....@news.shore.net>,
>Steve Whittet <whi...@shore.net> wrote:
>>In article <8856k7$6e2$1...@nntp.Stanford.EDU>, jona...@DSG.Stanford.EDU
says...
>>>
>>>In article <QEop4.203$Rh4....@news.shore.net>,
>>>Steve Whittet <whi...@shore.net> wrote:
>
>[snip quoted text]
>>>Uh, oh. Warning! Warning! Squink expected!
>>
>>Come on. 1.9... gives insufficient information to expand it
>>no ordered series has been established.
>
>
>Do you mean _sequence_ or _series_? I may have misued the terms.
The sequence is the order of the terms in the series. Without an
ordered series there is no ordered sequence.
>
>Consider: 1.9, 1.99, 1.999, 1.9999, 1.99999, ... to the number
>usually written as: . 1.9
>
>(1.9 with a dot over the nine).
>
>[snip more quoted text]
>
>>Brian claims that the correct interpretation, the one that is eactly right,
>>requires limits. First he is telling us the process is "assumed" to Be...
>>repeated indefinitely.
>
>
>Yes of course. That's what the ellipsis is shorthand _for_. The
>numerals 1.999999999..... _denote_ the number you get at the end of
>the infinite series. Or more formally, the limit of the sum I and
>others have quoted. Repeatedly.
>
>
>>That in and of itself ought to clue you in that it's an indefinite answer.
>
>No, Steve. Not at all. It depends on whether the limit converges.
You say its not indefinite, but it depends...?
Whether the limit diverges or converges it may still be boundable
but that isn't infinitely precise, its only precise up to its limit.
>See the references.
>
>
>>According to what Brian said the dots are supposed to be a shorthand way
>>for describing a limit to a sequence.
>
>Yes, Brian said that, but that's also the standard definition.
>
>>Now you come along and talk about
>>Brians sum where what Brian was talking about was the limit to a sequence
>>and while what you claim is that there are no limits I expect you actually
>>think is that you agree with what Brian said.
>
>As I said before, are we talking about a sequence or series?
I think we are now talking about both. First we need to establish
that there is an ordered array. That gives us a sequence.
>
>
>>A sequence is defined as an ordered array of infinitely many elements
>>chosen by a given rule.
>
>
>Okay, I'll accept that. So consider the _sequence_ of ever longer
>summations, 1.9, 1.99, 1.999, 1.9999, 1.9999. What is the _limit_ of
>that sequence?
>
>>An infinite series is obtained when the terms
>>of a sequence are summed. The sequence of partial sums converges or
>>diverges depending upon whether or not the correspondening sequence
>>has a limit. With no limit it diverges.
>
>From the previous discussion, it seems quite clear you copied this
>without understanding it, let alone attributing it.
This is the sum of the sequence you were talking about.
>
>>>>As you ought to know by now Brian, I find really basic
>>>>things that don't quite work but get accepted as if they do
>>>>interesting.
>>>
>>>Which part of limit theory do you think doesn't work?
>>
>>Well first you need to establish an ordered sequence, 1.9...
>>doesn't establish an ordered sequence, let alone one that
>>converges to 2.
>
>Oh _really_?
yes really. 1.916583954756869847832... rounded to 1.9 for example
there are an infinite number of possible ways to fill in the
blanks in that series.
> I described the sequence perfectly well. the n'th
>element is exactly { 1 + Sum (i = 1, n) 9*10^-i }.
>The limit is then
>
> Lim (n -> oo) of { 1 + Sum (i = 1, n) 9*10^-i }
Is it? I said a sequence is defined as an ordered array of
infinitely many elements chosen by a given rule. You said
you accepted that.
I said 1.9...doesn't establish an ordered sequence.
You said it does.
Show me the given rule 1.9...establishes. I think you just
made a bad assumption that 1.9... means 1.999... It doesn't
and that is exactly the point Eric was making.
>
>The formal definition of the limit L is (I think) that given
>_any_ epsilon e, we can always find some j such that
>
> abs(L - sum (i=1,j) ) < \epsilon
>
>Let L=2. Proof is left as an exercise for the reader.
Aw come on Jonathan, I want to see you prove that 1.9...= 2
>
>
>>>Sounds like gobble-de-gook to me. Brian and Alan Dunsmuir are making
>>>completley unexceptional tatemntes.
>>
>>I guess they don't teach the classics at Stanford anymore. Jonathan
>>has never even heard of Ontology, sounds like gobble-de-gook to him...:)
>
>No, they teach the classics to undergraduates. I wasn't an
>undergraduate at Stanford. But perhaps it's news to you that
>philosphy has moved on since the metaphysical muddles of the ancients?
Its news to me that Ontology and Metaphysics are considered muddles
of the ancients. We live in a time when Metaphysical Engineers are
busy applying relative absolutes to the structure of non systems
and to you its just so much Alchemy...
>
>[snip]
>
>>That's better, now you have the ordered sequence
>>but you are still not there yet, as you can see you had to
>>introduce the concept of a limit. Without a limit it doesn't work
>>>
>>> limit (n-> oo) { 1+ sum [i=1,n] 9*10^-i }
>>>
>>>And the limit is defined formally and precisely.
>
>Steve, in mathematics, that is the definition of what those dots mean.
You mean all I have to do is take 1.9 and add three dots 1.9...
and I get 2? thats news to me.
>
>>with a limit you are forced to introduce an assumption.
>>
>>Didn't you say earlier
>>
>>>So Brian's sum is Infinite, Without Limits, and Free to be 2.
>
>Yes. I'm sure everyone else saw the humour.
So what you are now saying is that the points you were trying
so hard to make are really a joke?
>
>
>>I'll refer you to:
>>> "Concrete Mathematics: A Foundation for Computer Science",
>>> Graham, Knuth, and Patashnick, 2nd. Ed. Addison-Wesley,
>>> (my references says 1989, but I beleive that's the first edition).
>>>
>>>Heavyweight, but it has examples of series, sums, convergent and
>>>divergent infinite sums, usw. And Steve may even like the opening
>>>examples, the Towers of Hanoi and the Josephus problem.
>>
>>Gee, don't you have Pearson? "Handbook for Applied Mathematics?
>>How about "Sequences Combinations and Limits" S I Gelfand,
>>M L Gerver, A A Kirilov, N N Konstantinov and AG Kushnirenko
>>translated from the Russian by leslie Chn and Joan Teller,
>>University of Chicago 1969
>
>Oh very good, you can use a search engine.
Yeah, its called a book shelf...
>How about reading
>some of those and learning from them, hmmm?
>
>
>>>>1.9... is Becoming 2.
>>>
>>>>Are you aware there is an ontological difference?
>>>
>>>No, there isn't, not when it comes to infinite summations.
>>
>>1.9... is not an ordered series. It cannot Become 2.
>>All it can be is 1.9... something, rounded off.
>
>
>NO, Steve, NO. If it's rounded off, then it's not 1.9999..., it's some
>other number. Think, Steve: if you round off at digit j, how _much_
>less is the result than the whole infinite digit string? Try writing
>it as 2 - epsilon. That is, write out the epsilons.
Stop! Think about it. 1.9... can be simply 1.9 or any value up to 2.
>
>You get (2 - 0.1, 2-0.01, 2 - 0.001, 2 - 0.0001, 2 - 0.00001, ...,
>et cetera. See how the numbers get closer and closer to 2?
>If you could carry it on infinitely, you'd _get_ to 2.
>And that's what the limit is looking at.
To have an omega you have to have an alpha. to limit the series
you first have to get it started.
>
>Its' not like this is so terribly new, Steve: Archimedes was almost
>knocking on the doors of the calculus. The Greeks did formal proofs,
>using geometric arguments, that today we'd do with limit arguments or
>the calculus. Sheesh. I suppose you're also going to say that
>derivatives (or integrals) don't ever reach limits as well?
It's terribly new to me Jonathan, do you really want to insist 1.9...= 2
>
>>You think there is no difference between Being and Becoming?
>>>
>>>We're talking about mathematics, Steve. You're confounding bad
>>>philosophy with well-defined mathematical terms.
>>
>>You don't think mathematics has any philosophical underpinnings?
>
>Not in any way that relates to Being and Becoming.
I don't know why I'm bothering to talk to you. What do you
think you are doing when you engage in a process of mathematical
calculations which determine the limit of an ordered sequence
according to a given rule. You are seeing what the number Becomes.
You want to find out what its going to Be. ...Sequence...Consequence.
>
>
>>> This talk of `being' and `becoming' is utterly irrelevant.
>>
>>Is there a difference between something which is engaged in
>>a process it has yet to complete and something which has
>>completed the process?
>>
>>Are there such things as order, sequence, position?
>
>Pure squink. Of course there are.
>[...]
thank you, I am glad you can admit it when you are wrong.
steve
Alan M Dunsmuir
<whi...@shore.net> writes
>The sequence is the order of the terms in the series. Without an
>ordered series there is no ordered sequence.
No Steve. A sequence is list of entries - e.g. 1, 1.5, 1.75, 1.875, ...,
each of which stands alone. A series is a list of entries -
e.g. 1 + 0.5 + 0.25 + 0.125 + ...
each of which has to be added to all the preceding entries.
If you have a series you should be able to see that the PARTIAL SUMS of
its first n entries forms a sequence as n varies. The partial sums of
the series offered above are 1, 1.5, 1.75, 1.875, ...
A sequence S(n) is said to have a limit S as n tends to infinity if, for
any e > 0, there exists a N such than, for all n > N, |S(n) - S| < e.
A series is said to have a sum to infinity S if the sequence of its
partial sums has limit S as n tends to infinity.
I leave it as an exercise for the pupil to confirm that the series
indicated above has the sum to infinity 2.
--
Alan M Dunsmuir
PaulDanaher
I'm entering this discussion at a much lower level, so I'm snipping the main
thread.
First, I ran my head up against this a few years back with 0.999... = 1.
I had trouble with the same answers.
An illuminating example (sorry, I've lost the attribution) was this:
1/3 = 0.333...
1/3 + 1/3 = 0.666...
1/3 + 1/3 + 1/3 = 0.999...
- only it *also* and identically equals 1.
Regarded as an *infinite* decimal representation, 0.999... is identical with
one.
Now, if you go back into the history of mathematics - about 150 years will
do - you find yourself in a position where a lot of people are very unhappy
about "infinite" this and that (the Greeks had been, too). To get away from
this loose talk of "infinitesimals", Riemann came up with the idea of
partial sums which can be made as accurate as you like by decreasing the
size of the bits you add.
Remembering your Achilles and the tortoise paradox, Zeno was wrong but
right. He was wrong, because Achilles passes the tortoise rather quickly,
demonstrating in a humdrum, everyday way that the infinite series has a
limit. But the habit of thought that tripped him up is the notion of a
stepwise process - in this case, the notion of the infinite sum as a
*process* that takes the same time for each step. If Achilles had to take
the same amount of time for each time he "catches up" the tortoise, he never
would pass it. Similarly, if you think of yourself as plodding along and
stopping somewhere on the trail of 9s, you'll naturally have a number that's
less than 1. But if you let it rush along at an (infinitely) increasing
pace, you get to 1.
The point about the "infinite" series 0.999... and 1 is that the difference
is infinitely small The point about the "limit" idea is that you get away
from having to go to infinity by saying "This is what it looks like there,
and I can go as close as I want, so I don't need to worry about it any
more".
Naturally, being the sickos they are, mathematicians have delighted in
tweaking that notion ever since ...
I hope that my musings help. It really is one of those questions that can
hold you up because it feels so wrong - like negative numbers, zero,
imaginary numbers and infinite numbers. It helps to delve a bit into the
history of mathematics, when you find out that these aren't "intuitive"
ideas, and were rejected by some very clever mathematicians in their day!
Steve Leibel
<umg...@nospamattglobal.net> wrote:
> Ok, how hard can this be to figure out:
>
> The ellipsis notation means to take the repeating digits, and continue
to repeat
> them without bound:
> 1.999... = 1.99999999..... = 1.999999999999999999... , etc.
>
> Let x = 1.999...
>
> 100x = 199.999...
> - 10x = 19.999...
> --------------------
> 90x = 180
>
> x = 2
>
> 1.999... = 2
>
> QED
>
> Forget the metaphysics, calculus and such. The above algebra is pretty
straight
> forward.
Eric,
This is a commonly given "proof," in quotes . . . but in fact it's bogus,
since you don't actually know for sure that it's legal to multiply an
infinite decimal expression by a constant.
To make this legit you first have to define a repeating decimal as a
particular infinite sum; and then you need to prove that you can multiply
a convergent sum term by term by a constant, and that the resulting series
sums to the constant times the sum of the original series.
Steve L
Adam Russell
Steve Leibel <ste...@coastside.net> wrote in message
news:stevel-1402...@192.168.100.2...
I guess first you need to define 1.999... as a real number or not. If it is
a real number then you can multiply by a constant and etc..., and if it is
not a real number then there is no use trying to decide if it is less than 2
or equal to 2 since it is neither.
Bob Silverman
Huh? Of course it is if you accept 1.999..... as a real number!
It is legal to multiply ANY real number by a constant. The reals form
a field. A field is closed under multiplication. End of discussion.
The only issue through all of these nonsensical threads is whether
one accepts an infinite repeating decimal as a real number. Most of
the crank responses do not seem to accept this.
--
Bob Silverman
"You can lead a horse's ass to knowledge, but you can't make him think"
Sent via Deja.com http://www.deja.com/
Before you buy.
