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'[OT]: Brain Burp Rounding??'
2001\05\30@115223 by Roman Black

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I was tabulating some data tonight, ie, doing
a division on my pocket calculator and then rounding
the number to three decimal places, then writing
the figure down on paper.

How do people round numbers to the nearest decimal
place on their pocket calculators?

My whole life I have checked if the last digit is
5, then if so check the next digit is 5, etc,
then once you get to a non-5 round it up or down.
Hope that makes sense.

So:
3.12471 = 3.125
3.12441 = 3.124
3.1245551 = 3.124
3.1245556 = 3.125

This just can't be right!
How were you taught? I started using pocket
calculators when they had led digits, probably
before I was 10 years old. I'm sure in school we
were taught to round digits like that. ??
-Roman

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2001\05\30@115901 by Alan B. Pearce

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>How do people round numbers to the nearest decimal
>place on their pocket calculators?

Easy, just set my TI-83 to 3 dec places. It keeps higher precision
internally if I need to change it. Yeah, I know, not quite what you wanted
to hear :)

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2001\05\30@120447 by Nick Taylor

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Roman Black wrote:
>
> I was tabulating some data tonight, ie, doing
> a division on my pocket calculator and then rounding
> the number to three decimal places, then writing
> the figure down on paper.
>
> How do people round numbers to the nearest decimal
> place on their pocket calculators?
On my calulator I round to three places by setting the display to only
show three places .... less brain strain.
{Quote hidden}

You're right, it's not right.  Just drop the digits past the third
decimal, then only consider the third decimal in your rounding.
>
> This just can't be right!
> How were you taught? I started using pocket
> calculators when they had led digits, probably
> before I was 10 years old. I'm sure in school we
> were taught to round digits like that. ??
> -Roman

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2001\05\30@121940 by David W. Gulley

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Roman Black wrote:
{Quote hidden}

I have used the following:

 3.12350 = 3.124 (if exactly .xxx50000, round up if previous digit odd)
 3.12450 = 3.124 (if exactly .xxx50000, round down if previous digit
even)

The idea here is that if you are tallying a series of values, the
"random" nature of rounding some up, some down will "tend" to average
out the errors. (This is a very debatable subject!)


 3.12451 = 3.125 (greater than .xxx50 so round up)
 3.12449 = 3.124 (less than .xxx50 so round down)

following the previous two examples:
 3.1245000000001  = 3.125 (greater than .xxx50 so round up)
 3.1244999999999  = 3.124 (less than .xxx50 so round down)

for the examples you used:
3.1245551 = 3.125   (greater than .xxx50 so round up)
3.1245556 = 3.125   (greater than .xxx50 so round up)

David W. Gulley
Destiny Designs

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2001\05\30@122943 by eter William Green
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roman,

i was taught to round up if it's >= 1/2.  little different than you system - little easier

> 3.12471 = 3.125
> 3.12441 = 3.124
> 3.1245551 = 3.124
> 3.1245556 = 3.125

3.1247 = 3.125   7 >= 5 round up
3.1244 = 3.124   4 <  5 truncate
3.1245 = 3.125   5 >= 5 round up
3.1245 = 3.125   5 >= 5 round up

ie: i only check the 4th digit. if it's >= 5 round up.  if not truncate.

-pete



On Thu, 31 May 2001, Roman Black wrote:

{Quote hidden}

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2001\05\30@124335 by Cris Wilson

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At 01:48 AM 5/31/01 +1000, you wrote:
{Quote hidden}

I first look and see how many digits my answer needs to have. I then carry
one extra digit
than needed and just drop the other digits. Unless I have a function
involved (like sin, cos,
e, LN, pi, etc), I carry all digits for the number associated with the
function. When all of
the calculations are done I use that last extra digit to decide whether or
not to round up
the final answer. If the last digit is 5 or greater I round up the next digit.

Of course there are tons of books on precision, accuracy, and error
calculations if you
really want something set down in writing as a reference. ASTM I know has a
section
on proper significant figure use. Measurement theory textbooks usually
contain a chapter
on proper rounding as well.

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2001\05\30@125600 by Kevin Blain

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Roman said.....

> How do people round numbers to the nearest decimal
> place on their pocket calculators?

and....

>
> So:
> 3.12471 = 3.125
> 3.12441 = 3.124
> 3.1245551 = 3.124
> 3.1245556 = 3.125
>
> This just can't be right!

It's not.

3.1245551 goes to 3.124 because it's greater than halfway (3.1425)

It's the old, add 5, 0.5, 0.05, etc, in your case, add 0.0005, and then
truncate it to n d.p.(in your case 3)

rule: add 5 x 10 exp -(decimal_places) then truncate to (decimal_places)
places.

That's almost how I was taught, but it's how I do it.

Regards, Kevin

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2001\05\30@130538 by Barry Gershenfeld

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We were taught: "Round off to make even".  It sounded outrageous
then and I'll admit to not doing it that way much.  But I think
the reasoning is sound.  Other posters have already said so.
Paraphrasing the example:
3.12471 = ?
3.12441 = ?
3.1245551 = ?
3.1245556 = ?

Nick Taylor said:
>You're right, it's not right.  Just drop the digits past the third
>decimal, then only consider the third decimal in your rounding.

