0.33(bar) * 3 = 0.99(bar) IS NOT 1/3 * 3.
John Pimentel
OK, let's take 0.33(bar) for a moment.
Let X= 0.33(bar)
X * 2 = 0.66(bar)
X * 3 = 0.99(bar) <===========
10X= 3.33(bar)
- X= - 0.33(bar)
----------------
9X= 3.00
X= 3/9 = 1/3
1/3 + 1/3 = 2/3
1/3 + 1/3 + 1/3 = 3/3 = 1 <============
Hey, wait a minute these two don't equal the same thing.
I converted the 0.33(bar) to it's equivalent approximation and when I dealt
with each on their own, they did not match after doing the same number and
type of actions. I multiplied both by two and even multiplied both by three.
All the "experts" here claim that the two are equal, even the phrase "exact"
has been used. Perhaps, the word "approximation" does not exist in your
vocabulary. It would appear that an orange is still an orange and not an
apple. Gee, and I don't even have a PhD. in mathematics.
BTW: To my last challenge to find a number between 0.88(bar) and 0.9 which
to my surprise, given all the "experts" here, one answer is 0.899(bar) which
is higher than 0.88(bar), but none of you offered that. Perhaps, because
word problems are difficult to grasp by the "experts".
---
John Pimentel
pime...@roots.ultranet.com
----
John Pimentel
pime...@roots.ultranet.com
be...@pacbell.net
John Pimentel wrote:
>
> OK, let's take 0.33(bar) for a moment.
>
> Let X= 0.33(bar)
>
> X * 2 = 0.66(bar)
>
> X * 3 = 0.99(bar) <===========
>
> 10X= 3.33(bar)
> - X= - 0.33(bar)
> ----------------
> 9X= 3.00
> X= 3/9 = 1/3
>
> 1/3 + 1/3 = 2/3
>
> 1/3 + 1/3 + 1/3 = 3/3 = 1 <============
>
> Hey, wait a minute these two don't equal the same thing.
>
Why do you call it an approximation? As has been pointed out in this
thread, it's a matter of definition. I.e., how do you represent 1/3 as a
decimal? The question is not "what is it's decimal approximation?" The
answer, of course, is .3(bar); there is no approximation. .3(bar) EQUALS
1/3. THAT'S WHAT THE BAR MEANS! A decimal point followed by a gazillion
3's is an approximation of 1/3; .3(bar) is not.
Alan Beban
Albert Yang
: I converted the 0.33(bar) to it's equivalent approximation and when I dealt
0.33.... is *exactly* 1/3, not an approximation.
:
: BTW: To my last challenge to find a number between 0.88(bar) and 0.9 which
: to my surprise, given all the "experts" here, one answer is 0.899(bar) which
: is higher than 0.88(bar), but none of you offered that. Perhaps, because
: word problems are difficult to grasp by the "experts".
OK,0.89 is between 0.88(bar) and 0.9. 0.899(bar) is not because it is
*exactly* equal to 0.9.
--
Albert Yang |"Reports of my assimilation have
Internet: apy...@ucdavis.edu | been greatly exaggerated."
http://dcn.davis.ca.us/~albert/ | - Jean-Luc Picard, ST:FC
Earle D. Jones
In article <1997020618...@cinna.ultra.net>,
pime...@roots.ultranet.com wrote:
======Clipcrap Here=======
Pimentel, I am through with you.
You are either a troll or a shithead.
Tell us which.
earle
=====
__
__/\_\
/\_\/_/
\/_/\_\ earle
\/_/ jones
Robin E. Brown
John Pimentel wrote:
>
> OK, let's take 0.33(bar) for a moment.
>
>
> X * 2 = 0.66(bar)
>
> X * 3 = 0.99(bar) <===========
>
> 10X= 3.33(bar)
> - X= - 0.33(bar)
> ----------------
> 9X= 3.00
> X= 3/9 = 1/3
>
> 1/3 + 1/3 = 2/3
>
> 1/3 + 1/3 + 1/3 = 3/3 = 1 <============
>
> Hey, wait a minute these two don't equal the same thing.
>
> type of actions. I multiplied both by two and even multiplied both by three.
> All the "experts" here claim that the two are equal, even the phrase "exact"
> has been used. Perhaps, the word "approximation" does not exist in your
> vocabulary. It would appear that an orange is still an orange and not an
> apple. Gee, and I don't even have a PhD. in mathematics.
>
> to my surprise, given all the "experts" here, one answer is 0.899(bar) which
> is higher than 0.88(bar), but none of you offered that. Perhaps, because
> word problems are difficult to grasp by the "experts".
>
You seem to imply that .333... does not equal 1/3. Every decimal number
that terminates or repeats can be expressed as a rational number. By
definition a rational number is a number that can be expressed as the
fraction of two integers. Similarly, every fraction of two integers can
be expressed as a decimal number. And the decimal representation of 1/3
is???? Well, that is 3rd grade work.
Robin E. Brown
reb...@epix.net
BTW the ... is called an ellipsis - it is used in ASCII instead of the
cumbersome (bar).
Irving Lubliner
> BTW: To my last challenge to find a number between 0.88(bar) and 0.9 which
> to my surprise, given all the "experts" here, one answer is 0.899(bar) which
> is higher than 0.88(bar), but none of you offered that. Perhaps, because
> word problems are difficult to grasp by the "experts".
----------------------------
And some things are difficult for those who poke fun at the "experts."
The trouble here is that 0.899(bar) is not between 0.88(bar) and 0.9.
0.899(bar) is actually equal to 0.9.
It is, however, possible to find numbers between 0.88(bar) and 0.9.
0.89 is an example.
--
Irving Lubliner, Mathematics Enthusiast
Josh Boyd
John Pimentel wrote:
>
> OK, let's take 0.33(bar) for a moment.
>
> Let X= 0.33(bar)
>
> X * 2 = 0.66(bar)
>
> X * 3 = 0.99(bar) <===========
>
> 10X= 3.33(bar)
> - X= - 0.33(bar)
> ----------------
> 9X= 3.00
> X= 3/9 = 1/3
>
> 1/3 + 1/3 = 2/3
>
> 1/3 + 1/3 + 1/3 = 3/3 = 1 <============
>
> Hey, wait a minute these two don't equal the same thing.