Alan M Dunsmuir
<adamr...@msn.com> writes
>I guess first you need to define 1.999... as a real number or not.
1.999... HAS a well-specified standard definition.
It is 1.0 + sum(n=1 to oo){9/(10^n)}, which in turn can easily be shown
to be an absolutely convergent series with sum 2.
--
Alan M Dunsmuir
Martin Keller-Ressel
intuitive):
For two *distinct* real numbers a and b there's always a real number c
'between' a and b, meaning a < c < b or b < c < a.
(Take for example c := a + (b-a)/2 )
Now let a := 1,999... and b:=2.
c, which has to be greater then a must at least be 2, thus c is greater or
equal b
=> a < c < b cannot be true for any real c; b < c < a neither.
thus a and b are not distinct, they are equal.
--Martin Keller-Ressel
Bruce Grubb
>Ok, how hard can this be to figure out:
>
>The ellipsis notation means to take the repeating digits, and continue to
>repeat
>them without bound:
>1.999... = 1.99999999..... = 1.999999999999999999... , etc.
>
>Let x = 1.999...
>
> 100x = 199.999...
>- 10x = 19.999...
>--------------------
> 90x = 180
>
>x = 2
>
>1.999... = 2
>
>QED
This commonly given 'proof' is totally bogus as can be shown by a Reduco ad
Absurdum argument using another repeating decimal that can be expressed as a
fraction.
1) Any fraction which is represented by a repeating decimal can be multiplied
or divided in that form.
2) any such fraction -must- equal 1 when multiplied by its inverse.
x = 1/3 or .333333333333.
Since 3x in decimal is .9999999999999999 and 3/3 = 1 one of the above
statements is false. Since statement 2) can be shown to be true it is
statement 1) that is false. Therefore 1.999 can not be manipulated via
multiplication until converted into a fraction.
Furthermore since 1.9999 can be espressed as 1 + 3x .33333 the decimal is not
rational.
E. Weddington
Bruce Grubb wrote:
> In article <38A89420...@nospamattglobal.net>, "E. Weddington"
> <umg...@nospamattglobal.net> wrote:
>
> >Ok, how hard can this be to figure out:
> >
> >The ellipsis notation means to take the repeating digits, and continue to
> >repeat
> >them without bound:
> >1.999... = 1.99999999..... = 1.999999999999999999... , etc.
> >
> >Let x = 1.999...
> >
> > 100x = 199.999...
> >- 10x = 19.999...
> >--------------------
> > 90x = 180
> >
> >x = 2
> >
> >1.999... = 2
> >
> >QED
>
> This commonly given 'proof' is totally bogus as can be shown by a Reduco ad
> Absurdum argument using another repeating decimal that can be expressed as a
> fraction.
>
> 1) Any fraction which is represented by a repeating decimal can be multiplied
> or divided in that form.
>
> 2) any such fraction -must- equal 1 when multiplied by its inverse.
>
> x = 1/3 or .333333333333.
>
> Since 3x in decimal is .9999999999999999 and 3/3 = 1 one of the above
> statements is false. Since statement 2) can be shown to be true it is
> statement 1) that is false. Therefore 1.999 can not be manipulated via
> multiplication until converted into a fraction.
>
> Furthermore since 1.9999 can be espressed as 1 + 3x .33333 the decimal is not
> rational.
Looks to me like you just solved it again!
1. x = 1/3 = .333...
2. 3x = .999...
3. 1.999... = 1 + 3x
Substitute x in equation 3:
4. 1.999 = 1 + 3(1/3) = 1+1 = 2
What's the problem?
Eric
E. Weddington
Bob Silverman wrote:
> In article <stevel-1402...@192.168.100.2>,
> ste...@coastside.net (Steve Leibel) wrote:
> > In article <38A89420...@nospamattglobal.net>, "E. Weddington"
> > <umg...@nospamattglobal.net> wrote:
> >
> > > Ok, how hard can this be to figure out:
> > >
> > > The ellipsis notation means to take the repeating digits, and
> continue
> > to repeat
> > > them without bound:
> > > 1.999... = 1.99999999..... = 1.999999999999999999... , etc.
> > >
> > > Let x = 1.999...
> > >
> > > 100x = 199.999...
> > > - 10x = 19.999...
> > > --------------------
> > > 90x = 180
> > >
> > > x = 2
> > >
> > > 1.999... = 2
> > >
> > > QED
> > >
> > > Forget the metaphysics, calculus and such. The above algebra is
> pretty
> > straight
> > > forward.
> >
> > Eric,
> >
> > This is a commonly given "proof," in quotes . . . but in fact it's
> bogus,
> > since you don't actually know for sure that it's legal to multiply an
> > infinite decimal expression by a constant.
>
> Huh? Of course it is if you accept 1.999..... as a real number!
>
> It is legal to multiply ANY real number by a constant. The reals form
> a field. A field is closed under multiplication. End of discussion.
>
> The only issue through all of these nonsensical threads is whether
> one accepts an infinite repeating decimal as a real number. Most of
> the crank responses do not seem to accept this.
>
What I don't understand is why can't one accept that an infinite repeating
decimal is not only in R, but you only have to go so far as Q to find one:
1/3 = .333...
Eric
Tapio Hurme
For example Rudy Rucker in his book "Infinity and the mind - The science and
philosophy of the infinite" presents a "counter" proof, which is also weak
as he well writes:
1) FAQ - proof:
10K=9.999...
- K=0.999...
___________
9K=9 and K=1 , but
2) Ruckers example:
Assuming K=1-(1/w)
10K=10K-10/w
- K= 1- 1/w
____________
9K=9-1/w and K=1-1/w
which proofs only that "counting" works, but nothing about 0.999...
BTW: It is amasing that the same thing rises like a sun. A while ago 0.999=1
Now 1.999..=2 and I guess very soon 2.999...=3 . Infinitely
I like humour but...
there are two kinds of people : Those who understand the joke immediately
and those who need the joke told twice - need the joke told twice.
Do not repeat the same "facts" on items once told under title 0.999.... Try
to lift the level of discussion for and against-too. New argumentations are
worth to discuss and wellcome.
Thanks
Tapio
Adam Russell <adamr...@msn.com> kirjoitti
viestissä:OTQeKK3d$GA.122@cpmsnbbsa05...
>
> Steve Leibel <ste...@coastside.net> wrote in message
> news:stevel-1402...@192.168.100.2...
> > In article <38A89420...@nospamattglobal.net>, "E. Weddington"
> > <umg...@nospamattglobal.net> wrote:
> >
> > > Ok, how hard can this be to figure out:
> > >
> > > The ellipsis notation means to take the repeating digits, and continue
> > to repeat
> > > them without bound:
> > > 1.999... = 1.99999999..... = 1.999999999999999999... , etc.
> > >
> > > Let x = 1.999...
> > >
> > > 100x = 199.999...
> > > - 10x = 19.999...
> > > --------------------
> > > 90x = 180
> > >
> > > x = 2
> > >
> > > 1.999... = 2
> > >
> > > QED
> > >
> > > Forget the metaphysics, calculus and such. The above algebra is pretty
> > straight
> > > forward.
> >
> > Eric,
> >
> > This is a commonly given "proof," in quotes . . . but in fact it's
bogus,
> > since you don't actually know for sure that it's legal to multiply an
> > infinite decimal expression by a constant.
> >
Ken Cox
> 2) Ruckers example:
>
> Assuming K=1-(1/w)
>
> 10K=10K-10/w
> - K= 1- 1/w
> ____________
> 9K=9-1/w and K=1-1/w
This last line should read 9K = 9-1/w and K = 1-1/(9w).
> which proofs only that "counting" works, but nothing about 0.999...
Actually, when you apply this to the 0.999... case, this
informally shows that there is no non-zero real number W
such that 0.999... = 1-W (take the above with W = 1/w).
Otherwise you get the contradiction that W is non-zero
and W = W/9.
--
Ken Cox k...@research.bell-labs.com
Virgil
<bgr...@zianet.com> wrote:
>Because x is _also_ 1/3 and 3 * 1/3 =1 3x cannot both equal 1 and .9999
>ergo
>you canNOT multiply or divide a fraction that result in a repeating
>decimal
>in decimal form.
By your type of argument, 2/4 cannot be equal to 1/2 because they look
different.
If 0.999... is an abreviation for the number represented by a decimal
point followed by infinitely many nines, then the assumption that it
represents a number means that the number that it represents must be 1.
If it does not represent 1, then it cannot represent a number at all,
and this entire thread is a waste of bandwidth.
If it is a number, then it must be 1 and this entire thread is a waste
of bandwidth.
--
Virgil
vm...@frii.com
Bruce Grubb
<vm...@frii.com> wrote:
>In article <bgrubb-F0CBCF....@www.zianet.com>, Bruce Grubb
><bgr...@zianet.com> wrote:
>
>>Because x is _also_ 1/3 and 3 * 1/3 = 1 3x cannot both equal 1 and .9999
>>ergo you canNOT multiply or divide a fraction that result in a repeating
>>decimal in decimal form.
>
>By your type of argument, 2/4 cannot be equal to 1/2 because they look
>different.
Totally different situation. Both the above faction result in a finite value
0.5. If not just that the value look different but -behave- differently.
This is why fractions that result in repeating decimals need to put back into
>If 0.999... is an abreviation for the number represented by a decimal
>point followed by infinitely many nines, then the assumption that it
>represents a number means that the number that it represents must be 1.
The assumption makes no sence. The reality is no matter how far you take the
0.9999... it NEVER is equal to 1 using normal mathmatics. Now if one is
talking -limits- than one can make the arugement but limits mathmatic is a
different type of math that what has been used so far.
>If it does not represent 1, then it cannot represent a number at all,
>and this entire thread is a waste of bandwidth.
This is more nonsence. 0.9999... is distict number from 1. At best is it a
limit that aproaches 1 but NEVER gets there. Trying to claim otherwise
produces some really silly results.
The prove is of the same order of nonsence of saying 0/0 = a is a valid
equation. The standard 'if c/b = a then a * b = c' shows all other divitions
by 0 to be invalid (c cannot be any value but 0) but 0/0 = a produces a
prefectly valid equation (a * 0 = 0). The problem is that if one fallows
though with the logic then 1 = 2.
Given: if c/b = a then a * b = c; a/a = 1
0/0 = 1 (1 * 0 = 0)
0/0 = 2 (2 * 0 = 0)
ergo 1 = 2
This total nonsence results because a rule in mathmatics is ignored. The
proof similarlly ignores a similar rule (treating a limit as a finate value)
Torkel Franzen
> For two *distinct* real numbers a and b there's always a real number c
> 'between' a and b, meaning a < c < b or b < c < a.
> (Take for example c := a + (b-a)/2 )
> Now let a := 1,999... and b:=2.
> c, which has to be greater then a must at least be 2, thus c is greater or
> equal b
Your comments don't prove anything, and indeed the "let a:=1,999..."
doesn't define anything in the absence of limits.
I think we just have to acquiesce in the fact that 1,999... is a
great mystery, just like 0,999... It is impossible for mathematicians
to dissolve this mystery by their crude art.
So the question then is only what to do with those who merely and
genuinely want to understand what "0,999...=1" means. These might
best be served by an email reply referring them to some standard site.
The rest is mystery.
david_w_...@my-deja.com
"E. Weddington" <umg...@nospamattglobal.net> wrote:
> What I don't understand is why can't one accept that an infinite
> repeating decimal is not only in R, but you only have to go so far as
> Q to find one: 1/3 = .333...
Who said that this is not valid in Q? It certainly is. Although it does
seem a bit strange to think of summing an infinite series strictly in Q
(without ever having developed R first), it seems to me that such could
indeed be done. So yes, in Q, 1/3 = .333...
(Of course, you've just got to be aware that infinite series of
rational terms which would have converged to an irrational number in R
will not converge in Q.)
Cheers,
David Cantrell
Eric Stevens
wrote:
>In article <38A89420...@nospamattglobal.net>, "E. Weddington"
><umg...@nospamattglobal.net> wrote:
>
>>Ok, how hard can this be to figure out:
>>
>>The ellipsis notation means to take the repeating digits, and continue to
>>repeat
>>them without bound:
>>1.999... = 1.99999999..... = 1.999999999999999999... , etc.
>>
>>Let x = 1.999...
>>
>> 100x = 199.999...
>>- 10x = 19.999...
>>--------------------
>> 90x = 180
>>
>>x = 2
>>
>>1.999... = 2
>>
>>QED
>
>This commonly given 'proof' is totally bogus as can be shown by a Reduco ad
>Absurdum argument using another repeating decimal that can be expressed as a
>fraction.
>
>1) Any fraction which is represented by a repeating decimal can be multiplied
>or divided in that form.
>
>2) any such fraction -must- equal 1 when multiplied by its inverse.
>
>
>x = 1/3 or .333333333333.
>
>Since 3x in decimal is .9999999999999999 and 3/3 = 1 one of the above
>statements is false. Since statement 2) can be shown to be true it is
>statement 1) that is false. Therefore 1.999 can not be manipulated via
>multiplication until converted into a fraction.
>
>Furthermore since 1.9999 can be espressed as 1 + 3x .33333 the decimal is not
>rational.
Apatr from the fact that the above looks like a circular argument to
me, the matter is not helped by your presentation of it as a finite
series of digits
Now 1/0.3 = 3.3333...
1/0.33 = 3.030303...
1/0.333 = 3.003003003...
In other words, when there are 'n' digits in the original decimal, the
reciprocal is of the form 3.[(n-1) zeros]3... .