Thus, in each example you consider just 3.124   =:-o

The premise is that since you are rounding to 3 digits, that the
4th digit is not reliable anyway (even though it looks like it
should be!)  Then, as David Gulley said:

>The idea here is that if you are tallying a series of values,
> the "random" nature of rounding some up, some down will
> "tend" to average out the errors.

And so they divide all possible digits into two groups:
0,2,4,6,8  and 1,3,5,7,9  .  So if digits occur randomly
then 50% of the time you round up and 50% of the time
you don't.

And in the example the answer is 3.124 no matter what
follows.

3.124x = 3.124
3.124xxxx = 3.124    (even if the x's are 9's  =:-o)

3.125x -> 3.126
3.126x -> 3.126
3.127x -> 3.128
3.128x -> 3.128   et c.

Well, I went to school in New Jersey :)  You wanted to know
what we were taught.  Take it or leave it. :)

Barry

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2001\05\30@131341 by Bob Ammerman

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I use "banker's rounding". This is also called "round to even" rounding.
When the digit being eliminated is a 5 you go up or down as needed to get to
an even number. Unlike conventional rounding (where you round up on 5 or
higher) banker's rounding does not have a slight positive bias in it.

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)


{Original Message removed}

2001\05\30@131542 by Nick Taylor

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ooops!  I just read my post, and I didn't type what I meant to type.  I
should have said:  Just drop the digits past the fourth decimal, then
only consider the fourth decimal in your rounding.
 -N.

Nick Taylor wrote:
{Quote hidden}

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2001\05\30@133808 by Barry Gershenfeld

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>ooops!  I just read my post, and I didn't type what I meant to type.  I
>should have said:  Just drop the digits past the fourth decimal, then
>only consider the fourth decimal in your rounding.
>  -N.

Yeah and it was just enough to convince me that my flawed
explanation was ok!  I thought something looked strange. So
I guess we did use that one digit past where you round to,
and then, as Bob just said, do the "make even" thing if
that digit is a 5.

Enough of this, I don't even use it...

Barry

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2001\05\30@160440 by Harold M Hallikainen

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       If the digit beyond what you are rounding is higher than 5, round up,
lower than 5, round down. If equal to 5, round to the even digit. For
example,

3.4 rounds to 3
3.5 rounds to 4
3.6 rounds to 4

4.4 rounds to 4
4.5 rounds to 4
4.6 rounds to 5

       Using this "even digit" method (though an odd will also work), errors do
not accumulate as multiple random rounded numbers are added.

Harold


On Thu, 31 May 2001 01:48:00 +1000 Roman Black <RemoveMEfastvidEraseMEspamEraseMEEZY.NET.AU>
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2001\05\30@203042 by Alexandre Domingos F. Souza

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>How do people round numbers to the nearest decimal
>place on their pocket calculators?

       HP people have the Round(F) function.

>How were you taught? I started using pocket
>calculators when they had led digits, probably
>before I was 10 years old. I'm sure in school we
>were taught to round digits like that. ??

       Mee too.

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2001\05\30@203244 by michael brown

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> roman,
>
> i was taught to round up if it's >= 1/2.  little different than you
system - little easier
{Quote hidden}

Finally some one got it right!  ;-D  Way to go Pete.  If you want to round
to N decimal places then look at position N+1.  If position (N+1) >4 then
increment position(N).  That's all there is to it.  You don't have to look
at anything else.   Hope this helps.  TTYL

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2001\05\31@014757 by Dale Botkin

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On Wed, 30 May 2001, Peter William Green wrote:

> roman,
>
> i was taught to round up if it's >= 1/2.  little different than you system - little easier
>
> > 3.12471 = 3.125
> > 3.12441 = 3.124
> > 3.1245551 = 3.124
> > 3.1245556 = 3.125
>
> 3.1247 = 3.125   7 >= 5 round up
> 3.1244 = 3.124   4 <  5 truncate
> 3.1245 = 3.125   5 >= 5 round up
> 3.1245 = 3.125   5 >= 5 round up
>
> ie: i only check the 4th digit. if it's >= 5 round up.  if not truncate.
>
> -pete

Yup, we were taught that as "artillery roundoff" in the Army.  Carry it
out as many digits as you want...

3.124555551 = 3.125
3.124555556 = 3.125

Dale
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2001\05\31@053859 by Roman Black

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Hi Michael, yep, darn right, half the people in
the world (including me!) have been rounding wrongly
their entire lives. Very scary. There is and can only
be one correct way of rounding decimal numbers...

Here is an example, assuming rounding to 3 decimal
places, that is 1000 combinations, from 000 to 999.

000 to 499 (first half), 500 to 999 (second half).
There are the first 500 combinations in the first
half, and the second 500 combination in the second
half.

All my life I have been rounding at 555, under the
false impression that 5 is "half way" in decimal
terms... Wow.

I asked my 26 yr old science-degreed girlfriend how
to round and she looked at me like I was stupid.
"5 or more, round up". She was taught correctly.

Now I'm wondering if it is mainly us old-timers
from the dawn of pocket calculators age that were
taught wrong?? Any thoughts, older people??
:o)
-Roman


michael brown wrote:
{Quote hidden}

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2001\05\31@083724 by michael brown

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I'm pretty old myself (39), but I have seen calculators that have something
called 5/4 rounding.  Being uneducated (no degree) I'm not sure what that is
about.  Remember the good ole RPN calculators?  If you could understand how
to use one of those, you were well on your way to being a FORTH programmer.