>
> I converted the 0.33(bar) to it's equivalent approximation and when I dealt
> with each on their own, they did not match after doing the same number and
> type of actions. I multiplied both by two and even multiplied both by three.
> All the "experts" here claim that the two are equal, even the phrase "exact"
> has been used. Perhaps, the word "approximation" does not exist in your
> vocabulary. It would appear that an orange is still an orange and not an
> apple. Gee, and I don't even have a PhD. in mathematics.
>
> to my surprise, given all the "experts" here, one answer is 0.899(bar) which
> is higher than 0.88(bar), but none of you offered that. Perhaps, because
> word problems are difficult to grasp by the "experts".
So this heinous poorly mathematical explination makes you an "expert?"
If you look at your proof, you have ---
3 * 1/3 = .99999(bar)
But you also have.
3 * 1/3 = 1/3 + 1/3 + 1/3 = 3/3 = 1
So obviously
1 = .99999(bar)
by substitution.
Josh
Lee Jaap
|>OK, let's take 0.33(bar) for a moment.
|>
|> Let X= 0.33(bar)
|>
|> X * 2 = 0.66(bar)
|>
|> X * 3 = 0.99(bar) <===========
|>
|> 10X= 3.33(bar)
|> - X= - 0.33(bar)
|> ----------------
|> 9X= 3.00
|> X= 3/9 = 1/3
|>
|> 1/3 + 1/3 = 2/3
|>
|> 1/3 + 1/3 + 1/3 = 3/3 = 1 <============
|>
|>Hey, wait a minute these two don't equal the same thing.
It isn't at all obvious to me that 3/3 is not the same as 1.
Please elucidate.
|>BTW: To my last challenge to find a number between 0.88(bar) and 0.9 which
|>to my surprise, given all the "experts" here, one answer is 0.899(bar) which
|>is higher than 0.88(bar), but none of you offered that. Perhaps, because
|>word problems are difficult to grasp by the "experts".
Perhaps, because .89bar is precisely the same number as .9, therefore
it is not between (by the common definition of between).
|>---
|>John Pimentel
|>pime...@roots.ultranet.com
|>----
|>John Pimentel
|>pime...@roots.ultranet.com
|>
--
J Lee Jaap <Jaa...@ASMSun.LaRC.NASA.Gov> +1 757/865-7093
employed by, not necessarily speaking for,
AS&M Inc, Hampton VA 23666-1340
Dave B.
John Pimentel <pime...@roots.ultranet.com> wrote
>To my last challenge to find a number between 0.88(bar) and 0.9 which
> to my surprise, given all the "experts" here, one answer is 0.899(bar)
which
> is higher than 0.88(bar), but none of you offered that. Perhaps, because
> word problems are difficult to grasp by the "experts".
Check the postings and you'll find that I said that 0.8999999... is equal
precisely to 0.9
Dave B.
(and yes, a good number of people on this newsgroup are indeed experts)
Dave Weisbeck
That fact is blatantly obvious. I really suggest you go to the local
university and visit someone who does have his/her PhD in math and
ask them to show you that in fact .9 bar == 1. Although I suspect
that no matter how many proofs you are shown you will continue
to disbelieve what you can't see.
Dave Weisbeck
z2...@ugrad.cs.ubc.ca
John Pimentel
> You seem to imply that .333... does not equal 1/3. Every decimal number
> that terminates or repeats can be expressed as a rational number. By
> definition a rational number is a number that can be expressed as the
> fraction of two integers. Similarly, every fraction of two integers can
> be expressed as a decimal number. And the decimal representation of 1/3
> is???? Well, that is 3rd grade work.
This is all very nice but is that representation an EXACT and EQUAL
representation?
Ah! There is the key. IF 1/3 is exactly 0.33(bar) then if I multiple
both sides by the same number, in this case three, I shall arrive at
the exact same number on both sides. I should NOT have to run one
through another calculation to get to the same number as the other
side. That by defination is called mathematical fudging.
A challenge to all the whizzes and "experts" out there.
Take 1.0 and show me the work to get this number to equal 0.99(bar).
Taking 0.99(bar) to 1.0 is not the same thing. I want to see 1.0
calculated into 0.99(bar).
BTW, whizzes and "experts" here's a hint, it will save countless hours
terminating decimals have an EQUAL and EXACT representation as a fraction.
non-terminating, repeating decimals do not and never will. If they do
then you MUST be able take all convertings and go back to the original
problem mathematically, which from what I have seen none of you have.
Basically, admit it you are fudging the results.
CASE CLOSED.
> Robin E. Brown
> reb...@epix.net
And you paid good money for your education?
----
John Pimentel
pime...@roots.ultranet.com
Tore August Kro
John Pimentel wrote:
>
> > You seem to imply that .333... does not equal 1/3. Every decimal number
> > that terminates or repeats can be expressed as a rational number. By
> > definition a rational number is a number that can be expressed as the
> > fraction of two integers. Similarly, every fraction of two integers can
> > be expressed as a decimal number. And the decimal representation of 1/3
> > is???? Well, that is 3rd grade work.
>
> This is all very nice but is that representation an EXACT and EQUAL
> representation?
>
> Ah! There is the key. IF 1/3 is exactly 0.33(bar) then if I multiple
> both sides by the same number, in this case three, I shall arrive at
> the exact same number on both sides. I should NOT have to run one
> through another calculation to get to the same number as the other
> side. That by defination is called mathematical fudging.
>
> A challenge to all the whizzes and "experts" out there.
>
> Take 1.0 and show me the work to get this number to equal 0.99(bar).
>
> Taking 0.99(bar) to 1.0 is not the same thing. I want to see 1.0
> calculated into 0.99(bar).
>
> BTW, whizzes and "experts" here's a hint, it will save countless hours
> terminating decimals have an EQUAL and EXACT representation as a fraction.