It follows that when n = oo the reciprocal will be 3.[infinite series
of zeros] = 3.00000000000000000000000000000....
The important point is that value of the reciprocal of the decimal
approaches 3 only as the number of decimal 3s in the expression
approaches infinity and the limit of 3/0.333[to n digits] tends to 1
only when n = oo.
Your conclusion is correct when 'n' is finite but that is not the case
we have been talking about. I expect you to be severely jumped upon.
Eric Stevens
There are two classes of people. Those who divide people into
two classes, and those who don't. I belong to the second class.
Andrew Hackard
>3x cannot both equal 1 and .9999[...]
Sure it can, if you assume they're the same number to begin with.
Transitive property of equality, don'tcha know.
(Even if you don't assume it, it's a trivial matter of thirty seconds to sum
the geometric series and *prove* they're the same number, and then you may
as well treat them the same, even if it's a minor abuse of notation.)
--
"As for rumor and reputation, let us consider them as
matters that must not guide, but follow our actions."
======================== --L. Annaeus Seneca Minor
<*> ahac...@uvic.ca
Adam Russell
Bruce Grubb <bgr...@zianet.com> wrote in message
news:bgrubb-9D8A68....@www.zianet.com...
> In article <vmhjr-4607EF....@news.frii.com>, Virgil
> <vm...@frii.com> wrote:
>
> >In article <bgrubb-F0CBCF....@www.zianet.com>, Bruce Grubb
> ><bgr...@zianet.com> wrote:
> >
Limits ARE finite values (except when they increase (in value) without
bound).
Dave Seaman
Bruce Grubb <bgr...@zianet.com> wrote:
>In article <vmhjr-4607EF....@news.frii.com>, Virgil
><vm...@frii.com> wrote:
>
>>In article <bgrubb-F0CBCF....@www.zianet.com>, Bruce Grubb
>><bgr...@zianet.com> wrote:
>>
>>>Because x is _also_ 1/3 and 3 * 1/3 = 1 3x cannot both equal 1 and .9999
>>>ergo you canNOT multiply or divide a fraction that result in a repeating
>>>decimal in decimal form.
>>
>>By your type of argument, 2/4 cannot be equal to 1/2 because they look
>>different.
>
>Totally different situation. Both the above faction result in a finite value
>0.5. If not just that the value look different but -behave- differently.
Both 1 and 0.999... result in a finite value 1.0. They may look
different, but they do not behave differently. Assuming that both 1 and
0.999... are real numbers, show me how they behave differently.
>The assumption makes no sence. The reality is no matter how far you take the
>0.9999... it NEVER is equal to 1 using normal mathmatics. Now if one is
>talking -limits- than one can make the arugement but limits mathmatic is a
>different type of math that what has been used so far.
Looks like it's time for me to dig out my four questions again.
1. Do you agree that 0.999... is a real number?
2. Do you agree that 0.999... > 1 - 1/10^n for every n > 0?
3. Do you agree that 0.999... is not greater than 1?
4. Do you agree that the reals are a complete ordered field?
In particular, #4 implies the least upper bound property: every nonempty
set of reals that is bounded above has a least upper bound.
If you agree to these four points, then I can prove that 0.999... = 1.
Which do you not accept, and why?
It sounds as if you do not accept #1, because you think 0.999... is
somehow a moving target and does not have a fixed value, and therefore
can't be a real number at all. If this is the case, then there is
nothing to argue about. Obviously, we can't have 0.999... = 1 if the
object on the right is a real number and the object on the left is not.
It's like asking whether an apple can equal an orange.
>>If it does not represent 1, then it cannot represent a number at all,
>>and this entire thread is a waste of bandwidth.
>
>This is more nonsence. 0.9999... is distict number from 1. At best is it a
>limit that aproaches 1 but NEVER gets there. Trying to claim otherwise
>produces some really silly results.
You misunderstand limits. A limit (in the present context, at least) is
a definite value and does not approach anything. It is incorrect to say
that the limit of 1-1/10^n as n -> oo *approaches* 1. The correct
statement is that the limit *is* 1.
>This total nonsence results because a rule in mathmatics is ignored. The
>proof similarlly ignores a similar rule (treating a limit as a finate value)
Wrong. A limit is indeed a finite value. If { a_n } is a sequence and L
is a real number, then the following statements have the same meaning.
a. a_n -> L as n -> oo.
b. lim_{n->oo} a_n = L.
c. For every epsilon > 0, there exists N > 0 such that
| a_n - L | < epsilon for every n > N.
Notice that statement (b) contains an equal sign. Yet it has precisely
the same meaning as statements (a) and (c), which do not contain equal
signs. In fact, (c) is the definition of (a) and (b).
You might also notice that statement (b) is the only one of the three
that actually mentions the limit. Those two facts are related. You are
confusing "the limit equals L" with "a_n equals L for some n". Those two
phrases have nothing to do with each other.
--
Dave Seaman dse...@purdue.edu
U.S. State Department Lies about Mumia Abu-Jamal
<http://mojo.calyx.net/~refuse/mumia/111999statedept.html>
Richard Tobin
>The assumption makes no sence. The reality is no matter how far you take the
>0.9999... it NEVER is equal to 1 using normal mathmatics. Now if one is
>talking -limits- than one can make the arugement but limits mathmatic is a
>different type of math that what has been used so far.
0.9999... is *defined* as meaning the infinite sum
9/10 + 9/100 + 9/1000 + ...
and an infinite sum is defined to mean the limit as the number of
terms goes to infinity. And in this case that limit is 1.
>This is more nonsence. 0.9999... is distict number from 1. At best is it a
>limit that aproaches 1 but NEVER gets there. Trying to claim otherwise
>produces some really silly results.
Limits don't approach anything. A sequence approaches a limit. The
sequence
0.9 0.99 0.999 0.999 ...
never reaches its limit, of course, but 0.9999... means the limit, not
the sequence.
-- Richard
--
Spam filter: to mail me from a .com/.net site, put my surname in the headers.
"The Internet is really just a series of bottlenecks joined by high
speed networks." - Sam Wilson
Bruce Grubb
Hackard) wrote:
>Bruce Grubb <bgr...@zianet.com> said...
>>3x cannot both equal 1 and .9999[...]
>
>Sure it can, if you assume they're the same number to begin with.
But they are not. While 0.9999... is close to 1 is it not 1. The number
-limit- is 1 not the number itself.
Iain Davidson
> The assumption makes no sence. The reality is no matter how far you take
the
> 0.9999... it NEVER is equal to 1 using normal mathmatics. Now if one is
> talking -limits- than one can make the arugement but limits mathmatic is a
> different type of math that what has been used so far.
> This is more nonsence. 0.9999... is distict number from 1. At best is it
a
> limit that aproaches 1 but NEVER gets there. Trying to claim otherwise
> produces some really silly results.
If you are talking about a number then a number has a
constant, unchanging value.
If 0.99.. is a number then it must have a constant value.
You seem to think 0.99. is some continuously growing quantity
that increases as the decimal expansion gets "longer and longer".
But the decimal expansion contains the same number of 9's
as there are counting numbers and the "number" of counting
numbers is the same today as it was yesterday.
You can truncate the decimal expansion of 0.99.. at any point
0.9
0.99
0.999
and so on
and each of these values is indeed less than 1.
0.99.. is the limit of this sequence and is greater than
any decimal with a finite number of 9's after the decimal
point.
You are confusing the value of increasing terms in the
sequence 0.9, 0.99, with the limit of the sequence
0.999... which is also equal to 1.
If you claim that 0.999.... <> 1
then |1 - 0.999...| > 0 = D or |1 - 0.999...| = 0
you should, therefore, be able to state some number D>0 that meets this
condition.
It can't be 0.1 as |1 - 0.99| = 0.01 and 0.99.. > 0.99
No matter how small you make D you can always
find a decimal with a finite number of 9's after the
decimal point which will give a smaller difference D' and so
0.999.. must give a even smaller difference.
So D must be the smallest number greater than 0.
But no such number D exists because D/2 would
be smaller.
So |1 - 0.999...| = 0
E. Weddington
Bruce Grubb wrote:
> In article <88bpjn$fko$1...@nnrp1.deja.com>, Bob Silverman <bo...@rsa.com>
> wrote:
>
> >In article <stevel-1402...@192.168.100.2>,
> > ste...@coastside.net (Steve Leibel) wrote:
> >> In article <38A89420...@nospamattglobal.net>, "E. Weddington"
> >> <umg...@nospamattglobal.net> wrote:
> >>
> >> > Ok, how hard can this be to figure out:
> >> >
> >> > The ellipsis notation means to take the repeating digits, and
> >continue
> >> to repeat
> >> > them without bound:
> >> > 1.999... = 1.99999999..... = 1.999999999999999999... , etc.
> >> >
> >> > Let x = 1.999...
> >> >
> >> > 100x = 199.999...
> >> > - 10x = 19.999...
> >> > --------------------
> >> > 90x = 180
> >> >
> >> > x = 2
> >> >
> >> > 1.999... = 2
> >> >
> >> > QED
> >> >
> >> > Forget the metaphysics, calculus and such. The above algebra is
> >> > pretty straight forward.
> >>
> >> Eric,
> >>
> >> This is a commonly given "proof," in quotes . . . but in fact it's
> >> bogus, since you don't actually know for sure that it's legal to multiply an
> >> infinite decimal expression by a constant.
> >
> >Huh? Of course it is if you accept 1.999..... as a real number!
>
> It is not a question of 1.999..... being a real number but of it being a
> -rational- number. The multiplication and subtraction hinge on 1.999.....
> being a rational number. If it is not then the subtraction is invalid.
>
> >It is legal to multiply ANY real number by a constant. The reals form
> >a field. A field is closed under multiplication. End of discussion.
>
> But is is NOT legal to manipulate an irrational number as if it was a
> rational number nor is is legal to manipulate a fraction that results in a
> repeting decimal in decimal form.
>
> For example x = 1/7 = .142857... but if you multiply the decimal by 7 you do
> not get 1. By the rules of Reduco ad Absurdum then it is only legal to
> manipulate a rational number that is a repeating decimal in factional form.
>
> Furthermore if 1.999... is not a rational number than it cannot be added or
> subtracted as if it was rational which is a key assuption of the 'proof'.
>
> >The only issue through all of these nonsensical threads is whether
> >one accepts an infinite repeating decimal as a real number.
>
> No the issues are accually is whether one accepts a partical infinite
> repeating decimal as a -rational- number and if manipulating it in decimal
> form is valid.
>
> As the case of 1/3, 1/7, 1/11, 1/13 or about any 1/(prime number or multiple)
> shows this is nonsence.
I've got a calculus textbook here, says that every rational number can be written
as a repeating decimal and that every repeating decimal can be written as a
rational number.
Examples:
13/11 = 1.18181818...
3/8 = 0.3750000000...
Take your example:
x = 1/7 = .142857142857...
1000000x = 142857.142857142857...
- x = 0.142857142857...
-----------------------------------------
999999x = 142857
x = 142857 / 999999
which can be reduced to... guess what? 1/7!
Another example that might not be to farfetched:
x = 3/8 = 0.375000000...
10000x = 3750.000000...
- 1000x = 375.000000...
----------------------------
9000x = 3375
x = 3375 / 9000
Divide the numerator and denominator by 1125, whaddya get? 3/8
Tell me what the difference is between the two infinitely repeating decimals.
.999999... = 1
.999... = 1
.999999999999999999999999999999999...9 not= 1, because the decimal number has a
finite number of digits.
Eric Weddington
Alan M Dunsmuir
<bgr...@zianet.com> writes
>While 0.9999... is close to 1 is it not 1.
That claim would need to imply that you can find another number between
0.9999... and 1. What number would that be?
Just how close to 1 do you think it is?
--
Alan M Dunsmuir
Ken Cox
> It is not a question of 1.999..... being a real number but of it being a
> -rational- number. The multiplication and subtraction hinge on 1.999.....
> being a rational number. If it is not then the subtraction is invalid.
When did it become illegal to multiply and subtract with
real numbers? I have some computer programs I will have
to modify...
--
Ken Cox k...@research.bell-labs.com
Ken Cox
> But they are not. While 0.9999... is close to 1 is it not 1. The number
> -limit- is 1 not the number itself.
0.999... does not have a limit, since no single number has a
limit (what is the limit of 1? Does 1.000... have a limit?).
The notation 0.999... *means* the limit of the series 0.9,
0.99, 0.999, 0.9999, ... and that limit is 1. So 0.999... = 1.
--
Ken Cox k...@research.bell-labs.com
Tapio Hurme
Adam Russell <adamr...@msn.com> kirjoitti
viestissä:uOqWsUBe$GA.318@cpmsnbbsa03...
>
> Tapio Hurme <hurm...@dlc.fi> wrote in message
> news:88cbog$385$1...@tron.sci.fi...
> > 2) Ruckers example:
> >
> > Assuming K=1-(1/w)
> >
> > 10K=10K-10/w
> > - K= 1- 1/w
> > ____________
> > 9K=9-1/w and K=1-1/w
> Actually you made 2 mistakes here. The last line ends up being 9k=9-9/w
> which is obviously true.
Thanks for correcting typo. Should be 9K=9-9/w , but as professor Rucker
wrote it does not proof anything about 0.999... as well as FAQ example.
Tapio
Tapio Hurme
Ken Cox <k...@research.bell-labs.com> kirjoitti
viestissä:38A9B677...@research.bell-labs.com...