Michael Brown
Instant Net Solutions
http://www.KillerPCs.net




{Quote hidden}

truncate.
> > >
> > > -pete
> >
> > Finally some one got it right!  ;-D  Way to go Pete.  If you want to
round
> > to N decimal places then look at position N+1.  If position (N+1) >4
then
> > increment position(N).  That's all there is to it.  You don't have to
look
> > at anything else.   Hope this helps.  TTYL
>
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2001\05\31@102928 by David W. Gulley

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Roman Black wrote:
{Quote hidden}

Cept for one minor detail,

There are 999 values BETWEEN 0 and 1000.
  500 IS the halfway point.
    500 values below it (0 - 499) and
    500 values above it (501 - 1000).
Therefore, if accumulating a large series of "random" numbers and
rounding using the 5 or greater rule, the result will "tend" to be too
large since statistically you are using 500 possibilities below the
number and 499 above the number.

For most things this error would probably not be a problem, but the
"round to even if 5" rule will "tend" to average out the error
accumulation at the expense of complicating the rule slightly.

The validity of the least significant digits does play an important role
in the process, so when absolute accuracy is required use a tolerance
(+/-) to exactly specify the values, and the error accumulation.
 For example (round to nearest 1/100):
   Value   Tolerance    Round >=5     Round to Even if 5
   1.125   +/- 0.005      1.13          1.12
   1.135   +/- 0.005      1.14          1.14
    Repeat above 1000 times and Total
   2260                   2270          2260

when the tolerances say:
   2260 +/- 10 or somewhere in the range of 2250 to 2270 so yes the
"Round >=5" is a 'correct' answer (since it is in the possible range
defined by the tolerances, even if it is at the upper limit), but the
2260 provided by the "Round to even if 5" is a 'better' answer (in some
cases).

 Obviously this is a contrived scenerio, but if a particular sensor
reported to 0.5 units of precision with +/- 0.5 units of accuracy, and
you wanted to accumulate readings over a long interval, the "Round >=5"
will provide an answer that is "typically" too large.

If you were buying gasoline, and the pumps all rounded using the "Round
>=5" rule, would you mind that you were charged for 2270 gallons instead of 2260? :-)

As I stated in an earlier message in this thread,
   This is a very debatable subject!


David W. Gulley
Destiny Designs

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2001\05\31@103822 by Harold M Hallikainen

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       Good o' RPN calculators!  I still have my HP35 from 1972 and use an
HP15C (I think that's the number) daily. I still agree with the old T
shirts that [ENTER] > = .
       I teach electronics at night at the local community college. My students
all have very fancy calculators that use algebraic notation. And they
consistently get the wrong answer. They're always leaving out parenthesis
and having the calculator solve a problem other than the one they want.
RPN solves this by having you work the problem the way you would by hand:
from the inside out. Also, you see all intermediate results so you can
catch errors if you have a feel for the numbers.
       On rounding, here's an example:

1.5+2.5 = 4.0 without rounding

With rounding using "round to even"

2 + 2 = 4

With "rounding up"

2 + 3 = 5

Round to even tends to minimize error.

Harold


On Thu, 31 May 2001 07:31:01 -0500 michael brown <spamBeGonen5qmgspamKILLspamAMSAT.ORG>
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2001\05\31@110645 by Kevin Blain

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NO! becasue 1000 is the same as 0. Er - that needs explaining! i.e. there
are 500 numbers 0-499 inclusive and 500 numbers 500-999 inclusive. So the
1000 is the 0 in the next set.

Regards, Kevin


{Original Message removed}

2001\05\31@111700 by Patrik Husfloen

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I only used an RPN calc for a few days but I got hooked imediatly,
those things are great, think I used an HP 32 II or something like that.
I normally use an TI83 though. I like the RPN calc much better.

{Original Message removed}

2001\05\31@113549 by David W. Gulley

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To say "NO!" is a bit emphatic, since I could just as well say that we
will only truncate numbers, which would provide an answer which may be
fine for some conditions.

I understand your logic; but this is still including 500 numbers below
the "rounded to" value and 499 above, therefore (and I repeat)
statistically, you are favoring the round up when given a random set of
values.

 Again looking at the results of the scenario I presented below,
The actual answer is                 2260 +/- 10

The "Round >=5" result is            2270  (+0/-20)
The "Round Even if 5" result is      2260  (+/- 10)
And the "Truncate" result is         2250  (+20/-0)

So which answer is right?
  All of them (or none of them) depending on the requirements...

David W. Gulley
Destiny Designs


Kevin Brain wrote:
>
> NO! becasue 1000 is the same as 0. Er - that needs explaining! i.e. there
> are 500 numbers 0-499 inclusive and 500 numbers 500-999 inclusive. So the
> 1000 is the 0 in the next set.
>
> Regards, Kevin
>
> {Original Message removed}

2001\05\31@113952 by Bill Westfield

face picon face
   Now I'm wondering if it is mainly us old-timers
   from the dawn of pocket calculators age that were
   taught wrong?? Any thoughts, older people??

Hmm.  I'm 41, learned to use a slide rule in high school, and learned "5 or
more; round up" (and also "add .5 and truncate", on ASR33.)  Actually,
number lines were very popular in the "new math" cirricula, and it was/is
awfully clear that EXACTLY halfway is the only sticky point.  I'd never
heard of the odd/even rounding mentioned here before, but that makes sense.