> non-terminating, repeating decimals do not and never will.
This is why we write 0.33(bar) for the non-terminating decimalnumber
0.33333333.. were there are infinitely many 3's. Or 0.99(bar) for the
number that contains infinitely many 9's:
0.9999999999999999999999999999999999999999999999999999999999999999999999
<encoded_portion_removed>
If you get the idea.
> If they do
> then you MUST be able take all convertings and go back to the original
> problem mathematically, which from what I have seen none of you have.
> Basically, admit it you are fudging the results.
>
> CASE CLOSED.
>
> > Robin E. Brown
> > reb...@epix.net
>
> And you paid good money for your education?
> ----
> John Pimentel
> pime...@roots.ultranet.com
Tore August Kro
tor...@ifi.uio.no
Stan Armstrong
In article <ejones-ya0235800...@206.15.64.37>, "Earle D.
Jones" <ejo...@hooked.net> writes
>
>Pimentel, I am through with you.
>
>You are either a troll or a shithead.
>
>Tell us which.
That doesn't sound as if you're through with him, Earle.
>
>earle
>=====
>
> __
> __/\_\
> /\_\/_/
> \/_/\_\ earle
> \/_/ jones
--
Stan Armstrong
John Pimentel
> Date: Fri, 7 Feb 1997 06:45:07 -0800
> To: pime...@roots.ultranet.com
> From: irv...@grfn.org (Irving Lubliner)
> Subject: Re: 0.33(bar) * 3 = 0.99(bar) IS NOT 1/3 * 3.
> >
> >Oh you're funny. Now show me how to calculate 0.899(bar) and arrive at the
> >0.90 and then go from 0.90 go back to 0.899(bar). When you cannot, take
> >your degree, crossout your name and mail it to me, as it's not doing you
> >any good. You cannot mathematically go from 0.90 back to 0.899(bar).
> >Now prove me wrong.
> >
> ----------------------------------------
> OK, you asked for it......
>
> Let X = 0.8999(bar).
>
> Multiply both sides by 10.
> 10X = 8.9999(bar)
>
> Since X and 0.8999(bar) name the same number, we can subtract X from
> the left and 0.8999(bar) from the right side of this last equation:
> 9X = 8.1000(bar)
>
> Divide both sides by 9:
> X = 0.9
>
> Therefore, X is equal to both 0.8999(bar) and 0.9.
> What does that tell us about these two numbers?
> They're equal!
Nice try, but you failed to do the second part of this.
Take 0.90 and go back to 0.899(bar).
Even I can do the first part, but I have not proved anything until I
arrive back at the original problem. Don't you proof any of your
work or do you just check the back of the book for the answers?
> ---------------------------------------------
>
> Method #2:
> Consider this addition problem: 17/30 + 1/3.
> The sum is 9/10. (I hope this requires no explanation.)
Nope with you so far.
> Now, convert each fraction to its decimal equivalent:
Now we convert (adding a degree of error, that some have no problem
disregarding)...
> 17/30 is equal to 0.5666(bar) and 1/3 is equal to 0.3333(bar).
17/30 is equivalent to 0.566(bar) and 1/3 is equivalent to 0.33(bar)
and if I take, say, 0.33(bar) and multiple it by three I get 0.99(bar)
and 1/3 and mulitple it by three I get 3/3 which reduces to 1/1 or 1.
But I am left with a problem 0.99(bar) on one side and 1 on the
other, but if I take 0.99(bar) through an added step (to introduce
an error) I get 1.00. So by converting with a formula I can
take a number that continues on to infinity and round it up to
the nearest number. Ok, I am with so far.
As to the 17/30 if I take it and double it I get 34/30 and I
add itself to it I get 51/30 which has a decimal equivalent of
1.7 (non-repeating and terminating).
If I take 0.566(bar) and double it and add itself to it I get
1.699(bar), oh but then there is that error I can introduce through
a formula to round up to the next digit.
Yep, I am with you. Now comes the question what is that error
that when included in this that allows the two become equal or
unequal?
> What is the sum of these two numbers? It is 0.8999(bar)
0.566(bar) + 0.33(bar) is indeed 0.899(bar) but the componets
used do not represent exact equivalents.
> What does this tell us about the numbers 9/10 and 0.8999(bar)?
> They're equal!
No, it tells me that they are almost equal.
> ---------------------------------------------
>
> Method #3:
> Consider 0.9000(bar).
> You can exchange one of the tenths for ten hundredths.
> You will be left with 8 tenths.
> You can exchange one of those hundredths for ten thousandths.
> You will be left with 9 hundredths.
> You can exchange one of those thousandths for ten ten-thousandths.
> You will be left with 9 thousandths.
> Since you can always exchange one from a given place for ten of
> whatever's in the next place to the right, we find that 0.9 is equal to
> 0.89999(bar).
So what you are saying is that you agree there is an error in the
converstation. Well, thank you very much for agreeing with me.
I am very pleased.
> If this is difficult to follow, see methods 2 and 3.
Not in the least bit. I knew there was an error and that we were dealing
with equivalents that were not quite exact.
> If they are difficult to follow, I'll leave it to others to show you the way.
The others need help to be shown the way.
> Irving Lubliner, Mathematics Enthusiast
----
John Pimentel
pime...@roots.ultranet.com
John Pimentel
> 17/30 is equivalent to 0.566(bar) and 1/3 is equivalent to 0.33(bar)
> and if I take, say, 0.33(bar) and multiple it by three I get 0.99(bar)
> and 1/3 and mulitple it by three I get 3/3 which reduces to 1/1 or 1.
> But I am left with a problem 0.99(bar) on one side and 1 on the
> other, but if I take 0.99(bar) through an added step (to introduce
> an error) I get 1.00. So by converting with a formula I can
> take a number that continues on to infinity and round it up to
> the nearest number. Ok, I am with so far.
>
> As to the 17/30 if I take it and double it I get 34/30 and I
> add itself to it I get 51/30 which has a decimal equivalent of
> 1.7 (non-repeating and terminating).