> Tapio Hurme wrote:
> > 2) Ruckers example:
> >
> > Assuming K=1-(1/w)
> >
> > 10K=10K-10/w
> > - K= 1- 1/w
> > ____________
> > 9K=9-1/w and K=1-1/w
>
Well, I wrote a typo - the last line should read 9K=9-9/w and thus K=1-1/w
as prof. Rucker wrote it.
>
> > which proofs only that "counting" works, but nothing about 0.999...
>
> Actually, when you apply this to the 0.999... case, this
> informally shows that there is no non-zero real number W
> such that 0.999... = 1-W (take the above with W = 1/w).
> Otherwise you get the contradiction that W is non-zero
> and W = W/9.
Should be. W=1/w
A)
But now assuming 0.999...=1=1-1/w it follows assuming 1=1-1/w:
1) 0=-1/w or
writing 1= (w-1)/w implies w*1= w-1 -> w=w-1 ->
2) 0=-1 Impossible or not true!
Can we conclude?:
1) Case is OK
2) Case is not OK
It follows: multiplication is not valid operation. If this is true, then it
is not allowed to multiply with infinite number "notation". If that is true,
then the multiplication of 10*0.999... is not allowed as in the case of "FAQ
proof".
Or shall we ask, because multiplication and division are inverse operations,
that neither is allowed? And nothing is proofed.
Tapio
> --
> Ken Cox k...@research.bell-labs.com
Ken Cox
> Well, I wrote a typo - the last line should read 9K=9-9/w and thus K=1-1/w
> as prof. Rucker wrote it.
Whoops, didn't catch that.
> It follows: multiplication is not valid operation. If this is true, then it
> is not allowed to multiply with infinite number "notation".
If 0.999... is a real number, then multiplication is most
certainly a valid operation on it, regardless of notation.
Since it is easy to show that 0.999... is a real number[*],
we can use it in multiplication. The only question is
*what* real number it is, and again it's pretty easy to
show that it's 1.
Personally I've never understood why some people get upset
that there are numbers that have more than one representation
in our notational system. To me, it's like getting upset
because 0.5, 1/2, (2+8)/(4*5), etc. all represent the same
number.
[*] Indeed, it is pretty easy to show that any digit string
0.(d1)(d2)(d3)... is a real number, since it pretty much
defines its own Cauchy sequence of rationals.
--
Ken Cox k...@research.bell-labs.com
Doug Norris
>But they are not. While 0.9999... is close to 1 is it not 1. The number
>-limit- is 1 not the number itself.
Number limit? NUMBER LIMIT?!?! It's either a number, or a limit. If
it's a limit, then why the @!@!@# are you talking about it like a number?
If it's a number, which it IS, then it's equal to 1.
Explain just WTF a number limit is.
Doug
----------------------------------------------------------------------------
Douglas Todd Norris (norr...@euclid.colorado.edu) "The Mad Kobold"
Hockey Goaltender Home Page:http://ucsu.colorado.edu/~norrisdt/goalie.html
----------------------------------------------------------------------------
"Maybe in order to understand mankind, we have to look at the word itself.
Mankind. Basically, it's made up of two separate words---"mank" and "ind".
What do these words mean? It's a mystery, and that's why so is mankind."
- Deep Thought, Jack Handey
Tapio Hurme
Alan M Dunsmuir <al...@moonrake.demon.co.uk> kirjoitti
viestissä:esSEPCAx...@moonrake.demon.co.uk...
>
> 0.9999... and 1. What number would that be?
That is also discussed question and the counter question is:
What number is between 10 and 9 if decimal notation or fractions are not
allowed or invented ?
What is their mean value if decimal notation or fractions are not allowed or
invented.
Do you claim that 10=9 because you cannot find anything between them if
decimal notation or fractions are not allowed or invented?
See the answers in dejanews.
Tapio
E. Weddington
Tapio Hurme wrote:
> Ken Cox <k...@research.bell-labs.com> kirjoitti
> viestissä:38A9B677...@research.bell-labs.com...
> > Tapio Hurme wrote:
> > > 2) Ruckers example:
> > >
> > > Assuming K=1-(1/w)
> > >
> > > 10K=10K-10/w
> > > - K= 1- 1/w
> > > ____________
> > > 9K=9-1/w and K=1-1/w
> >
> > This last line should read 9K = 9-1/w and K = 1-1/(9w).
>
> Well, I wrote a typo - the last line should read 9K=9-9/w and thus K=1-1/w
> as prof. Rucker wrote it.
> >
> > > which proofs only that "counting" works, but nothing about 0.999...
> >
> > Actually, when you apply this to the 0.999... case, this
> > informally shows that there is no non-zero real number W
> > such that 0.999... = 1-W (take the above with W = 1/w).
> > Otherwise you get the contradiction that W is non-zero
> > and W = W/9.
>
> Should be. W=1/w
>
> A)
> But now assuming 0.999...=1=1-1/w it follows assuming 1=1-1/w:
>
> 1) 0=-1/w or
>
> writing 1= (w-1)/w implies w*1= w-1 -> w=w-1 ->
> 2) 0=-1 Impossible or not true!
>
> Can we conclude?:
> 1) Case is OK
> 2) Case is not OK
> It follows: multiplication is not valid operation. If this is true, then it
> is not allowed to multiply with infinite number "notation". If that is true,
> then the multiplication of 10*0.999... is not allowed as in the case of "FAQ
> proof".
> Or shall we ask, because multiplication and division are inverse operations,
> that neither is allowed? And nothing is proofed.
>
> Tapio
Your assumption in A) above is invalid. You can't assume that 0.999... =
1-(1/w).
That's like saying
f(x) = x
f(y) = y
Therefore x = y.
The rest of the logic following that assumption is then flawed.
Eric
Ken Cox
> Alan M Dunsmuir <al...@moonrake.demon.co.uk> wrote:
> > That claim would need to imply that you can find another number between
> > 0.9999... and 1. What number would that be?
> That is also discussed question and the counter question is:
> What number is between 10 and 9 if decimal notation or fractions are not
> allowed or invented ?
> What is their mean value if decimal notation or fractions are not allowed or
> invented.
> Do you claim that 10=9 because you cannot find anything between them if
> decimal notation or fractions are not allowed or invented?
The reals have a density property -- I do not recall the
correct name -- that the integers do not. This property
states that for any two reals r1 and r2, if r1 < r2 then
there exists some r3 such that r1 < r3 < r2.
Thus, if you say that 0.999... is not equal to 1, then you
must be saying 0.999... is less than 1. That means that
there must be some real r3 such that 0.999... < r3 < 1. As
you are the one claiming 0.999... is not equal to 1, it is
up to you to exhibit such a real r3.
Because the integers do not have the above property, there
is no requirement that there be an integer between any two
unequal integers. So there's no problem that there is no
integer between 9 and 10.
--
Ken Cox k...@research.bell-labs.com
Doug Norris
>> That claim would need to imply that you can find another number between
>> 0.9999... and 1. What number would that be?
>That is also discussed question and the counter question is:
>What number is between 10 and 9 if decimal notation or fractions are not
>allowed or invented ?
That's hardly a "counter-question". You might as well answer the above
question by covering your ears and yelling "la la la - I can't hear you"
at the top of your lungs.
Seriously, Tapio - what's your problem with the question?
Doug
Bruce Grubb
<Sttsc...@tesco.net> wrote:
>Bruce Grubb <bgr...@zianet.com> wrote in message
>news:bgrubb-9D8A68....@www.zianet.com...
>> In article <vmhjr-4607EF....@news.frii.com>, Virgil
>> <vm...@frii.com> wrote:
>
>> The assumption makes no sence. The reality is no matter how far you take
>the
>> 0.9999... it NEVER is equal to 1 using normal mathmatics. Now if one is
>> talking -limits- than one can make the arugement but limits mathmatic is a
>> different type of math that what has been used so far.
>
>> This is more nonsence. 0.9999... is distict number from 1. At best is it
>a
>> limit that aproaches 1 but NEVER gets there. Trying to claim otherwise
>> produces some really silly results.
>
>If you are talking about a number then a number has a
>constant, unchanging value.
Right and to say 0.9999... = 1 change is vallu and supports my possition
Bruce Grubb
(Richard Tobin) wrote:
>In article <bgrubb-9D8A68....@www.zianet.com>,
>Bruce Grubb <bgr...@zianet.com> wrote:
>
>>The assumption makes no sence. The reality is no matter how far you take
>>the
>>0.9999... it NEVER is equal to 1 using normal mathmatics. Now if one is
>>talking -limits- than one can make the arugement but limits mathmatic is a
>>different type of math that what has been used so far.
>
>0.9999... is *defined* as meaning the infinite sum
>
> 9/10 + 9/100 + 9/1000 + ...
>
>and an infinite sum is defined to mean the limit as the number of
>terms goes to infinity. And in this case that limit is 1.
But the statement is that 0.99999 = 1 which is of course nonsence. At at
best it come as close to 1 without accually being 1. Its like traveling half
the distance between you and a destination. No matter how long you travel
you NEVER get to the destination.
Dave Seaman
Bruce Grubb <bgr...@zianet.com> wrote:
>But the statement is that 0.99999 = 1 which is of course nonsence. At at
>best it come as close to 1 without accually being 1. Its like traveling half
>the distance between you and a destination. No matter how long you travel
>you NEVER get to the destination.
Yes, it is nonsense, but you have the wrong statement. It is not 0.9999
= 1 that we are discussing. It is 0.9999... = 1. Do you see those three
dots? They mean that the string of nines goes all the way to infinity.
That makes all the difference.
Brett K. Heath
> -rational- number. The multiplication and subtraction hinge on 1.999.....
> being a rational number. If it is not then the subtraction is invalid.
You are mistaken.
First of all I do actually know for sure that it is "legal" to multiply an
infinite decimal expression by a constant. Secondly, even if 1.999... were
irrational (which it isn't) that would not invalidate the proof.
>
> >It is legal to multiply ANY real number by a constant. The reals form
> >a field. A field is closed under multiplication. End of discussion.
>
> But is is NOT legal to manipulate an irrational number as if it was a
> rational number nor is is legal to manipulate a fraction that results in a
> repeting decimal in decimal form.
Name one elementary operation that is valid when performed on
rational numbers but invalid when irrationals are used, or even one
that behaves differently depending on whether or not it's arguments are
rational or irrational. As for your objection to the repeating decimal
form, how's this.
1/2 * 2 = .500000... * 2.00000... = 1.00000..... = 1
EVERY REAL NUMBER has a repeating decimal representation, and this form can
be operated on in exactly the same way that the "shorthand" version can. In
fact, the fundamental requirement of the notation that the results are not
affected by the form in which the arguments are represented is one of the
reasons that we accept the "odd looking" 1=.9999999..., if this equation
were not true it would "break" the internal consistency of the logic, and
it wouldn't be math anymore.
>
> For example x = 1/7 = .142857... but if you multiply the decimal by 7 you do
> not get 1. By the rules of Reduco ad Absurdum then it is only legal to
> manipulate a rational number that is a repeating decimal in factional form.
You seem to be confusing round-off errors with mathematical inconsistency,
it is certainly more convenient in some cases to use the fractional form,
that's the entire point of having multiple representations of the same
number, sometimes one form is easier to work with than another. The price
we pay for this flexibility is that both (or all) representations of the
same number _must_ be equally valid in all contexts, if this isn't the case
then the representations aren't really equivalent and we have either
ambiguity or inconsistency, neither of which is acceptable in mathematical
notation.
>
> Furthermore if 1.999... is not a rational number than it cannot be added or
> subtracted as if it was rational which is a key assuption of the 'proof'.
I've covered this above, but what is different about the way you add,
subtract, multiply and divide rationals and the way you do the same
operations on irrationals (and 1.999... is definitely rational anyway).
>
> >The only issue through all of these nonsensical threads is whether
> >one accepts an infinite repeating decimal as a real number.
Again I'm repeating myself but *EVERY* real number has an infinite decimal
representation. Conversely every number with an infinite decimal
representation is real. Further every rational number has an infinite
repeating decimal representation, and every infinite repeating decimal
represents a rational number. Even further, infinite decimals are the only
commonly used notations that are capable of providing a unique
representation for each real number.
>
> No the issues are accually is whether one accepts a partical infinite
> repeating decimal as a -rational- number and if manipulating it in decimal
> form is valid.
Those aren't the issues, all repeating decimals *are* rational and all
rationals have an omniequivalent repeating decimal representation. If you
couldn't manipulate the decimal form there would be no reason to _have_ a
decimal form, the form was defined in such a way as to make sure arithmetic
manipulation was valid.
>
> As the case of 1/3, 1/7, 1/11, 1/13 or about any 1/(prime number or multiple)
> shows this is nonsence.
No, it shows you really aren't that comfortable with fractions and decimals
( no crime here, many aren't).
Brett K. Heath
Russell Easterly
It is easy to show that there is a "number" between
1.0... (base 10) and 0.999... (base 10) if you allow
numbers in bases other than 10.
For example, 0.AAA...AAA (base 11) can be shown
to be less than 1.0...000 (base 10) for any finite number of positions.
(A in base 11 = 10 in base 10)
It is also possible to show that
0.AAA...AAA (base 11) is greater than 0.999...999 (base 10)
for any finite number of positions.
n positions of .999... (base 10) = 1 - 10^(-n)
n positions of .AAA... (base 11) = 1 - 11^(-n)
So:
0.999...999 (base 10) < 0.AAA...AAA (base 11) < 1.000...000 (any base)
for any finite number of positions.