BillW

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2001\05\31@114618 by Bob Barr

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Roman Black <RemoveMEfastvidspamspamBeGoneEZY.NET.AU> wrote:

<snip>

{Quote hidden}

Early Bronze Age recollections:

I was taught to go just one decimal beyond the digit to be rounded. Four or
less rounds down, 5 or more rounds up.

We were specifically taught that it doesn't matter what's beyond that first
digit. Whether the digit being rounded is odd or even was never mentioned as
being significant.


Bob
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2001\05\31@122418 by Kevin Blain

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000-499 round down
500-599 round up
1000-1499 round down
1500-1999 round up

etc. etc.

where is the uneven bias? where is the 'favouring rounding up?'

if you consider 1000 to be part of the 500-999 group (i.e. 0-4999 and
500-1000) then you are including the same number twice.

the scale is 0-9 in decimal, not 0-10

here's a scale

0,1,2,3,4   |  5,6,7,8,9

the | has an equal number of numbers on each side. The | does not go through
the number 5, it IMMEDIATLY precedes it, so if the scale had decimals in
too....

0.0 through 4.9 | 5.0 through 9.9

again, exactly the same number of numbers. The bar is infinately thin, and
therefore once the data is quantised, there is no possibility of the data
landing on the bar.

Regards, Kevin



{Original Message removed}

2001\05\31@122429 by Paul Hutchinson

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>Hmm.  I'm 41, learned to use a slide rule in high school, and learned "5 or
>more; round up"

Me too, on all three items :-).

However, when I got to chemistry class the teacher took extra care to teach
us the slightly more accurate way of rounding (odd/even).

Found this nice web tutorial on significant figures and rounding by, Stephen
L. Morgan, Professor, Department of Chemistry & Biochemistry, The University
of South Carolina.
http://www.chem.sc.edu/faculty/morgan/sigfigs/index.html


Paul

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2001\05\31@122840 by Roman Black

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David W. Gulley wrote:
{Quote hidden}

No, actually 4.99999999999 (repeating) is the
half-way point.
5.0 is definitely in the top half.
Decimal has ten digits, 0-4 are the bottom 5 and
5-9 are the top 5. Anything under 500 is the
bottom half, and anything from 5.0 up is the
top half. Nothing debatable about it, just like
0-127 is the bottom half and 128-255 are the
top half in binary. :o)
-Roman

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2001\05\31@123502 by Roman Black

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David W. Gulley wrote:
>
> To say "NO!" is a bit emphatic, since I could just as well say that we
> will only truncate numbers, which would provide an answer which may be
> fine for some conditions.
>
> I understand your logic; but this is still including 500 numbers below
> the "rounded to" value and 499 above, therefore (and I repeat)
> statistically, you are favoring the round up when given a random set of
> values.


I don't understand your logic?? There are 500 values,
ie, 000 to 499, then there are 500 values in the top
half, ie, 500 to 999. The dividing point is *immediately*
before 500, ie, 499.999999999999999 (forever)

So if it starts with a 5 it absolutely must be in the
top half. There is no favoring, just accuracy.
-Roman

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2001\05\31@124342 by Doug Joel

picon face
Does this really have anything to do with statistics?

I thought it was representing a more accurate number with
fewer digits.

Doug

{Original Message removed}

2001\05\31@133343 by Paul Hutchinson

flavicon
face
>No, actually 4.99999999999 (repeating) is the half-way point.
>5.0 is definitely in the top half.

4.99999999999 (repeating) = 5.00000000 (repeating)
Remember, repeating decimals are equal to an integer divided by 9.
1/9 = 0.1(repeating)
2/9 = 0.2(repeating)
3/9 = 0.3(repeating) = 1/3
4/9 = 0.4(repeating)
5/9 = 0.2(repeating)
6/9 = 0.6(repeating) = 2/3
7/9 = 0.7(repeating)
8/9 = 0.8(repeating)
9/9 = 0.9(repeating) = 1


Maybe this example will help illustrate the bias in >=5 rounding versus
odd/even rounding.

You have four numbers to sum, 1.5, 2.5, 3.5, 4.5
Using the decimal values:
1.5 + 2.5 + 3.5 + 4.5 = 12.

If we round using >=5 rounding:
2 + 3 + 4 + 5 = 14

If we round using odd/even rounding:
2 + 2 + 4 + 4 = 12

Is it any wonder that on Massachusetts State Income Taxes you must (by law)
use the >=5 rounding method, more revenue for the state :-).
Actually, because you use the same rounding for exemptions/deductions as for
income the revenue isn't any better for the state but, the simpler >=5
rounding is easier on the tax preparer.

BTW - The CINT() rounding function in all flavors of Microsoft Basic
languages uses odd/even rounding.

Paul

=========================================
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Chief Engineer
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2001\05\31@140430 by Roman Black

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face
Paul Hutchinson wrote:

> Maybe this example will help illustrate the bias in >=5 rounding versus
> odd/even rounding.
>
> You have four numbers to sum, 1.5, 2.5, 3.5, 4.5
> Using the decimal values:
> 1.5 + 2.5 + 3.5 + 4.5 = 12.
>
> If we round using >=5 rounding:
> 2 + 3 + 4 + 5 = 14


This is what I would expect. The numbers x.5 are
all on the TOP half of the decimal spread, so of
course the average or total of them will be greater
than the half way mark. 0.00 to 0.49 is in the first
half, 0.50 to 0.99 is in the second half.
Just like 0 to 127 is the first half and 128 to 255
is the second half in binary.