>
> If I take 0.566(bar) and double it and add itself to it I get
> 1.699(bar), oh but then there is that error I can introduce through
> a formula to round up to the next digit.
While we are on this topic of converting it to and fro.
To double the 0.566(bar) I had to convert it to it's fractional
form because I could not find the last digit to multiple by two,
then after I converted it back to it's new decimal equivalent
with two doses of errors in the number. If 0.566(bar) could
be doubled without converting (and that's a big if) the last
digit would be a two, but then it would not be 1.133(bar). Just
an example of the error that is thrown in from using the convertion
formula. Taking the 1.133(bar) and adding 0.566(bar) then becomes
1.699(bar), but as shown here this is not a true representation of
0.566(bar) doubled which can be readily done by using it's
fractional equivalent, but is impossible to do with it's decimal
equivalent.
I think mathematics is fun.
And as a calculus professor once said "You only need to find one
case where it is not true". Done.
----
John Pimentel
pime...@roots.ultranet.com
Haran Pilpel
John Pimentel wrote:
>
> > Date: Fri, 7 Feb 1997 06:45:07 -0800
> > >0.90 and then go from 0.90 go back to 0.899(bar).
For some reason you seem to believe that equality is not commutative,
i.e.:
a = b does not imply b = a.
I'm afraid I have to break the horrible news: the equality relation is
transitive, reflexive and - what is important in this case - symmetric.
So indeed a = b DOES imply b = a. This is part of the definition of
equality relations. If you want a REALLY formal statement of this take
a discrete math course.
So if we have shown that 0.89(9bar) equals 0.9 then also 0.9 equals
0.89(9bar).
I will not go into 'going from 0.89(9bar) to 0.9' as this has been shown
several times already.
Haran
P.S. I highly recommend to everyone on this group to stop posting to
this thread. It is obviously getting nowhere: as Parkinson put it, "one
side will never be convinced and the other is convinced already". My
apologies if this is a breach of netiquette...
--
e^(Pi*i) + 1 = 0 (Euler)
he...@dorsai.org
On Thu, 06 Feb 1997 18:17:22 -0800, irv...@grfn.org (Irving Lubliner)
wrote:
consider .88(bar)=8/9
.9 =9/10
a number between them would be the average (80+81)/180.
please explain the difference between .899.... and .9
>In article <1997020618...@cinna.ultra.net>,
>pime...@roots.ultranet.com wrote:
>
>> BTW: To my last challenge to find a number between 0.88(bar) and 0.9 which
>> to my surprise, given all the "experts" here, one answer is 0.899(bar) which
>> is higher than 0.88(bar), but none of you offered that. Perhaps, because
>> word problems are difficult to grasp by the "experts".
Lee Jaap
|>To double the 0.566(bar) I had to convert it to it's fractional
|>form because I could not find the last digit to multiple by two,
|>then after I converted it back to it's new decimal equivalent
|>with two doses of errors in the number. If 0.566(bar) could
|>be doubled without converting (and that's a big if) the last
|>digit would be a two, but then it would not be 1.133(bar). Just
|>an example of the error that is thrown in from using the convertion
|>formula. Taking the 1.133(bar) and adding 0.566(bar) then becomes
|>1.699(bar), but as shown here this is not a true representation of
|>0.566(bar) doubled which can be readily done by using it's
|>fractional equivalent, but is impossible to do with it's decimal
|>equivalent.
Define "error". How much error are you introducing with the
conversions? By standard definitions, there is zero error
because the fractional representation and the decimal representation
are merely two names for the same number. By your definitions,
there seems to be an error. Tell me how big the error is.
10^-1? 10^-100? 10^-googolplex?
I begin to think that you and a couple other posters are using
different definitions from the rest of us.
|>I think mathematics is fun.
So do I.
|>And as a calculus professor once said "You only need to find one
|>case where it is not true". Done.
And I don't understand what your "case" claims to prove.
Dave Weisbeck
In article <1997020715...@cinna.ultra.net>, pime...@roots.ultranet.com wrote:
>
>To double the 0.566(bar) I had to convert it to it's fractional
>form because I could not find the last digit to multiple by two,
>then after I converted it back to it's new decimal equivalent
>with two doses of errors in the number. If 0.566(bar) could
>be doubled without converting (and that's a big if) the last
>digit would be a two, but then it would not be 1.133(bar). Just
>an example of the error that is thrown in from using the convertion
>formula. Taking the 1.133(bar) and adding 0.566(bar) then becomes
>1.699(bar), but as shown here this is not a true representation of
>0.566(bar) doubled which can be readily done by using it's
>fractional equivalent, but is impossible to do with it's decimal
>equivalent.
>
What is this error you speak of?
So if I understand what you are claiming then:
17/30 = .566... + e
where e is some error term.
Could you now tell me what e is?
I think this is where this is all falling apart. The definitions of
mathematics I am using define e to be 0. Are you using some other definitions
and if so, could you then clarify what they are?
[snip]
>----
>John Pimentel
>pime...@roots.ultranet.com
>
Dave Weisbeck
z2...@ugrad.cs.ubc.ca
Dave B.
he...@dorsai.org wrote
> please explain the difference between .899.... and .9
Alas, they are one and the same.
Dave B.
Dave Weisbeck
In article <1997020713...@cinna.ultra.net>, pime...@roots.ultranet.com wrote:
>
>Taking 0.99(bar) to 1.0 is not the same thing. I want to see 1.0
>calculated into 0.99(bar).
>
I think you have confused something here once again John.
Are you thinking about "if and only if" here and somehow trying
to say that therefore we haven't proven 1 = 0.99... since we haven't
gone both ways?
Well, in case you are, it really isn't necessary but since doing the
work is easier than explaining to you why it isn't necessary I will
do it.
1 = 9/9 = 0.9/0.9 = 0.9 *( 1 / (1 - 1/10)) =
= 0.9 * Sigma[n=0..infinity](1/10)^n = 0.999....
Happy now?