Russell
-2 many 2 count
NightGaunt J Smith
>
> This is more nonsence. 0.9999... is distict number from 1. At best
> is it a limit that aproaches 1 but NEVER gets there.
Bruce,
First, your grammar: it is incorrect to say that a limit "approaches
1." A sequence can approach 1 [in which case 1 is the "limit,"] but
a number can not "approach" 1 or anything else. A number just sits put
in one spot on the number line; its value does not vary, nor is its
value mysteriously indefinite.
That being said: The string "0.999..." is shorthand for the _limit_
of the sequence { 0, 0.9, 0.99, 0.999, ... }. The _limit_ of this
sequence is 1, and there are plenty of ways to prove this, with
whatever level of rigor you desire.
You're correct that, if you start going through this sequence, or
start adding 9s after a decimal point, you'll never finish, and
after any finite amount of time have a number less than 1
(I'm assuming you can only move a pencil so quickly.) However,
0.999... doesn't mean that; it represents the limit approached by
this process.
In sum (hee hee,) 0.9999... is a number. It does not "have a limit"
or "approach" a value. Rather the sequence {0, 0.9, 0.99, ...}
approaches IT. 0.999... IS the limit. That limit is 1.
NG
NightGaunt J Smith
>
> This is a commonly given "proof," in quotes . . . but in fact it's
> multiply an infinite decimal expression by a constant.
Sure I know for sure. We've known this for some time.
> To make this legit you first have to define a repeating decimal as a
> particular infinite sum;
This is already the way it is defined. The decimal (not necessarily
repeating!!) 0.(a_1)(a_2)(a_3)(a_4)...(a_i)... is defined as:
SUM (a_i) / 10^i
i>0
This always converges, no matter what the values of the sequence
{ a_i } are (given, of course, that each a_i is between 0 and 9.)
When someone provides a finite sequence of digits and an ellipsis,
the intention is to define the entire sequence { a_i } through a
finite pattern. For instance, "0.99..." means a_i = 9 for all i.
> and then you need to prove that you can multiply
> a convergent sum term by term by a constant, and that the
> resulting series sums to the constant times the sum of the original
> series.
You don't think this has been proven yet??
> Steve L
Eric Stevens
<k...@research.bell-labs.com> wrote:
>Bruce Grubb wrote:
>> It is not a question of 1.999..... being a real number but of it being a
>> -rational- number. The multiplication and subtraction hinge on 1.999.....
>> being a rational number. If it is not then the subtraction is invalid.
>
>When did it become illegal to multiply and subtract with
>real numbers? I have some computer programs I will have
>to modify...
Thats the long and the short of it.
Eric Stevens
There are two classes of people. Those who divide people into
two classes, and those who don't. I belong to the second class.
Richard Tobin
>But the statement is that 0.99999 = 1 which is of course nonsence. At at
>best it come as close to 1 without accually being 1. Its like traveling half
>the distance between you and a destination. No matter how long you travel
>you NEVER get to the destination.
I'm beginning to think you're trolling.
The "statement" is that 0.999... = 1, and it's true whether you can
understand it or not.
0.999... means the limit. To use your analogy, it *is* the destination.
0.9 isn't the destination, nor is 0.99, nor is 0.999. But 0.999...
is, that's what we mean by 0.999...
If you mean something else by 0.999... then it might not be equal
to 1, but in that case we can't help you. Go off and develop your
own mathematics based on it, but it won't be much use for anything.
Doug Norris
>But the statement is that 0.99999 = 1 which is of course nonsence. At at
>best it come as close to 1 without accually being 1. Its like traveling half
>the distance between you and a destination. No matter how long you travel
>you NEVER get to the destination.
How long you travel? WTF!? 0.999.... is a fixed number (we're not talking
about 0.99999, we're talking about 0.999...) and doesn't travel anywhere.
Get it straight, folks - this damn thread shows up two or three times a
month these days.
Doug
Jonathan Stone
>http://work.ucsd.edu:5141/cgi-bin/http_webster?isindex=sarcasm
*D'oh*. Sorry, I didn't get the sarcasm at all. Flew right over my
head. I thought perhaps you really were talking about symbolic math
systems. I guess I've spent so long staring at bugs in software IEEE
emulation and inadequate NA that i'm oblivious to sarcasm about it.
Steve Whittet
>>best it come as close to 1 without accually being 1. Its like traveling half
>>the distance between you and a destination. No matter how long you travel
>>you NEVER get to the destination.
>
>= 1 that we are discussing. It is 0.9999... = 1. Do you see those three
>dots? They mean that the string of nines goes all the way to infinity.
>That makes all the difference.
Since infinity is unbounded they can go all the way to infinity
and still have infinitely far to go.
That's why the series needs an ordered array with a limit.
Once you establish the limit you have in fact given up and
ceased to try to go all the way to infinity.
The reason 1.9... isn't definable as 1 is because even with its idots
it doesn't establish an ordered sequence.
>
>--
>Dave Seaman
steve
Andrew Hackard
>ahac...@UVic.CA (Andrew Hackard) wrote:
>>Bruce Grubb <bgr...@zianet.com> said...
>>>3x cannot both equal 1 and .9999[...]
>>
>>Sure it can, if you assume they're the same number to begin with.
>
>But they are not.
Oh, but they *are*.
And my proof has precisely as much validity as yours, which is to say
*none*. You're attempting a proof by superior firepower, unaware that
you're shooting blanks.
--
"As for rumor and reputation, let us consider them as
matters that must not guide, but follow our actions."
======================== --L. Annaeus Seneca Minor
<*> ahac...@uvic.ca
Doug Norris
>> >> That claim would need to imply that you can find another number between
>> >> 0.9999... and 1. What number would that be?
>>
>> >That is also discussed question and the counter question is:
>> >What number is between 10 and 9 if decimal notation or fractions are not
>> >allowed or invented ?
>>
>> That's hardly a "counter-question". You might as well answer the above
>> question by covering your ears and yelling "la la la - I can't hear you"
>> at the top of your lungs.
>>
>> Seriously, Tapio - what's your problem with the question?
>It is easy to show that there is a "number" between
>1.0... (base 10) and 0.999... (base 10) if you allow
>numbers in bases other than 10.
>For example, 0.AAA...AAA (base 11) can be shown
>to be less than 1.0...000 (base 10) for any finite number of positions.
>(A in base 11 = 10 in base 10)
For any *finite* number of positions. The number 0.AAA... (base 11) EQUALS
0.999... (base 10) EQUALS 1.0 (base 10).
Russell, you're literally killing me here - you've been arguing the
same argument for years now, and you've been disproven several times.
Doug
Miguel Carrasquer Vidal
Whittet) wrote:
>The reason 1.9... isn't definable as 1 is because even with its idots
>it doesn't establish an ordered sequence.
The reason 1.9... isn't definable as 1 is that it's 2.
=======================
Miguel Carrasquer Vidal
m...@wxs.nl
Alan M Dunsmuir
<whi...@shore.net> writes
>Since infinity is unbounded they can go all the way to infinity
>and still have infinitely far to go.
>
>That's why the series needs an ordered array with a limit.
>Once you establish the limit you have in fact given up and
>ceased to try to go all the way to infinity.
>
>it doesn't establish an ordered sequence.
Oh! Do be quiet, Steve. Go back and play in your sand-pit.
(Have you worked out the difference between grammatical and biological
gender yet, and whether possessive pronouns have to agree in gender with
the OWNED, or the OWNING object, in Egyptian?)
--
Alan M Dunsmuir
Bruce Grubb
>In article <bgrubb-9D8A68....@www.zianet.com>,
>Bruce Grubb <bgr...@zianet.com> wrote:
>
>>The assumption makes no sence. The reality is no matter how far you take
>>the 0.9999... it NEVER is equal to 1 using normal mathmatics. Now if one is
>>talking -limits- than one can make the arugement but limits mathmatic is a
>>different type of math that what has been used so far.
>
>0.9999... is *defined* as meaning the infinite sum
>
> 9/10 + 9/100 + 9/1000 + ...
>
>and an infinite sum is defined to mean the limit as the number of
>terms goes to infinity. And in this case that limit is 1.
But the statement is that 0.99999... = 1 which is of course nonsence. At at
Steve Whittet
>>it doesn't establish an ordered sequence.
>
It could be; or perhaps 1.9,-u1,-u2,...-un,...=1
with an ordered array that sums up to a limit at -.9
for a sequence you need an ordered array which follows a rule.
With 1.9...Where is the ordered array? What's the rule?
9 could just rounded off and then followed by any random
sequence of digits. Its in the form u1....
what you need to get to any limit
is a sequence in the form:
u1,u2,...un,...
1,1,1,...1,...
un=1
1,3,5...
un=2n-1
1,3,6,10,15,21,28
un = n(n+1)/2
...
>Miguel Carrasquer Vidal
regards,
steve
Bruce Grubb
Whittet) wrote:
>In article <88fh0t$a...@seaman.cc.purdue.edu>, a...@seaman.cc.purdue.edu
>says...
>>
>>In article <bgrubb-8484A4....@www.zianet.com>,
>>Bruce Grubb <bgr...@zianet.com> wrote:
>>>But the statement is that 0.99999 = 1 which is of course nonsence. At at
>>>best it come as close to 1 without accually being 1. Its like traveling
>>>half
>>>the distance between you and a destination. No matter how long you travel
>>>you NEVER get to the destination.
>>
>>Yes, it is nonsense, but you have the wrong statement. It is not 0.9999
>>= 1 that we are discussing. It is 0.9999... = 1. Do you see those three
>>dots? They mean that the string of nines goes all the way to infinity.
>>That makes all the difference.
>
>Since infinity is unbounded they can go all the way to infinity
>and still have infinitely far to go.
Which is my point. No matter how far you represent 0.9999... it is not 1.
Using similar 'secial conditions' math I can 'prove' 1 = 2 = 3.
Given: if c/b = a then a * b = c; a/a = 1; 0/0 = 0/0 = 0/0
0/0 = 1 (1 * 0 = 0); 0/0 = 2 (2 * 0 = 0); 0/0 = 3 (3 * 0 = 0)
Therefore 1 = 2 = 3.
This is of course total nonsence but this is exactly the sort of nonsence
'special conditions' math results in.
Besides there was a book on logic which used 0.9999... = 1 as an example of
how an impropper assuption at the beginning (0.999... in that form is a
rational number) results in really nutty comcutions.
I should point out that the same logic used to show 0.9999,,, = 1 proves they
must be values between 0.9999... and 1.
If 0.9999... is so close to 1 then it -is- 1 then epsilon is so close to be
zero that it is 0; In other words if we say 0.9999,,, = 1 then by the same
logic epsilon = 0.
Thefore 0.9999... + epsilon = 1; 0.9999... + 2 epsilon = 1, 0.9999... + x
epsilon = 1 (X being any non zero value)
But epsilon accually has a value no matter how small so if 0.9999... + x
epsilon = 1 then there must be values between 0.9999... and 1 otherwise
0.9999... + x epsilon would be -greater- than 1 rather than equal to it.
Since epsilon is a value that aproches but never reaches 0
(For those who don't know)
Mike Thomas
says...
>
>In article <88edln$6lv$1...@pc-news.cogsci.ed.ac.uk>, ric...@cogsci.ed.ac.uk
>(Richard Tobin) wrote:
>
>>Bruce Grubb <bgr...@zianet.com> wrote:
>>
>>>The assumption makes no sence. The reality is no matter how far you take
>>>the 0.9999... it NEVER is equal to 1 using normal mathmatics. Now if one is
>>>talking -limits- than one can make the arugement but limits mathmatic is a
>>>different type of math that what has been used so far.
>>
>>0.9999... is *defined* as meaning the infinite sum
>>
>> 9/10 + 9/100 + 9/1000 + ...
>>
>>and an infinite sum is defined to mean the limit as the number of
>>terms goes to infinity. And in this case that limit is 1.
>
>half the distance between you and a destination. No matter how long you
>travel you NEVER get to the destination.
Oh my god...
Utter, utter nonsense.
You have a lot to learn about infinity, fields, and by the looks of this reply
_definitions_. Half of infinity is infinity, the rationals are a subfield of
the reals, 0.9999... is a real, multiplication in the reals is well-defined,
and the limit = 1. Saying that the limit might be 1, but the number itself
isn't is complete rubbish, because the number _is_ the limit by definition. The
dots mean "to infinity", not "to a very big finite number."
This thread turns up all the time, and the only thing it achieves it
demonstrating how little most people know about the real numbers and rigorous
mathematics in general.
Adam Russell
at
> best it come as close to 1 without accually being 1. Its like traveling
> half the distance between you and a destination. No matter how long you
> travel you NEVER get to the destination.
?!?
I always get where I'm going. What are you talking about? Are you trying
to prove your point by using an obviously false paradox? If you are trying
to be sarcastic you might put in a <g> so we can tell what you intended.
Dave Seaman
Bruce Grubb <bgr...@zianet.com> wrote:
>But the statement is that 0.99999... = 1 which is of course nonsence. At at
>best it come as close to 1 without accually being 1. Its like traveling
>half the distance between you and a destination. No matter how long you
>travel you NEVER get to the destination.
But 0.999... does not represent a journey; it represents the destination.
Dave Seaman
Bruce Grubb <bgr...@zianet.com> wrote:
>In article <mXKq4.606$Rh4.1...@news.shore.net>, whi...@shore.net (Steve
>Whittet) wrote:
>
>>In article <88fh0t$a...@seaman.cc.purdue.edu>, a...@seaman.cc.purdue.edu
>>says...