Where you might be confused is that 0.50 is not
the half way point between 0 and 1. The half way
point is AFTER 0.4999999 and BEFORE 0.5.
So 0.5 is past the half way mark.

:o)
-Roman

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2001\05\31@163350 by Paul Hutchinson

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>This is what I would expect. The numbers x.5 are
>all on the TOP half of the decimal spread, so of
>course the average or total of them will be greater
>than the half way mark. 0.00 to 0.49 is in the first
>half, 0.50 to 0.99 is in the second half.
>Just like 0 to 127 is the first half and 128 to 255
>is the second half in binary.
>
>Where you might be confused is that 0.50 is not
>the half way point between 0 and 1. The half way
>point is AFTER 0.4999999 and BEFORE 0.5.
>So 0.5 is past the half way mark.

Absolutely not true!

I think I see where the confusion is occurring. You are talking about
numbers limited to ranges of 2^x-1 (computer representations of numbers). In
the first paragraph you are talking about a range of 0.00 -> 0.99 (not
1.00).
In the second paragraph you change the range to 0 -> 1 but don't take into
account the change.

The mid way point of any range of numbers is exactly equal to:
(highest value - lowest value) / 2 + lowest value.

For the normal one byte range (0->255) this is:
(255 - 0) / 2 + 0 = 127.5

For the range of 0.00->0.99 it's:
(0.99 - 0.00) / 2 + 0.00 = 0.495

For the range of 0->1 it's:
(1 - 0) / 2 + 0 = 0.5

Here's all the possible 1 decimal place numbers in the range of 0 to 1.
They are: 0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9 and, 1.0.

There are 11 values and 0.5 is exactly the half way point of a 0 to 1
sequence. There are five numbers above and, five numbers below the mid way
point.

Paul

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2001\05\31@173739 by Bill Westfield

face picon face
   I don't understand your logic?? There are 500 values,
   ie, 000 to 499, then there are 500 values in the top
   half, ie, 500 to 999. The dividing point is *immediately*
   before 500, ie, 499.999999999999999 (forever)

   So if it starts with a 5 it absolutely must be in the
   top half. There is no favoring, just accuracy.

No, because we're not talking about "number of values" when rounding, we're
talking about a "distance."  There are (always) an infinite number of values
between any two numbers, but .5 is exactly half-way between 0 and 1, so
going consistantly either up or down introduces a slight bias.

Or, looking at all the values of precision N+1 that round to a particular
value, you'll find that there are 5 values that round up to that value, and
only 4 values that round down to it (plus the value itself.)

A trivial pieces of software ought to be able to calculate the accumulated
error over a large number of samples, which IS the value you want to
minimize.  Right?

BillW

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'[OT]: Brain Burp Rounding??'
2001\06\01@051140 by Roman Black
flavicon
face
William Chops Westfield wrote:
>
>     I don't understand your logic?? There are 500 values,
>     ie, 000 to 499, then there are 500 values in the top
>     half, ie, 500 to 999. The dividing point is *immediately*
>     before 500, ie, 499.999999999999999 (forever)
>
>     So if it starts with a 5 it absolutely must be in the
>     top half. There is no favoring, just accuracy.
>
> No, because we're not talking about "number of values" when rounding, we're
> talking about a "distance."  There are (always) an infinite number of values
> between any two numbers, but .5 is exactly half-way between 0 and 1, so
> going consistantly either up or down introduces a slight bias.


Alright, so you are saying it looks more like
this:

0 to 499 = bottom half (500 units)
500 = dead centre
501 to 999 = top half. (499 units)

This still makes no sense to me. You have a very
obvious bias, one half is bigger than the other.
We are not working with distances, instead there
are a very finite number of possible combinations
in our 3 digits. 0 to 999. There are 1000 total
possible numerical combinations.

1000 can't be included as it is the first combination
of the next 1000 units. And 2000 is the first of the
next block after that.

If you don't believe me write a program that generates
a 3-digit random number and then round it... 500 is
past half way and must be rounded up. :o)
-Roman

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2001\06\01@063639 by Bob Ammerman

picon face
Roman,

This 'distance' thing really is the way to look at it.

Since n+0.500 is right in the middle between 'n' and 'n+1', the error
introduced is equal (but of opposite sign), whether you go down to 'n' or up
to 'n+1'. So, to have a statistically average error of zero you just go up
half the time and down half the time (ie: round to even).

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)



{Original Message removed}

2001\06\01@064942 by Kevin Blain

flavicon
face
No it doesn't because 9/9 == 1, not 0.9(recurring) (and 5/9 is not 0.2
recurring, but I guess that was a typo!)

4.99999999999 (repeating) is not equal to 5.00000000 (repeating), that's why
we have different numbers to express it. However, 4.99999999999 (repeating)
is nearly equal to 5

This reminds me of the thing where :-

- There is 1 cake
- Each person takes half of the amount of cake that is left, indefinatly

Therefore there is always some cake left......

Regards, Kevin



{Quote hidden}

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2001\06\01@070637 by Kevin Blain

flavicon
face
but n+1 is in the next set. the range should be from n to
n+0.999999999999(recurring)

example.