Dave Weisbeck
z2...@ugrad.cs.ubc.ca
Raymond E. Griffith
========================
Robin E. Brown wrote:
You seem to imply that .333... does not equal 1/3. Every decimal number
that terminates or repeats can be expressed as a rational number. By
definition a rational number is a number that can be expressed as the
fraction of two integers. Similarly, every fraction of two integers can
be expressed as a decimal number. And the decimal representation of 1/3
is???? Well, that is 3rd grade work.
========================
John Pimentel replied:
This is all very nice but is that representation an EXACT and EQUAL
representation?
Ah! There is the key. IF 1/3 is exactly 0.33(bar) then if I multiple
both sides by the same number, in this case three, I shall arrive at
the exact same number on both sides. I should NOT have to run one
through another calculation to get to the same number as the other
side. That by defination is called mathematical fudging.
A challenge to all the whizzes and "experts" out there.
Take 1.0 and show me the work to get this number to equal 0.99(bar).
Taking 0.99(bar) to 1.0 is not the same thing. I want to see 1.0
calculated into 0.99(bar).
BTW, whizzes and "experts" here's a hint, it will save countless hours
terminating decimals have an EQUAL and EXACT representation as a
fraction.
non-terminating, repeating decimals do not and never will. If they do
then you MUST be able take all convertings and go back to the original
problem mathematically, which from what I have seen none of you have.
Basically, admit it you are fudging the results.
CASE CLOSED.
>
>Robin E. Brown
>reb...@epix.net
>
And you paid good money for your education?
----
John Pimentel
pime...@roots.ultranet.com
===============================
John seems to feel that he can insult his way into acceptance...
John Pimentel wrote to me in an e-mail:
>
> Yawn... all these math whizzes are putting me to sleep....
>
> Show me how to calculate 0.899(bar) and arrive at the 0.90.
>
> Now comes the tricky part take 0.90 and calculate that this
> is 0.899(bar).
>
> When you fail to do this take your degree, cross out your name
> and mail it to me, it's doing you no good.
> ----
> John Pimentel
> pime...@roots.ultranet.com
My reply:
As to your challenge:
0.90 = 0.80 + 0.10
= 0.80 + 0.09 + 0.01 = 0.89 + 0.01
= 0.80 + 0.09 + 0.009 + 0.001 = 0.899 + 0.001
etc. You may repeat the process infinitely (never ending, as the "bar"
indicates), obtaining 0.899(bar).
Sir, I have tried to be polite. I do not know what math degree you have
(or don't have), but your understanding is seriously flawed, as well as
your manners.
Your lack of manners merely shows that you are ignorant, and willing to
stay that way. I hope that you will mature out of your arrogance.
When you decide to be reasonable and willing to learn, I would like to
hear from you. Until then I will hope for you.
Raymond E. Griffith
________________________________________________
The educated man possesses six characteristics: He must be able to
1. communicate;
2. calculate;
3. see the world as it really is;
4. appropriately manipulate his environment;
5. accurately predict the effects of that manipulation;
6. admit the limits of his knowledge.
Raymond E. Griffith
Lee Jaap
In article <32FB96...@vnet.net> "Raymond E. Griffith" <rgri...@vnet.net> writes in response to John Pimentel:
|>Your lack of manners merely shows that you are ignorant, and willing to
|>stay that way. I hope that you will mature out of your arrogance.
I'd go a step beyond "willing" to "intending". I think John likes
using definitions different from the rest of the mathematical
community.
A limit (if it exists) is a number, not a process. (Definition)
.9bar is shorthand for a limit of a sequence. (Definition)
.9bar is a number. (From the two statements above)
If .9bar != 1, then there exists a number between .9bar and 1.
(Property of real numbers)
There is no such number. (It's easy to prove this assertion
wrong: FIND SUCH A NUMBER.)
Therefore .9bar, 1, and even 9/9 are different ways of exactly
representing precisely the same quantity.
Robin E. Brown
> stay that way. I hope that you will mature out of your arrogance.
>
> hear from you. Until then I will hope for you.
>
> Raymond E. Griffith
> ________________________________________________
>
> The educated man possesses six characteristics: He must be able to
> 1. communicate;
> 2. calculate;
> 3. see the world as it really is;
> 4. appropriately manipulate his environment;
> 5. accurately predict the effects of that manipulation;
> 6. admit the limits of his knowledge.
>
> Raymond E. Griffith
The December 1971 edition of The Mathematics Teacher has an article that
takes .999... to 1 and also takes 1 to .999...
Robin
reb...@epix.net
Albert Y.C. Lai
In article <1997020618...@cinna.ultra.net>,
"John Pimentel" <pime...@roots.ultranet.com> wrote:
>I converted the 0.33(bar) to it's equivalent approximation and when I dealt
What is the point of converting to "equivalent approximation"? What do
you mean by "equivalent approximation"? Equivalence is equivalence,
approximation is approximation. Your newly invented term is an
oxymoron. Talk about silent noise.
>with each on their own, they did not match after doing the same number and
>type of actions.
Yeah you dealt with each on their own. No you did not really prove
that they are really different. My friends on their own deal with me
as Albert. My parents on their own deal with me as Y.C. My school on
its own deals with me as Albert Y.C. Lai. Does that make me three
distinct persons. Oh my ISP deals with me as trebla, I guess that
makes me four.
>All the "experts" here claim that the two are equal,
Save this jealousy against experts. If a person X is called an
expert, it is not because X self-proclaims to be an expert, it is
because other people can see clearly X's competence and decide to call
X and expert. If you can't win a place from other people, it is your
own problem.
And besides we do not consider ourselves experts. We just argue with
logic.
>even the phrase "exact"
>has been used. Perhaps, the word "approximation" does not exist in your
>vocabulary. It would appear that an orange is still an orange and not an
>apple.
When we say "exact" we mean "exact", and we are ready to support it.
The entire topic does not belong to numerical analysis so approximation
isn't relevant. You want to show that your vocabulary is larger than
ours? You will only show that you don't know which part of your
vocabulary is useless for the problem at hand. Like the "equivalent
approximation" fiasco.