>>>
>>>Bruce Grubb <bgr...@zianet.com> wrote:
>>>>But the statement is that 0.99999 = 1 which is of course nonsence. At at
>>>>half
>>>>the distance between you and a destination. No matter how long you travel
>>>>you NEVER get to the destination.
>>>
>>>= 1 that we are discussing. It is 0.9999... = 1. Do you see those three
>>>dots? They mean that the string of nines goes all the way to infinity.
>>>That makes all the difference.
>>
>>Since infinity is unbounded they can go all the way to infinity
>>and still have infinitely far to go.
>
>Which is my point. No matter how far you represent 0.9999... it is not 1.
>Using similar 'secial conditions' math I can 'prove' 1 = 2 = 3.
That may be true for all the ways that *you* can think of to represent
0.999..., but it is not true of the way *I* represent 0.999....
There is more than one way to represent the real numbers, but what
follows is all standard textbook stuff. I am not making any of this up.
Let S be the sequence { 0.9, 0.99, 0.999, ... } Notice that a sequence,
by definition, is a function whose domain is the natural numbers. There
is a value s_n for each natural number n, given by s_n = 1 - 1/10^n.
Do you agree that this function is defined for every natural number n?
S, however, is not the same as the real number 0.999.... To understand
what a real number is, you first need to understand the concept of an
equivalence relation.
The objects we will discuss are all sequences of rational numbers. More
than that, we require that each sequence satisfy the Cauchy property (or,
to be precise, a modified version of the Cauchy property that applies
only to rational numbers). We say that a sequence S = { s_n } is Cauchy
if, for every rational epsilon > 0, there exists N > 0 such that
| s_m - s_n | < epsilon for every m, n > N. Loosely speaking, the Cauchy
sequences are the ones that look as if they ought to converge to
something. For example, the sequence S above obviously converges to 1,
but the sequence { 1, 1.4, 1.41, 1.414, 1.4142, ... } does not converge
to any rational number. We want to say that it "converges" in some sense
to the square root of 2, but that is not a rational number, and the
rationals are all we have to work with so far.
Given two Cauchy sequences U = { u_n } and V = { v_n } of rational
numbers, we say that U ~ V if, for every rational epsilon > 0, there
exists N > 0 such that | u_n - v_n | < epsilon for every n > N.
The relation ~ is easily seen to have the following properties for all
Cauchy sequences U, V, and W:
1. ~ is reflexive ( U ~ U ).
2. ~ is symmetric ( U ~ V => V ~ U ).
3. ~ is transitive ( U ~ V & V ~ W => U ~ W ).
A relation that is reflexive, symmetric, and transitive is called an
equivalence relation. It is well known that an equivalence relation ~ on
a set A partitions that set into mutually disjoint subsets, called
equivalence classes, such that, for all x, y, in A, we have x ~ y iff x
and y belong to the same equivalence class.
Now I am ready to tell you how I represent 0.999.... This number is the
unique equivalence class (within the set of Cauchy sequences of rational
numbers) that contains the sequence S = { 0.9, 0.99, 0.999, ... }.
Notice that this equivalence class happens to be the same one that also
contains the constant sequence S' = { 1, 1, 1, 1, ... }. That's because
S ~ S' according to the definition of the equivalence relation ~.
The notation [S] is often used to represent the equivalence class
containing S. Therefore we have S ~ S' and thus [S] = [S']. The latter
equation means that [S] and [S'] are sets that have the same members, and
therefore they are equal.
Thus, the statement 0.999... = 1 is a statement that two particular sets
have exactly the same members. That is what it means to say that those
two real numbers are identical.
In case you find it uncomfortable to think of real numbers as sets, you
may be interested to know that other kinds of numbers (natural numbers,
integers, rational numbers) are also defined as particular sets.
Tapio Hurme
what happens with that.
E. Weddington <umg...@nospamattglobal.net> kirjoitti
viestissä:38AB00BD...@nospamattglobal.net...
>
>
> Tapio Hurme wrote:
>
> > Ken Cox <k...@research.bell-labs.com> kirjoitti
> > viestissä:38A9B677...@research.bell-labs.com...
> > > Tapio Hurme wrote:
> > > > 2) Ruckers example:
> > > >
> > > > Assuming K=1-(1/w)
> > > >
> > > > 10K=10K-10/w
> > > > - K= 1- 1/w
> > > > ____________
> > > > 9K=9-1/w and K=1-1/w
> > >
> > > This last line should read 9K = 9-1/w and K = 1-1/(9w).
> >
> > Well, I wrote a typo - the last line should read 9K=9-9/w and thus
K=1-1/w
> > as prof. Rucker wrote it.
> > >
> > > > which proofs only that "counting" works, but nothing about 0.999...
> > >
> > > Actually, when you apply this to the 0.999... case, this
> > > informally shows that there is no non-zero real number W
> > > such that 0.999... = 1-W (take the above with W = 1/w).
> > > Otherwise you get the contradiction that W is non-zero
> > > and W = W/9.
> >
> > Should be. W=1/w
> >
> > A)
> > But now assuming 0.999...=1=1-1/w it follows assuming 1=1-1/w:
> >
> > 1) 0=-1/w or
> >
> > writing 1= (w-1)/w implies w*1= w-1 -> w=w-1 ->
> > 2) 0=-1 Impossible or not true!
> >
> > Can we conclude?:
> > 1) Case is OK
> > 2) Case is not OK
> > It follows: multiplication is not valid operation. If this is true, then
it
> > is not allowed to multiply with infinite number "notation". If that is
true,
> > then the multiplication of 10*0.999... is not allowed as in the case of
"FAQ
> > proof".
> > Or shall we ask, because multiplication and division are inverse
operations,
> > that neither is allowed? And nothing is proofed.
> >
> > Tapio
>
> Your assumption in A) above is invalid. You can't assume that 0.999... =
> 1-(1/w).
> That's like saying
> f(x) = x
> f(y) = y
> Therefore x = y.
>
> The rest of the logic following that assumption is then flawed.
>
> Eric
>
Brian M. Scott
wrote:
>In article <88bpjn$fko$1...@nnrp1.deja.com>, Bob Silverman <bo...@rsa.com>
>wrote:
>>In article <stevel-1402...@192.168.100.2>,
>> ste...@coastside.net (Steve Leibel) wrote:
>>> In article <38A89420...@nospamattglobal.net>, "E. Weddington"
>>> <umg...@nospamattglobal.net> wrote:
[...]
>>> > Let x = 1.999...
>>> > 100x = 199.999...
>>> > - 10x = 19.999...
>>> > --------------------
>>> > 90x = 180
>>> > x = 2
>>> > 1.999... = 2
>>> > QED
>>> > Forget the metaphysics, calculus and such. The above algebra is
>>> > pretty straight forward.
>>> Eric,
>>> This is a commonly given "proof," in quotes . . . but in fact it's
>>> bogus, since you don't actually know for sure that it's legal to multiply an
>>> infinite decimal expression by a constant.
>>Huh? Of course it is if you accept 1.999..... as a real number!
>It is not a question of 1.999..... being a real number but of it being a
>-rational- number. The multiplication and subtraction hinge on 1.999.....
>being a rational number. If it is not then the subtraction is invalid.
>>It is legal to multiply ANY real number by a constant. The reals form
>>a field. A field is closed under multiplication. End of discussion.
>But is is NOT legal to manipulate an irrational number as if it was a
>rational number nor is is legal to manipulate a fraction that results in a
>repeting decimal in decimal form.
>For example x = 1/7 = .142857... but if you multiply the decimal by 7 you do
>not get 1.
In fact you do: you get 0.999999..., which of course is 1. The
'proof' given at the top is indeed bogus, but not for the reason that
you give. It is perfectly possible to prove that the calculations in
question make sense and are done correctly, but the proof requires a
knowledge of limits and an understanding of exactly what is meant by a
non-terminating decimal (repeating or not). Anyone who understands
enough mathematics to understand those proofs already understands
exactly why 1.999... = 2. In short, the calculation is correct, and
it may help to convince a beginner that the equation 1.999... = 2 is
plausible, but it's not a proof.
[...]
Note followups.
Brian M. Scott
Brian M. Scott
wrote:
[...]
>But they are not. While 0.9999... is close to 1 is it not 1. The number
>-limit- is 1 not the number itself.
This is nonsense: a single number does not have a limit, and 0.999...
is a single number. The *sequence* of numbers <0.9, 0.99, 0.999, ...>
has a limit, which is the number 0.999..., also written 1.
E. Weddington
number into a quotient, f(x).
I showed that f(x) = x, i.e. 0.999... = 1.
Let y = 1-(1/w)
You showed that f(y) = y.
You then proceeded to say x=y, i.e. 1 = 1-(1/w).
Where is the logic in saying f(x)=x, f(y)=y, therefore x=y?
Eric
E. Weddington
1. Get out any calculus text book and review Infinite Series and look up Zeno's
Paradox.
2. Use a spell-check or dictionary. The word is "sense" and "nonsense" not
'sence' and 'nonsence'.
Bruce Grubb wrote:
> In article <88edln$6lv$1...@pc-news.cogsci.ed.ac.uk>, ric...@cogsci.ed.ac.uk
> (Richard Tobin) wrote:
>
> >In article <bgrubb-9D8A68....@www.zianet.com>,
> >Bruce Grubb <bgr...@zianet.com> wrote:
> >
> >>The assumption makes no sence. The reality is no matter how far you take
> >>the 0.9999... it NEVER is equal to 1 using normal mathmatics. Now if one is
> >>talking -limits- than one can make the arugement but limits mathmatic is a
> >>different type of math that what has been used so far.
> >
> >0.9999... is *defined* as meaning the infinite sum
> >
> > 9/10 + 9/100 + 9/1000 + ...
> >
> >and an infinite sum is defined to mean the limit as the number of
> >terms goes to infinity. And in this case that limit is 1.
>
> But the statement is that 0.99999... = 1 which is of course nonsence. At at
Iain Davidson
Bruce Grubb <bgr...@zianet.com> wrote in message
news:bgrubb-BE07E8....@www.zianet.com...
> In article <88dghf$2rhe$1...@uvaix7e1.comp.UVic.CA>, ahac...@UVic.CA (Andrew
> Hackard) wrote:
>
> >Bruce Grubb <bgr...@zianet.com> said...
> >>3x cannot both equal 1 and .9999[...]
> >
> >Sure it can, if you assume they're the same number to begin with.
>
> -limit- is 1 not the number itself.
What do you mean by close ?
If two numbers are not the same, like 2.4 and 2.51,
then | 2.4 -2.51| = 0.11
The absolute value of their difference is some number
greater than zero.
If you claim that
0.999... <> 1 then what is |0.999... - 1| ?
The absolute difference between any two distinct real numbers
is some number > 0.
If you cannot exhibit a difference of this kind then
the only alternative is that the difference is zero
and this means 0.999... = 1
Do you think the difference is 0.0000000000067 ?
Or could it be a decimal with an unending string of zeros
after the decimal point 0.000.. ?
Are 0 and 0.000... different numbers ?
You seem unable to answer these questions.
E. Weddington
"Brian M. Scott" wrote:
> Note followups.
>
> Brian M. Scott
And in my defense, I never said it *was* a proof. I had hoped, as mentioned in
Brian's post, that it is an accurate and convincing method to show how 1.999... =
2, .999 = 1, and how any infinitely repeating decimal number can be converted to
it's equivalent quotient.
Eric
Tapio Hurme
E. Weddington <umg...@nospamattglobal.net> kirjoitti
viestissä:38AC3E2C...@nospamattglobal.net...
> Let's call the algorithm I posted that converts a repeating infinite
decimal
> number into a quotient, f(x).
> I showed that f(x) = x, i.e. 0.999... = 1.
> Let y = 1-(1/w)
> You showed that f(y) = y.
> You then proceeded to say x=y, i.e. 1 = 1-(1/w).
> Where is the logic in saying f(x)=x, f(y)=y, therefore x=y?
Vaude, It is told that 1-(1/w) =1 for any epsilon, i.e. (1/w). And it is
told that there is no such an epsilon and therefore 0.999... =1. <g>
Actually prof. Rucker´s point was that FAQ proof is valid also for 1-epsilon
and therefore the FAQ proof does not tell anything about the nature of
0.999...
BTW: I personally prefer mostly the Cauchy sequence proof. That is hard
stuff to claim 0.999...=1.
FYI: I do understand both sides of problem. The real problem - IMHO- is to
explain why there is nothing between quantitative jump from P_i, =9 to
P_i=0 for all decimal placeholders P_i.
I think math does not treat empty jumps between reals like 0.999 ... and 1
and also 0.999...8 and 0.999... etc.
Tapio
Phunda Mental
In article <bgrubb-81977D....@www.zianet.com>,
Bruce Grubb <bgr...@zianet.com> wrote:
>But the statement is that 0.99999... = 1 which is of course nonsence. At at
>best it come as close to 1 without accually being 1. Its like traveling
>half the distance between you and a destination. No matter how long you
>travel you NEVER get to the destination.
Uhm. I'm afraid that I cannot make heads nor tails of the above.
For a rigorous proof that .999... = 1, just look at the limit
convergence.
The following observations are meant to help build an intuitive
feel for what is happening.
Let .9999... = Z.
Now, you say that Z is as close to one as you can get without
actually being 1.
So, you claim that Z is not equal to 1. By the trichotomy
theorem, you must therefore hold that Z is either greater
than, or less than 1. Given the notation that you use,
I would assume that you are approaching 1 from the bottom,
or equivalently, that you believe Z is less than 1.