Here are 10 numbers ( fairly equally distributed between 0 and 1, i.e they
are not biased toward either end) , round them to the nearet integer, then
take the average. Also take the average of the numbers - they should be the
same if the rounding method works


0.05  0.15  0.25  0.35  0.45  0.55  0.65  0.75  0.85  0.95

rounded intergers are: 0  0  0  0  0  1  1  1  1  1 (my method)
average of integers is 0.5

average of actual numbers is 0.5

Regards, Kevin



{Original Message removed}

2001\06\01@070653 by Alan B. Pearce

face picon face
My brain burps from all this rounding... It just seems to go round and round
and round and round.... are we having fun yet? that would make it a merry go
round :)

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2001\06\01@072306 by michael brown

flavicon
face
> Is it any wonder that on Massachusetts State Income Taxes you must (by
law)
> use the >=5 rounding method, more revenue for the state :-).
> Actually, because you use the same rounding for exemptions/deductions as
for
> income the revenue isn't any better for the state but, the simpler >=5
> rounding is easier on the tax preparer.

I was wondering when this was going to be brought up.  I can only imagine
trying to explain to the IRS at an audit that you were only rounding the
"other" way in the interest of national accuracy.  BTW if accuracy is so
important that even/odd rounding is needed, then the (IMHO) sensible way to
deal with the math is to carry more digits of precision than required and
then (and only then) round the final total.  IIRC (its been a long time)
this is the way the COBOL compiler did math, I suspect other compilers do
the same.

michael

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2001\06\01@082303 by Thomas McGahee

flavicon
face
Sorry, but .99999 repeating is EXACTLY equal to 1.00000

They are just two different ways of EXPRESSING the same
number. Don't believe me? 1/9=.11111 repeating. 2/9=.22222 repeating.
3/9=.3333333 repeating. 4/9=.4444 repeating, 5/9=.55555 repeating.
6/9=.6666 repeating. 7/9=.7777 repeating. 8/8=.888 repeating.
So 9/9=.99999 repeating. The "error" between .999 repeating and
1.000 (repeating, by the way!) is .0000 repeating. Get it?

.999 repeating and 1.000 (repeating the 000 part) are simply two
different ways of EXPRESSING the value. They may LOOK different,
but mathematically the difference is ZERO.

.999 repeating, 9/9, x/x (where x is non-zero), 1,  1.000 repeating
all REPRESENT the same VALUE. One.

Fr. Tom McGahee

{Original Message removed}

2001\06\01@092542 by Paul Hutchinson

flavicon
face
>No it doesn't because 9/9 == 1, not 0.9(recurring)

You missed the point, I was saying that 0.9(repeating) is absolutely equal
to 1. This is a basic principal of mathematics that I think was taught to me
in junior high school.

Here is the algebraic proof:

Given:
 x = 0.9(repeating)
Multiply both sides by 10:
 10x = 9.9(repeating)
Subtract x from both sides:
 10x - x = 9.9(repeating) - x
Substitute value for x on right side:
 10x - x = 9.9(repeating) - 0.9(repeating)
Simplify right side:
 10x - x = 9
Simplify left side:
 9x = 9
Divide both sides by 9:
 x = 1
Substitute value for x on left side:
0.9(repeating) = 1


>(and 5/9 is not 0.2 recurring, but I guess that was a typo!)

Yep, a typo should have been 0.5(repeating)

>4.99999999999 (repeating) is not equal to 5.00000000 (repeating), that's
why
>we have different numbers to express it. However, 4.99999999999 (repeating)
>is nearly equal to 5

As above 4.99999999999 (repeating) is exactly equal to 5.

Paul

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2001\06\01@093431 by Alan B. Pearce

face picon face
>Given:
>  x = 0.9(repeating)
>Multiply both sides by 10:
>  10x = 9.9(repeating)
>Subtract x from both sides:

The fallacy in this argument is that at the end x->1, as any repeating
number is not exact.

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2001\06\01@110639 by Roman Black

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face
Alan B. Pearce wrote:
>
> My brain burps from all this rounding... It just seems to go round and round
> and round and round.... are we having fun yet? that would make it a merry go
> round:)


Ha ha!!! That has got to be a 3 bourbon answer if I
ever heard one. I am feeling the same way. I never
thought rounding a simple decimal number would spark so
much controversy...
:o)
-Roman

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2001\06\01@112711 by Doug Joel

picon face
The there is .00001 difference between .99999 and
1.00000....



----- Original Message -----
From: Thomas McGahee <RemoveMEtom_mcgahee@spam@spamspamBeGoneSIGMAIS.COM>
To: <.....PICLIST@spam@spamEraseMEMITVMA.MIT.EDU>
Sent: Friday, June 01, 2001 5:24 AM
Subject: Re: [OT]: Brain Burp Rounding??