By refusing that 0.9(bar) is another way to write 1, you are
effectively refusing that homo sapiens is human.
>Gee, and I don't even have a PhD. in mathematics.
I tell you what, a PhD in Math would rather tell you that you are wrong
and have you throw stones, than to agree with you and get you pleased.
What makes you more special than any of us? If anything your reasoning
is the sloppiest; heck you actually never presented any reasoning.
First we have presented at least two kinds of proofs of our claims, and
you never succeeded in scrutinizing them. Second everytime you post
someone (in fact everyone else) can find a flaw in it, and as a result
you keep posting new attempts and we keep finding new flaws. You
didn't even reason, you just did trial-and-error.
I am glad no one granted you a PhD in Math.
>BTW: To my last challenge to find a number between 0.88(bar) and 0.9 which
>to my surprise, given all the "experts" here, one answer is 0.899(bar) which
>is higher than 0.88(bar), but none of you offered that. Perhaps, because
>word problems are difficult to grasp by the "experts".
BTW 0.899(bar) is the same as 0.9. Perhaps the definitions of real
numbers are difficult to grasp by you.
--
Albert Y.C. Lai tre...@vex.net http://www.vex.net/~trebla/
David Tarkowski
I just wanted to put my last two cents in on this topic before I quit
this thread for good. While the whole topic seems to be way over done,
it does seem to bring out a few points that need illustrating, and I
though that I might try and do that. Hopefully my discourse will clear
up a few things for at least one person. Here I go:
1) Mathematical Proofs
First let's prove something.
Theory: .99bar = 1 where .99bar means .9999... with an infinte string of
nines.
Proof:
Let x = .99bar definition of x
10x = 9.99bar multiplication
10x - x = 9.99bar - .99bar (Just added the same thing to both
sides)
9x = 9 algebraic simplification
x = 9/9 = 1 Q.E.D
(Note to all those Ph.D math people out there: I am a physist (in
training) so I have never had a course on rigourous proofs. I am sure
that there is some error in my terminology, but I feel that it is
sufficent for the level of this group).
So now we have a mathematical proof that .999bar = 1. If you feel that
this is false, you must either show that one of my steps is invalid, and
why it is invalid, or you must "find a case where it is not true,"
namely find a number between .99bar and 1 (which is equivalent to
proving that one of my steps was not valid). Otherwise you must accept
the proof whether you like it or not. That is the problem with such
things as math and science; there are some things that we might not
like, but that has absolutely no bearing on their verity. Until
somebody can reasonable demonstrate why one of the steps in my proof is
wrong, I will belive that .99bar = 1. Of course, being an open minded
person, I will have already challenged every step of my proof with every
possible flaw that I can think of. If somebody wants to send me that
famous example of why 1=2 (I can't remember it off hand and don't want
to give it credit by deriving it) I can show how one would find flaw
with a proof.
2) Infinity:
Infinity is a hard topic to grasp, which probably accounts for much of
the problems with .99bar. One of my professors gave the defining
equation of infinty as infinity = infinity + 1. This is handy for some
applications of infinity, but not very clear in this case.
I believe that some of the current problems with infinty stem from our
digital, calculator powered world. An infinite series of nines can not
be represented on a TI- 85 calculator and since our educational system
(at least in the USA) has taken a "plug and chug" approach to teaching
math (less thinking, more typing) I think many students are not
introduced to many of the beauties of math.
Anyway, as an example of .99bar, imagine 1 billion Cray YMP(tm?)
supercomputers all dedicated to producing an endless stream of nines on
a series of extremely high speed video terminals. After 10 billion
years, you still would not have even .01% of the number of nines in
.99bar. In fact, you could never have .01%. In this list .99bar = 1
becomes less of an approximation than the identity that it is.
I am sorry if my statements have angered you. I have not meant to
preach/write a manifesto, but rather I wanted to illustrate a few points
to a few begining algebra students. I hope that I have done that. If
you feel that any of my points are in error, please post a reply or send
my an e-mail and we can discuss. Any irrational flames, attacks on my
character, or other negative remarks can be directed to /dev/null.
Sorry for the length, but I wanted to be complete.
Dave
P.S. How about this:
1 = x
10 = 10x
9 = 9x
x = 9/9
x = 9 * (1/9) = 9 * (.1bar) = .9bar
(Have I gone backwards?)
Josh Boyd
> This is all very nice but is that representation an EXACT and EQUAL
> representation?
Yes it is.
> Take 1.0 and show me the work to get this number to equal 0.99(bar).
Ok...
1.0 = 1/3 + 1/3 + 1/3
1.0 = .33(bar) + .33(bar) + .33(bar)
1.0 = .99(bar)
> BTW, whizzes and "experts" here's a hint, it will save countless hours
> terminating decimals have an EQUAL and EXACT representation as a fraction.
> non-terminating, repeating decimals do not and never will. If they do
Yes they do. 1/3 is EXACTLY equal to .33(bar) (this is assuming you
could write 3's forever, you will NEVER have it terminate with a 4...)
If you don't believe this, do the long division...
____
3|1.00
If you do this long division, you will see that you will have a non
terminating decimal...
= .33333(bar) I would write it out and actually do the math for you,
but that would be a little hard to do on this --- so why don't you? See
for yourself.
> CASE CLOSED.
Not yet...
Josh
BrianScott
Stan Armstrong <St...@stanleya.demon.co.uk> wrote (in part):
>You are either a troll or a shithead.
Pain in the arse, for me. (note the Anglo-Saxon spelling!)
I can't resist pointing out that the true Anglo-Saxon
spelling is 'ears', that being the Anglo-Saxon (i.e., Old
English) etymon of modern English 'arse'!
Brian M. Scott
Do Not Use: brian...@aol.com
Always Use: sc...@math.csuohio.edu
BrianScott
From: g_st...@titan.sfasu.edu
Date: 6 Feb 97 15:14:16 CST
Message-ID: <1997Feb...@titan.sfasu.edu>
From: g_st...@titan.sfasu.edu
Date: 6 Feb 97 15:14:16 CST
Message-ID: <1997Feb...@titan.sfasu.edu>
John Pimentel asserted that he couldn't double 0.5(6bar) without
converting it to its fractional equivalent. He went on to imply that
the task was impossible, stating that if it could be done, the 'last'
digit would be '2'.