So, how much less is Z? What does 1 - Z equal? Is Z real?
1 - .9 = 1/10
1 - .99 = 1/100
1 - .999 = 1/1000
1 - .999... = "1/inf" = 0; If 1 - Z = 0, then Z = 1
For any two reals, A and B, if there does not exist C
(A < C < B) then A = B. If you take Z to be real, and
1 to be real, with no numbers in between, then Z = 1.
-- Phundie (phu...@yahoo.com)
"Trust me kid, I've been hacking since your mom was
in diapers" -- overhead some old d00d at Beyond HOPE
Doug Norris
>But the statement is that 0.99999... = 1 which is of course nonsence. At at
>best it come as close to 1 without accually being 1. Its like traveling
>half the distance between you and a destination. No matter how long you
>travel you NEVER get to the destination.
Bruce, either explain exactly WTF you mean by a number "travelling", or
you're going to hit the kill-file hard enough that you'll bounce.
1.999.... is a fixed number. FIXED NUMBERS DON'T TRAVEL. Let me know
where you need an explanation.
Doug
Doug Norris
>(For those who don't know)
This, right here, pretty much sums the totality of your postings to date.
Doug
E. Weddington
Tapio Hurme wrote:
> Vaude, It is told that 1-(1/w) =1 for any epsilon, i.e. (1/w). And it is
> told that there is no such an epsilon and therefore 0.999... =1. <g>
> Actually prof. Ruckeræ„€ point was that FAQ proof is valid also for 1-epsilon
> and therefore the FAQ proof does not tell anything about the nature of
> 0.999...
> BTW: I personally prefer mostly the Cauchy sequence proof. That is hard
> stuff to claim 0.999...=1.
> FYI: I do understand both sides of problem. The real problem - IMHO- is to
> explain why there is nothing between quantitative jump from P_i, =9 to
> P_i=0 for all decimal placeholders P_i.
> I think math does not treat empty jumps between reals like 0.999 ... and 1
> and also 0.999...8 and 0.999... etc.
>
> Tapio
>
There is no "jump" from 0.999... and 1 they are the one and the same number.
There can be no jump in digits because in 0.999... the digits repeat forever.
You say the you prefer the Cauchy sequence proof and then that it is hard stuff
to claim 0.999...=1. What do you think the exact problem is with the Cauchy
sequence proof?
Tapio Hurme
today .. somewhere in the threads under this same problem. ( This thread
has expanded so rapidly - sorry) If you cannot find it, I post it for you
personally.
E. Weddington <umg...@nospamattglobal.net> kirjoitti
viestissä:38AC5057...@nospamattglobal.net...
>
>
> Tapio Hurme wrote:
>
> > Vaude, It is told that 1-(1/w) =1 for any epsilon, i.e. (1/w). And it
is
> > told that there is no such an epsilon and therefore 0.999... =1. <g>
> > Actually prof. Rucker´s point was that FAQ proof is valid also for
Eric Dew
>But the statement is that 0.99999 = 1 which is of course nonsence. At at
>the distance between you and a destination. No matter how long you travel
>you NEVER get to the destination.
Suppose you travel 10 mph. You are 20 miles away from your destination.
After 1 hour, you are 10 miles away. After another 30 minutes, you are
5 miles away. Another 15 minutes, you are 2.5 miles away. And so on.
Do you get to your destination?
Ans: yes. After a total of 2 hours travel, you wll get to your destination.
If you claim that you travel 10 mph the first hour, then 5 mph the secon
hour, then 2.5 mph the third, and so on, then yes, you will never get there.
By the 10th hour, you'll be moving slower than a truck with flat tires in
rush hour.
EDEW
Bruce Grubb
<adamr...@msn.com> wrote:
>> But the statement is that 0.99999... = 1 which is of course nonsence. At
>> at best it come as close to 1 without accually being 1. Its like traveling
>> half the distance between you and a destination. No matter how long you
>> travel you NEVER get to the destination.
>
>?!?
>I always get where I'm going. What are you talking about? Are you trying
>to prove your point by using an obviously false paradox?
It is not an obviously false paradox but rather a principal in logic. The key
part of the statement is *half the distance* between you and a destination.
Given: it takes 1 second to travel half the distance between you and a
destination.
Logical conclusion: you never reach the destination.
Proof:
original distance = x
In the 1st second one covers x/2
In the 2nd second the distance is now x/2 so one covers (x/2)/2 [half the
distance between you and a destination]
In the 3rd second the distance is now x/4 so one covers (x/4)/2.
So on and so forth. This is the problem with such proofs.
I have noticed no one wants to touch the 0/0 proof that 1 = 2.
, 2nd second (x/2)/2 [the remaining distance], 3rd second (x/4)/2, and so on.
Note that the key part of the statement is
Its like traveling
Note the highlighted part.
Not it says 'traveling
half the distance between you and a destination.
Bruce Grubb
Dew) wrote:
>>
>>But the statement is that 0.99999 = 1 which is of course nonsence. At at
>>best it come as close to 1 without accually being 1. Its like traveling
>>half the distance between you and a destination. No matter how long you travel
>>you NEVER get to the destination.
>
>Suppose you travel 10 mph. You are 20 miles away from your destination.
>After 1 hour, you are 10 miles away. After another 30 minutes, you are
>5 miles away. Another 15 minutes, you are 2.5 miles away. And so on.
>Do you get to your destination?
>
>Ans: yes.
You are purposely misreading the statement.
Kent
Bud the Real <the...@san.rxyzr.com> wrote in message
news:u70pas0vo1tgk6v36...@4ax.com...
> norr...@rintintin.colorado.edu (Doug Norris), speaking for the crew of
the Mothership,
> wrote... :
>
> Could someone, ANYONE, explain WHY this is crossposted to sci.archaeology,
> AND soc.history.ancient, when it CLEARLY isn't relevant, and should be
kept in
> sci.math only??
No, but don't feel too bad - it isn't relevant to sci.math, either. The fact
that the symbol 1.999... represents the number 2 is not a matter of dispute
among mathematicians. The thread should be consigned to sci.idiots.
NightGaunt J Smith
> Given: if c/b = a then a * b = c; a/a = 1; 0/0 = 0/0 = 0/0
Correct.
> 0/0 = 1 (1 * 0 = 0); 0/0 = 2 (2 * 0 = 0); 0/0 = 3 (3 * 0 = 0)
No, this doesn't follow from your previous assumptions. You take
a "given" of the form: if X then Y. Then you observe Y and
conclude X, a mistake.
> This is of course total nonsence but this is exactly the sort of
> nonsence 'special conditions' math results in.
Sadly, this "total nonsence" doesn't result from the givens or
bad "special conditions," but from your own misunderstanding of the
words "if" and "then."
> Besides there was a book on logic which used 0.9999... = 1 as an
> example of how an impropper assuption at the beginning (0.999... in
> that form is a rational number) results in really nutty comcutions.
Can you name the book so I can go look it up?
> If 0.9999... is so close to 1 then it -is- 1 then epsilon is so close
> to be zero that it is 0; In other words if we say 0.9999,,, = 1 then
> by the same logic epsilon = 0.
Yes. Absolutely. Spot-on. *IF* 0.999... = 1,
*THEN* epsilon, defined as (1-0.9999...), is equal to 0.
> Thefore 0.9999... + epsilon = 1; 0.9999... + 2 epsilon = 1,
> 0.9999... + x epsilon = 1 (X being any non zero value)
Yes. If epsilon is zero then x*epsilon is zero no matter what x is.
Thus 0.9999... + epsilon == 0.9999... + 2 epsilon == 0.9999... + 0.
> But epsilon accually has a value no matter how small
"A value?" A nonzero value? Why? Justify this claim.
I think this is the false step in your little proof. You suddenly
blurt out (sort of) that epsilon is nonzero, to reach your
contradiction. In fact, one can prove that epsilon is zero.
> so if 0.9999... + x epsilon = 1 then there must be values between
> 0.9999... and 1 otherwise 0.9999... + x epsilon would be -greater-
> than 1 rather than equal to it.
False. This would only be true if epsilon was not zero. If it
is zero, then 0.9999... + x epsilon = 0.9999... + 0. Not greater
than 1.
> Since epsilon is a value that aproches but never reaches 0
Values do not "approach" 0. 0.9999... is a specific value, just
like 1 is a specific value, just like epsilon==(0.999... - 1) is a
specific value. Only by a gut feeling, it seems, are you able to
insist that epsilon is non-zero.
Sent via Deja.com http://www.deja.com/
Before you buy.
Eric Stevens
wrote:
>In article <88edln$6lv$1...@pc-news.cogsci.ed.ac.uk>, ric...@cogsci.ed.ac.uk
>(Richard Tobin) wrote:
>
>>In article <bgrubb-9D8A68....@www.zianet.com>,
>>Bruce Grubb <bgr...@zianet.com> wrote:
>>
>>>The assumption makes no sence. The reality is no matter how far you take
>>>the 0.9999... it NEVER is equal to 1 using normal mathmatics. Now if one is
>>>talking -limits- than one can make the arugement but limits mathmatic is a
>>>different type of math that what has been used so far.
>>
>>0.9999... is *defined* as meaning the infinite sum
>>
>> 9/10 + 9/100 + 9/1000 + ...
>>
>>and an infinite sum is defined to mean the limit as the number of
>>terms goes to infinity. And in this case that limit is 1.
>
>But the statement is that 0.99999... = 1 which is of course nonsence. At at
>best it come as close to 1 without accually being 1. Its like traveling
>half the distance between you and a destination. No matter how long you
>travel you NEVER get to the destination.
You can manage it if you don't just walk but Kantor :-)
Eric Stevens
There are two classes of people. Those who divide people into
two classes, and those who don't. I belong to the second class.
Richard Tobin
Steve Whittet <whi...@shore.net> wrote:
>The reason 1.9... isn't definable as 1 is because even with its idots
>it doesn't establish an ordered sequence.
No, the reason 1.9... isn't definable as 1 is that it's 2.
-- Richard
--
Spam filter: to mail me from a .com/.net site, put my surname in the headers.
"The Internet is really just a series of bottlenecks joined by high
speed networks." - Sam Wilson
Brian M. Scott
wrote:
>Given: it takes 1 second to travel half the distance between you and a
>destination.
>Logical conclusion: you never reach the destination.
Not in finite time, anyway. This is true. It is also wholly
irrelevant to the fact that 1.999... = 2, since 1.999... is simply a
number, not a serial process of some kind.
Jonathan Stone
In article <bgrubb-81977D....@www.zianet.com>,
Bruce Grubb <bgr...@zianet.com> wrote:
[quoted text snipped]
>But the statement is that 0.99999... = 1 which is of course nonsence. At at
>best it come as close to 1 without accually being 1. Its like traveling
>half the distance between you and a destination. No matter how long you
>travel you NEVER get to the destination.
Nonsense. If you travel each interval in half the time of the
preceding interval (which of course was twice as large), you reach
the destination in exactly twice as long as it took you to traverse the
first interval. Zeno. Achilles and tortoise.
Known for what, some 2,500 years?
Are you a troll?
b_m_...@my-deja.com
Bruce Grubb <bgr...@zianet.com> wrote:
> In article <88hrpu$rlj$1...@nntp2.atl.mindspring.net>, ed...@netcom.com
> (Eric Dew) wrote:
> >>But the statement is that 0.99999 = 1 which is of course nonsence.
> >>At at best it come as close to 1 without accually being 1. Its
> >>like traveling half the distance between you and a destination.
> >>No matter how long you travel you NEVER get to the destination.
> >Suppose you travel 10 mph. You are 20 miles away from your
> >destination. After 1 hour, you are 10 miles away. After another
> >30 minutes, you are 5 miles away. Another 15 minutes, you are 2.5
> >miles away. And so on. Do you get to your destination?
> >Ans: yes.
> You are purposely misreading the statement.
He probably just didn't think that anyone could be so foolish as to
suppose that the statement was relevant to decimal expansions. It
isn't.
[...]
Brian M. Scott
Jonathan Stone
Bud the Real <the...@san.rxyzr.com> wrote:
>norr...@rintintin.colorado.edu (Doug Norris), speaking for the crew of the Mothership,
>wrote... :
>
>Could someone, ANYONE, explain WHY this is crossposted to sci.archaeology,
>AND soc.history.ancient, when it CLEARLY isn't relevant, and should be kept in
>sci.math only??
I think you'll find Steve Whittet started it, back in sci.arch.
The same Steve, btw, who claimed that a 1:1:2 rectangular prism has a
right-angle triange with sides sqrt(2), sqrt(5), and sqrt(7). (the
obvious face diagonals for sqrt(2) and sqrt(5), but with a magic
right angle between them, giving a sqrt(7) diagonal).
Steve Lord
[snipped the non-sci.math newsgroups]
On Tue, 15 Feb 2000, Bruce Grubb wrote:
> In article <88bpjn$fko$1...@nnrp1.deja.com>, Bob Silverman <bo...@rsa.com>
> wrote:
>
> >In article <stevel-1402...@192.168.100.2>,
> > ste...@coastside.net (Steve Leibel) wrote:
> >> In article <38A89420...@nospamattglobal.net>, "E. Weddington"
> >> <umg...@nospamattglobal.net> wrote:
> >>
> >> > Ok, how hard can this be to figure out:
> >> >
> >> > The ellipsis notation means to take the repeating digits, and
> >continue
> >> to repeat
> >> > them without bound:
> >> > 1.999... = 1.99999999..... = 1.999999999999999999... , etc.