> Sorry, but .99999 repeating is EXACTLY equal to 1.00000
>
> They are just two different ways of EXPRESSING the same
> number. Don't believe me? 1/9=.11111 repeating. 2/9=.22222
repeating.
> 3/9=.3333333 repeating. 4/9=.4444 repeating, 5/9=.55555
repeating.
> 6/9=.6666 repeating. 7/9=.7777 repeating. 8/8=.888
repeating.
> So 9/9=.99999 repeating. The "error" between .999
repeating and
> 1.000 (repeating, by the way!) is .0000 repeating. Get it?
>
> .999 repeating and 1.000 (repeating the 000 part) are
simply two
> different ways of EXPRESSING the value. They may LOOK
different,
> but mathematically the difference is ZERO.
>
> .999 repeating, 9/9, x/x (where x is non-zero), 1,  1.000
repeating
{Quote hidden}

5/9 is not 0.2
> > recurring, but I guess that was a typo!)
> >
> > 4.99999999999 (repeating) is not equal to 5.00000000
(repeating), that's
> why
> > we have different numbers to express it. However,
4.99999999999
> (repeating)
> > is nearly equal to 5
> >
> > This reminds me of the thing where :-
> >
> > - There is 1 cake
> > - Each person takes half of the amount of cake that is
left, indefinatly
> >
> > Therefore there is always some cake left......
> >
> > Regards, Kevin
> >
> >
> >
> > > 4.99999999999 (repeating) = 5.00000000 (repeating)
> > > Remember, repeating decimals are equal to an integer
divided by 9.
{Quote hidden}

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2001\06\01@114644 by Roman Black

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face
OK, so now I feel totally inadequate to do rounding
in my simple data table.

As everyone seems to have their own system for rounding
I would like to ask a simple question andd hopefully one
of the math geniuses can give a simple answer??

I do a divide calc (on pocket calculator) that gives a
0.xxxxx answer. I need to round this and make it a binary
number (to fit in one byte).

So 0.400 x 256 = 102.4 = 102 (binary).

Any suggestions for rounding these data figures??
:o)
-Roman

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2001\06\01@114846 by Bob Barr

picon face
Kevin Blain <kevinbEraseMEspam@spam@WOODANDDOUGLAS.CO.UK> wrote:

>Subject: Re: [OT]: Brain Burp Rounding??
>Date: Fri, 1 Jun 2001 11:44:44 +0100
>

<snip>

>
>This reminds me of the thing where :-
>
>- There is 1 cake
>- Each person takes half of the amount of cake that is left, indefinatly
>
>Therefore there is always some cake left......
>

A similar story is told of a mathematician and an engineer presented with a
problem concerning a beautiful woman located 10 feet away. Each is told that
he can halve the distance to her every 10 minutes and asked how long it will
be until he reaches her.

The mathematician states that he will never be able to reach her.

The engineer figures that, within the hour, it'll be close enough for him.
:=)

_________________________________________________________________
Get your FREE download of MSN Explorer at http://explorer.msn.com

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2001\06\01@115058 by Dipperstein, Michael

face picon face
> From: Alan B. Pearce [RemoveMEA.B.PearcespamspamBeGoneRL.AC.UK]
>
> >Given:
> >  x = 0.9(repeating)
> >Multiply both sides by 10:
> >  10x = 9.9(repeating)
> >Subtract x from both sides:
>
> The fallacy in this argument is that at the end x->1, as any repeating
> number is not exact.

I don't understand where the fallacy lies.  A repeating number *is* exact as
long as it isn't truncated.

1/3 is exactly .3 (repeating)

1/3 is not exactly .3333333 or .3333333333 or any truncated version.

Along that line here's another simple Junior High School algebra proof that 1 =
.9 repeating.

1/3 = .3 repeating

multiply both sides by 3

(1/3)*3 = 3 * (.3 repeating)

1 = .9 repeating

-Mike

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2001\06\01@121653 by Paul Hutchinson

flavicon
face
> The fallacy in this argument is that at the end x->1, as any repeating
> number is not exact.

Wrong, all repeating fractions have an exact representation (they are
rational numbers). Only irrational numbers like PI have no exact
representation.

<rant on>
I consider, a complete understanding of rounding and number systems (very
basic algebra) to be absolutely essential for good engineering. In my
opinion, anyone involved in engineering of any sort should understand basic
algebra completely. Now, I'm not saying that all engineers must have this
info memorized but, they must have a gut feel for it and know where to look
up the precise info (e.g. their old text books).
<rant off>

In an attempt to help members learn the basics of algebra I searched the web
for some good references. However most of what I found was written for
children 13 years and younger (the authors expect the student to accept some
math principals on faith and, we all know that good engineers never accept
anything on faith :-).

I did find that James Brennan, professor of physics and mathematics at Boise
State University in the College of Applied Technology, has posted the
complete contents of his book "Understanding Algebra" online at:
http://www.edteach.com/algebra/
Read the "Decimals" section of Chapter 1 for info on rounding.
BTW - the subject of repeating fractions is also covered in that section.

For those who may have missed it the first time I posted it, here's a nice
web tutorial on significant figures and rounding by, Stephen
L. Morgan, Professor, Department of Chemistry & Biochemistry, The University
of South Carolina.
http://www.chem.sc.edu/faculty/morgan/sigfigs/index.html

Paul

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2001\06\01@122822 by Wendy J Olend

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face
It seems that we have broken into two camps, the "Always Round Up-ers"
(ARU)  and the "Round to Even Crew" (REC).  Being a member of the latter
camp, I'm going to try my hand at explaining our stance.

I've been watching everyone go back and forth, arguing if 5 is the
mid-point between 0 and 9 (REC) or if it's part of the upper half of the
set (ARU).  Of course it is part of the upper half of the set
{0,1,2,3,4,5,6,7,8,9}.  The confusion comes in at this point:  The
round-to-even method is meant to minimize rounding error.  When 0 is the
digit under consideration, there IS no rounding error introduced.  Before
the flames head my way, let me elaborate.