While John's inability isn't terribly surprising, the task is entirely
possible, and of course the supposed difficulty of the 'last' digit,
like that digit itself, doesn't exist. 0.5(6bar) is the limit of the
sequence 0.5, 0.56, 0.566, 0.5666, ..., so 2 * 0.5(6bar) is the
limit of the sequence 2 * 0.5, 2 * 0.56, 2 * 0.566, 2 * 0.5666, ...,
i.e., the sequence 1.0, 1.12, 1.132, 1.1332, .... This limit, as
is easily demonstrated by techniques already well-rehearsed
here, is 1.1(3bar). Look, ma: no fractions!
John Pimentel
From: Haran Pilpel <har...@ibm.net>
> P.S. I highly recommend to everyone on this group to stop posting to
> this thread. It is obviously getting nowhere: as Parkinson put it, "one
> side will never be convinced and the other is convinced already". My
> apologies if this is a breach of netiquette...
You mean agree to disagree. So I'll be on the "convinced already" pile
and you'll be on the other "never be convinced". I can live with that :-)
----
John Pimentel
pime...@roots.ultranet.com
Vincent Johns
(posted & emailed)
Lee Jaap <Jaa...@ASMObj.LaRC.NASA.Gov> wrote:
> [...]
>
> A limit (if it exists) is a number, not a process. (Definition)
>
> .9bar is shorthand for a limit of a sequence. (Definition)
>
> .9bar is a number. (From the two statements above)
>
> If .9bar != 1, then there exists a number between .9bar and 1.
> (Property of real numbers)
>
> There is no such number. (It's easy to prove this assertion
> wrong: FIND SUCH A NUMBER.)
> [...]
This might only show the limits of my ability, rather than that
there is no such number. An unsuccessful search for such a number
might, however, lead me to a proof that none exists.
--
-- Vincent Johns
Please feel free to quote anything I say here.
Vincent Johns
(posted & emailed)
John Pimentel <pime...@roots.ultranet.com> wrote:
> [...]
> Basically, admit it you are fudging the results.
If you don't understand the notation, how can you tell?
One purpose of mathematical expressions is economy of
expression, so that we can discuss infinite series
without writing infinitely long expressions. Another
is communication, so that another person who knows
what the symbology means can understand what one has
written. You seem to be using non-standard terms.
> CASE CLOSED.
Does this mean you've lost interest?
> And you paid good money for your education?
The education can indeed be worth the money. Before you decide
to go to college (or decide on which college to attend), I
suggest that you talk with teachers, counselors, and other
college graduates (yes, the teachers, etc., have all graduated
from college!) to determine what benefits they feel they received
from it, and what they would suggest for you. Mathematics can
be rewarding, but there are many other choices, too, and you
might prefer something else. You don't have to decide right
away.
A major benefit of obtaining a degree is learning new ways of
thinking, largely obtained through study of what other people have
done through the ages. You will be able to answer for yourself
questions like what is the exact value of 0.99999..., without
depending on other people to tell you. (In high school, there's a
lot of "Here's what happened. This is how it's done." In college,
this changes to "Find out for yourself why ... and clearly tell
someone else about it".)
Stan Armstrong
In article <32FB96...@vnet.net>, "Raymond E. Griffith"
<rgri...@vnet.net> writes
>
>
>Your lack of manners merely shows that you are ignorant, and willing to
>stay that way. I hope that you will mature out of your arrogance.
>
Rage" is but one of many manifestations.
--
Stan Armstrong
Mike Housky
Perhaps, but how would you prove that the proof was correct? And
then, of course, you would have to show that the proof of the proof
was correct. Ad nauseum.
I'd like to say that this is from The Zen of Zeno, but then I'd have to
prove that. And then I'd have to prove...well...nevermind...I just made
it up. This is an easily-refutable assertion. All you have to do is
show that there are at least two people crazy enough to write this
post to establish the reasonable probability that I wasn't both of them.
Cheers,
Mike.
Todd Harris
In article <JAAPJL.97...@asmobj.larc.nasa.gov>, J Lee Jaap
<Jaa...@ASMObj.LaRC.NASA.Gov> wrote:
€In article <32FB96...@vnet.net> "Raymond E. Griffith"
<rgri...@vnet.net> writes in response to John Pimentel:
<comments about Mr. Pimento snipped>
€A limit (if it exists) is a number, not a process. (Definition)
€
€.9bar is shorthand for a limit of a sequence. (Definition)
€
€.9bar is a number. (From the two statements above)
€
€If .9bar != 1, then there exists a number between .9bar and 1.
€(Property of real numbers)
€
€There is no such number. (It's easy to prove this assertion
€wrong: FIND SUCH A NUMBER.)
Hey...that ones easy...
0.0(bar) of course!
(he said with his tongue planted firmly in his cheek)
but now that I think about it.....hmmmm
--
Signature-File deleted by:
NetCensors Incorporated
"Your ignorance is our business"
Avital Pilpel
Using, of course, the very simple rule, that lim(c*a(n)) = c*lim(a(n)),
when the limit exists (works for c<>0 when there is no limit too).
--
Avital Pilpel.
=====================================
The majority is never right.
-Lazarus Long
=====================================
Vincent Johns
(posted & emailed)
Mike Housky <mi...@webworldinc.com> wrote:
>
> Vincent Johns wrote:
> >
> > (posted & emailed)
> >
> > Lee Jaap <Jaa...@ASMObj.LaRC.NASA.Gov> wrote:
> > > [...]
> > > There is no such number. (It's easy to prove this assertion
> > > wrong: FIND SUCH A NUMBER.)
> > > [...]
> >
> > This might only show the limits of my ability, rather than that
> > there is no such number. An unsuccessful search for such a number
> > might, however, lead me to a proof that none exists.