> >> >
> >> > Let x = 1.999...
> >> >
> >> > 100x = 199.999...
> >> > - 10x = 19.999...
> >> > --------------------
> >> > 90x = 180
> >> >
> >> > x = 2
> >> >
> >> > 1.999... = 2
> >> >
> >> > QED
> >> >
> >> > Forget the metaphysics, calculus and such. The above algebra is
> >> > pretty straight forward.
> >>
> >> Eric,
> >>
> >> This is a commonly given "proof," in quotes . . . but in fact it's
> >> bogus, since you don't actually know for sure that it's legal to multiply an
> >> infinite decimal expression by a constant.
> >
> >Huh? Of course it is if you accept 1.999..... as a real number!
>
> It is not a question of 1.999..... being a real number but of it being a
> -rational- number. The multiplication and subtraction hinge on 1.999.....
> being a rational number. If it is not then the subtraction is invalid.
So now we cannot multiply a real number by a constant? That's strange,
sounds reasonable to me.
> >It is legal to multiply ANY real number by a constant. The reals form
> >a field. A field is closed under multiplication. End of discussion.
>
> But is is NOT legal to manipulate an irrational number as if it was a
> rational number nor is is legal to manipulate a fraction that results in a
> repeting decimal in decimal form.
?
> For example x = 1/7 = .142857... but if you multiply the decimal by 7 you do
> not get 1. By the rules of Reduco ad Absurdum then it is only legal to
> manipulate a rational number that is a repeating decimal in factional form.
.142857142857...
x 7
----------------
.999999999999...
The whole message of this thread is that 0.999... = 1 and when you
multiply the decimal expansion of 1/7 by 7 you expect to get 1, right? So
this makes sense only if 0.999... = 1.
If you're trying to use this to prove that 0.999... != 1 then you need to
assume what you're trying to prove -- and circular logic is usually not
too useful.
Another example, simpler than 1/7. Consider 1/3. You agree that this is
0.33333....., right? 3*(1/3) = 1 and if (1/3)=0.3333.... then
3*(0.3333....) must equal 1, right? 3*0.333... = 0.999...
> Furthermore if 1.999... is not a rational number than it cannot be added or
> subtracted as if it was rational which is a key assuption of the 'proof'.
You can add or subtract two real numbers. There is no assumption that
1.999... is rational in the statement of the proof (implicit or explicit).
> >The only issue through all of these nonsensical threads is whether
> >one accepts an infinite repeating decimal as a real number.
>
> No the issues are accually is whether one accepts a partical infinite
> repeating decimal as a -rational- number and if manipulating it in decimal
> form is valid.
No. It doesn't matter whether or not a particular infinite repeating
decimal is rational or not.
> As the case of 1/3, 1/7, 1/11, 1/13 or about any 1/(prime number or multiple)
> shows this is nonsence.
No, this proof is not nonsence.
Steve L
Xiccarph
limit concept from precalc. You cannot define an x value, and then use operands to
make it become something else. By definition, x is still 1.999... and your 90x=180
is an inequality. However, using the Limit concept, the answer would be 2 as you
carried the ...9's out to infinity; but still, the actual value would never reach
2, but grow infinitesimally closer towards 2.
Mike
Steve Whittet wrote:
> In article <38A89420...@nospamattglobal.net>, umg...@nospamattglobal.net
> says...
E. Weddington
see that you're close to the correct concept but not quite right. Also, the algorithm
is not a proof, but it is a valid algorithm. When you read the threads, you'll note
that:
1. one can define x=1.999... and operate on it as it is a real number
2. 90x=180 is not an inequality as it has an equals sign
3. 2 is equal to 1.999... and it *is* the limit
One does this kind of operation all the time:
2.00000000000000...
-1.00000000000000...
--------------------------
1.00000000000000...
Then one can do the same with repeating, non-terminating 9 digits.
Every rational number can be written as a repeating decimal and every repeating decimal
can be written as a rational number.
Eric
Petar Bukvic
Xiccarph
wants, someone will come up with a sequence to QED what they want. I understand the
concept and "algorithmic math" you were using. I guess I was looking at it from what I'd
call "pure mathematics", which makes the limit concept more valid than an algorithm which
is designed to force a desired value at infinity. Just because it states that 90x=180,
where x=1.9999...9, is misleading. Its not the equals sign that defines either side, it is
the functions on each side. Saying that 3=4 because of the equals sign makes it an
identity is not valid; saying that 90*(1.999...9)=180 is also not "valid", but its limit
is. It all depends on how one defines the parameters and what type of math one employs.
I myself, sort of prefer fuzzy math ;-)
Mike
Brian M. Scott
wrote:
>Well, the nice thing about modern mathematics, is that if identities fail to get where one
>wants, someone will come up with a sequence to QED what they want.
This makes no sense as English, let alone as mathematics.
> I understand the
>concept and "algorithmic math" you were using. I guess I was looking at it from what I'd
>call "pure mathematics", which makes the limit concept more valid than an algorithm which
>is designed to force a desired value at infinity.
Everything that Eric wrote was correct as pure mathematics. You
appear to have misunderstood his (correct) comment about the algorithm
not being a proof; certainly he said nothing about any algorithm
'forc[ing] a desired value at infinity'.
> Just because it states that 90x=180,
>where x=1.9999...9, is misleading.
It doesn't. You have confused 1.9999...9, with some unstated but
finite number of 9s, with 1.9999..., with an infinite number of 9s.
[...]
dc...@erinet.com
Bruce Grubb <bgr...@zianet.com> wrote:
> In article <vmhjr-4607EF....@news.frii.com>, Virgil
> <vm...@frii.com> wrote:
>
> >In article <bgrubb-F0CBCF....@www.zianet.com>, Bruce Grubb
> ><bgr...@zianet.com> wrote:
> >
> >>Because x is _also_ 1/3 and 3 * 1/3 = 1 3x cannot both equal 1 and
.9999
> >>ergo you canNOT multiply or divide a fraction that result in a
repeating
> >>decimal in decimal form.
> >
> >By your type of argument, 2/4 cannot be equal to 1/2 because they
look
> >different.
>
> Totally different situation. Both the above faction result in a
finite value
> 0.5. If not just that the value look different but -behave-
differently.
>
> This is why fractions that result in repeating decimals need to put
back into
>
> >If 0.999... is an abreviation for the number represented by a decimal
> >point followed by infinitely many nines, then the assumption that it
> >represents a number means that the number that it represents must be
1.
>
> The assumption makes no sence. The reality is no matter how far you
take the
> 0.9999... it NEVER is equal to 1 using normal mathmatics. Now if one
is
> talking -limits- than one can make the arugement but limits mathmatic
is a
> different type of math that what has been used so far.
>
> >and this entire thread is a waste of bandwidth.
>
> This is more nonsence. 0.9999... is distict number from 1. At best
is it a
> limit that aproaches 1 but NEVER gets there. Trying to claim
otherwise
> produces some really silly results.
>
> The prove is of the same order of nonsence of saying 0/0 = a is a
valid
> equation. The standard 'if c/b = a then a * b = c' shows all other
divitions
> by 0 to be invalid (c cannot be any value but 0) but 0/0 = a produces
a
> prefectly valid equation (a * 0 = 0). The problem is that if one
fallows
> though with the logic then 1 = 2.
>
> Given: if c/b = a then a * b = c; a/a = 1
>
>
> 0/0 = 2 (2 * 0 = 0)
>
>
> This total nonsence results because a rule in mathmatics is ignored.
The
> proof similarlly ignores a similar rule (treating a limit as a finate
value)
>
Please tell me what is wrong with the following logic:
Let x = 0.999+
Then 10 x = 9.999+
Therefore, 10x - x = 9x = 9.000+
Therefore, x = 1.
Don Cool
June R Harton
<dc...@erinet.com> wrote
> Please tell me what is wrong with the following logic:
> Let x = 0.999+
> Then 10 x = 9.999+
i.e. 10 times 0.999+ = 9.999+
> Therefore, 10x - x = 9x = 9.000+
i.e. 10 times 0.999+ - 0.999+ = 9 times 0.999+ = not 9.000
> Therefore, x = 1.
No. You already know x so equation is invalid.
from: Spirit
(using June's e-mail to communicate to you)
Martin Harlow
<JUNEH...@prodigy.net> writes
>
><dc...@erinet.com> wrote
>> Please tell me what is wrong with the following logic:
>> Let x = 0.999+
>> Then 10 x = 9.999+
>
>i.e. 10 times 0.999+ = 9.999+
>
>> Therefore, 10x - x = 9x = 9.000+
>
>i.e. 10 times 0.999+ - 0.999+ = 9 times 0.999+ = not 9.000
>
>> Therefore, x = 1.
>
>No. You already know x so equation is invalid.
I disagree. It looks ok to me. Rather cute, actually.
ttfn
Martin
--
"Any girl can be glamorous. All you have to do is stand still and look stupid."
- Hedy Lamarr
Martin Harlow mar...@freedonia.demon.co.uk
Brian M. Scott
<mar...@freedonia.demon.co.uk> wrote:
>In article <8adc2a$50uo$1...@newssvr04-int.news.prodigy.com>, June R Harton
><JUNEH...@prodigy.net> writes
>>
>><dc...@erinet.com> wrote
>>> Please tell me what is wrong with the following logic:
>>> Let x = 0.999+
>>> Then 10 x = 9.999+
>>
>>i.e. 10 times 0.999+ = 9.999+
>>
>>> Therefore, 10x - x = 9x = 9.000+
>>
>>i.e. 10 times 0.999+ - 0.999+ = 9 times 0.999+ = not 9.000
>>
>>> Therefore, x = 1.
>>
>>No. You already know x so equation is invalid.
>
>I disagree. It looks ok to me. Rather cute, actually.
There's nothing wrong with the calculation. In order to justify the
steps, however, one must already know enough to show much more
directly that 0.999... = 1, so it's not useful as a proof of that fact
(as it's often misrepresented). Followups set.
Brian M. Scott
Steve Whittet
mar...@freedonia.demon.co.uk says...
>
>In article <8adc2a$50uo$1...@newssvr04-int.news.prodigy.com>, June R Harton
><JUNEH...@prodigy.net> writes
>>
>><dc...@erinet.com> wrote
>>> Please tell me what is wrong with the following logic:
>>> Let x = 0.999+
>>> Then 10 x = 9.999+
>>
>>i.e. 10 times 0.999+ = 9.999+
>>
>>> Therefore, 10x - x = 9x = 9.000+
>>
>>i.e. 10 times 0.999+ - 0.999+ = 9 times 0.999+ = not 9.000
>>
>>> Therefore, x = 1.
>>
>>No. You already know x so equation is invalid.
>
>I disagree. It looks ok to me. Rather cute, actually.
isn't 10 times 0.999+ = 9.99+ ?
so that
10 times .999+ -0.999+ ~= 9 times 0.999+
therefore x ~=1
>
>ttfn
>
>Martin
>
>--
>"Any girl can be glamorous. All you have to do is stand still and look
stupid."
> - Hedy Lamarr
>
>Martin Harlow
steve
Jan Kristian Haugland
On Sat, 11 Mar 2000, June R Harton wrote:
>
>
> <dc...@erinet.com> wrote
> > Please tell me what is wrong with the following logic:
> > Let x = 0.999+
> > Then 10 x = 9.999+
>
> i.e. 10 times 0.999+ = 9.999+
>
> > Therefore, 10x - x = 9x = 9.000+
>
> i.e. 10 times 0.999+ - 0.999+ = 9 times 0.999+ = not 9.000
I think it should be 9x = 10x - x = 9.000+
>
> > Therefore, x = 1.
>
> No. You already know x so equation is invalid.
>
>
David W. Cantrell
wrote:
>What are you saying - that 0.999+ does NOT equal 1? Does anyone
>want to end this argument?
Yes, yes, a thousand times yes!
>I will send a cashier's check for $20 to anyone that
>can prove to me using mathematical logic that 0.999+ is NOT
>equal to 1.
Let me try to save you $20, Don.
To everyone: He surely meant to say that he would give $20 to
anyone that can prove that, taken as representations of
*real*numbers*, 0.999... and 1 represent different numbers.
Cheers,
David Cantrell
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Jered Boyd
But can one say that .999... goes on for infinity but never actually reaches 1,
so the two are not equal.
- Jered
cool...@my-deja.com
"June R Harton" <JUNEH...@prodigy.net> wrote:
>
>
> <dc...@erinet.com> wrote
> > Please tell me what is wrong with the following logic:
> > Let x = 0.999+
> > Then 10 x = 9.999+
>
> i.e. 10 times 0.999+ = 9.999+
>
> > Therefore, 10x - x = 9x = 9.000+
>
> i.e. 10 times 0.999+ - 0.999+ = 9 times 0.999+ = not 9.000
>
>
> No. You already know x so equation is invalid.
>
> from: Spirit
> (using June's e-mail to communicate to you)
>
>
can prove to me using mathematical logic that 0.999+ is NOT equal to 1.
Regards to all,
Alan M Dunsmuir
Cantrell <dwcantrel...@aol.com.invalid> writes
>Let me try to save you $20, Don.
>To everyone: He surely meant to say that he would give $20 to
>anyone that can prove that, taken as representations of
>*real*numbers*, 0.999... and 1 represent different numbers.
What nonsense!
--
Alan M Dunsmuir
Doug Norris
>What nonsense!
Great to see you adding to the discussion, Alan - I can safely filter you
out now. Chiming in just to say "What nonsense!" - nice work.
Thanks again!
Doug