>
> Perhaps, but how would you prove that the proof was correct? And
> then, of course, you would have to show that the proof of the proof
> was correct. Ad nauseum.
It's fun to think of it like this, but there really are ways to
prove proofs correct (without the infinite, ad nauseam regression),
but the metamathematical techniques used are not usually addressed
directly in high-school algebra and geometry classes.
What I was saying was that, before I need to show that my proof is
correct, I have to come up with a proof, and I can't always do that.
Sometimes I've found it helpful to try to *disprove* the statement
I'm trying to prove (for example, by finding a counterexample to it)
and, when I can't do that, try to show why not. The "why not" becomes
the basis for the proof I'm looking for, if I'm lucky.
> I'd like to say that this is from The Zen of Zeno, but then I'd have to
> prove that. And then I'd have to prove...well...nevermind...I just made
> it up. This is an easily-refutable assertion. All you have to do is
> show that there are at least two people crazy enough to write this
> post to establish the reasonable probability that I wasn't both of them.
I sort of prefer "wild and fun-loving" to "crazy", but what the hey,
I think it's interesting to look at why we do things. Isn't that what
Descartes did with his "Cogito, ergo sum" conclusion? He came up with
a statement that was evidently true, but not provable in any system
(such as deductive logic) known at that time -- he used a new type of
logic which was nevertheless convincing.
Then, too, I suppose there are people who would consider anyone who
thinks mathematics can be lots of fun is a little ... crazy. ;-b
Lee Jaap
(Somebody, please slap me. I'm going to try to communicate with
John Pimentel again.)
In article <1997020715...@cinna.ultra.net> "John Pimentel" <pime...@roots.ultranet.com> writes:
|>> Date: Fri, 7 Feb 1997 06:45:07 -0800
|>> To: pime...@roots.ultranet.com
|>> From: irv...@grfn.org (Irving Lubliner)
|>> Subject: Re: 0.33(bar) * 3 = 0.99(bar) IS NOT 1/3 * 3.
|>> Method #2:
|>> Consider this addition problem: 17/30 + 1/3.
|>> The sum is 9/10. (I hope this requires no explanation.)
|>
|>Nope with you so far.
|>
|>> Now, convert each fraction to its decimal equivalent:
|>
|>Now we convert (adding a degree of error, that some have no problem
|>disregarding)...
What error? Please quantify the error you claim to be introduced.
The numbers are exactly equal. The "(bar)" is a shorthand to
represent that all the 6's and 3's are there in the number being
referred to.
|>> 17/30 is equal to 0.5666(bar) and 1/3 is equal to 0.3333(bar).
|>
|>17/30 is equivalent to 0.566(bar) and 1/3 is equivalent to 0.33(bar)
|>and if I take, say, 0.33(bar) and multiple it by three I get 0.99(bar)
|>and 1/3 and mulitple it by three I get 3/3 which reduces to 1/1 or 1.
|>But I am left with a problem 0.99(bar) on one side and 1 on the
|>other, but if I take 0.99(bar) through an added step (to introduce
|>an error) I get 1.00. So by converting with a formula I can
|>take a number that continues on to infinity and round it up to
|>the nearest number. Ok, I am with so far.
Introduce *what* error?
|>As to the 17/30 if I take it and double it I get 34/30 and I
|>add itself to it I get 51/30 which has a decimal equivalent of
|>1.7 (non-repeating and terminating).
|>
|>If I take 0.566(bar) and double it and add itself to it I get
|>1.699(bar), oh but then there is that error I can introduce through
|>a formula to round up to the next digit.
You keep talking about "error". Please explain what the error is.
|>Yep, I am with you. Now comes the question what is that error
|>that when included in this that allows the two become equal or
|>unequal?
|>
|>> What is the sum of these two numbers? It is 0.8999(bar)
|>
|>0.566(bar) + 0.33(bar) is indeed 0.899(bar) but the componets
|>used do not represent exact equivalents.
|>
|>> What does this tell us about the numbers 9/10 and 0.8999(bar)?
|>> They're equal!
|>
|>No, it tells me that they are almost equal.
What's the difference?
I mean that literally. What's the difference between 9/10 and 8.9bar?
Subtract one from the other and tell us the result. Most of us get
zero.
|>> ---------------------------------------------
|>>
|>> Method #3:
|>> Consider 0.9000(bar).
|>> You can exchange one of the tenths for ten hundredths.
|>> You will be left with 8 tenths.
|>> You can exchange one of those hundredths for ten thousandths.
|>> You will be left with 9 hundredths.
|>> You can exchange one of those thousandths for ten ten-thousandths.
|>> You will be left with 9 thousandths.
|>> Since you can always exchange one from a given place for ten of
|>> whatever's in the next place to the right, we find that 0.9 is equal to
|>> 0.89999(bar).
|>
|>So what you are saying is that you agree there is an error in the
|>converstation. Well, thank you very much for agreeing with me.
|>I am very pleased.
No error introduced. Each step yields a number precisely equal to
the one before. Performing the step infinitely many times
introduces absolutely no error.
|>> If this is difficult to follow, see methods 2 and 3.
|>
|>Not in the least bit. I knew there was an error and th>Not in the least bit. I knew there was an error and that we were dealing
|>with equivalents that were not quite exact.
Aha. I detect a disagreement of definitions. In this context, many
of us are using "equivalent" to mean "different in appearance but
exactly equal in value". John is using the word to mean "approximately
equal", which I do seem to recall from somewhere way back there. But
most of us are using the first meaning, where "approximation" has no
relevance.
All of the following representations are equivalent (different in
appearance but equal in value):
1, 1*1, 1+0, 0+1, 1/1, 3/3, 1.1/1.1, 1.00, 1.000, 1.0bar, 0.9bar,
sqrt(1), 2-1, 0.5+0.5, 0.3+0.1+0.6, .3bar+.3bar+.3bar, 3*.3bar,
1.0, 0.9+0.1, 0.99+0.01, 0.999+0.001,
the solution of x-1=0
If two of these numbers are not precisely equal in value, tell
me which two they are and the difference between them (or a number
between them).