We all agree on this point:  when you decide to round a number you first
decide how many digits you want in the final answer, then you throw away
all but one plus that number of digits:

Example:   13.1032
If I decide that I want to report my answer with only one decimal place.  I
truncate, leaving 13.10  Now we have to ask, "What does this value mean?"
Putting this value into words, it means "I have exactly 13.1, and it what
appears to be 0 in the hundredths place."  When I "round" this number to
13.1, I have not introduced any further error, merely reduced the
precision.

Example:   13.1132 to the tenths place, I start with 13.11, which means "I
have exactly 13.1, and what appears to be 1 in the hundredths place."  When
I truncate this value, I have reported the value as less than I have
observed it to be.  Hence the "rounding error".  The same happens when you
round up.

Looking at the error inducing numbers: {1,2,3,4,5,6,7,8,9}, it becomes
obvious why we call 5 the mid-point of this set.  If we always round up at
five, we are putting a positive biasing on the data, because we are
rounding up more than half of the time.

The confusion occurs because for 0-4 it looks like you are doing the same
operation; truncation.  But for 0, you are actually just recasting the
value at a lower precision.

There's my $.02 worth,
Wendy

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2001\06\01@125249 by Alan B. Pearce

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>1 = .9 repeating

Except that the error is 0.0_repeating_1.

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2001\06\01@131046 by John Pfaff

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face
> >1 = .9 repeating
>
> Except that the error is 0.0_repeating_1.
which is exactly equal to zero.

For those who may have missed it:

1/9 = 0.111(repeating)
2/9 = 0.222(repeating)
...
8/9 = 0.888(repeating)
9/9 = 0.999(repeating) which is EXACTLY EQUAL TO ONE.

A math teacher explained this to me years ago (no Chris, not Mr. Zurick).

Also, I'm in the Round To Even crowd.  This method was explained in a college chemistry text and was to be used because it
statistically caused the least errors.

jp

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2001\06\01@132950 by Paul Hutchinson

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You got the rounding correct, by anyone's method :-).

But, if the input value can be > 0.99804 it will overflow the 8 bits.

To make it work for any value < 1 you should multiply by 255.

Paul

{Quote hidden}

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2001\06\01@170404 by Bill Westfield

face picon face
   <rant on>
   I consider, a complete understanding of rounding and number systems
   (very basic algebra) to be absolutely essential for good engineering. In
   my opinion, anyone involved in engineering of any sort should understand
   basic algebra completely. Now, I'm not saying that all engineers must
   have this info memorized but, they must have a gut feel for it and know
   where to look up the precise info (e.g. their old text books).
   <rant off>

To tie things in with previous discussions, I suspect that this sort of
thing is one of the differences between "degreed" engineers and "self
taught" engineers.  While repeating decimals are theoretically high-school
or earlier math, the sort of "gut feel" for non-intuitive values of limits
and such might be something that one gets drummed into them during the two
years of calculus and physics one endures as "perparation" to using Smith
charts and phasors for EE (not to mention that numerical analysis class.)

There'd be exceptions, of course.

OTOH, in software engineering, I've been rather shocked at how seldom
floating point ("real" numbers) actually gets used...

Is there a simple algorithm for doing the odd/even rounding (equivilent to
N+0.5 and truncate?)

(of course, we're now in big arguments over fine points.  Whether you
always round .5 up, or do te even/odd thing, your error introduced over
typical sample sizes is going to be pretty small.)

BillW

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2001\06\01@191009 by Andrew Warren

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William Chops Westfield <spamBeGonebillwKILLspamspam@spam@cisco.com> wrote:

> Is there a simple algorithm for doing the odd/even rounding
> (equivilent to N+0.5 and truncate?)

Bill:

There's a simple implementation in PIC assembly:

   ; Enter with integer in REG, fraction in the Carry flag.

       BNC     DONE    ;If the carry's clear, no rounding
                       ;is required, so jump.

       BTFSS   REG,0   ;If the integer is odd, skip ahead.
       INCF    REG     ;Otherwise, round up (i.e., round
                       ;to odd).

   DONE:

This implementation uses "round to odd" rather than "round to even"
in order to make the rounding of multi-byte integers easier; with
this method, only the least-significant byte is ever changed.

This introduces a tiny bit of bias, however, since only ONE value
(0.0) rounds to 0 while THREE values (254.5, 255.0, and 255.5) round
to 255.

If it were important for numbers close to 0 to round down to 0, you
could reverse things so that the bias was toward 0 rather than 255...
But then you'd non-intuitively be rounding 1.0 to 0.

-Andy


=== Andrew Warren --- aiwspam_OUTspam@spam@cypress.com
=== IPD Systems Engineering, CYSD
=== Cypress Semiconductor Corporation
===
=== Opinions expressed above do not
=== necessarily represent those of
=== Cypress Semiconductor Corporation

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2001\06\01@193126 by Alexandre Domingos F. Souza

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>I only used an RPN calc for a few days but I got hooked imediatly,
>those things are great, think I used an HP 32 II or something like that.
>I normally use an TI83 though. I like the RPN calc much better.

       I never liked RPN calculators. Had a HP20 and it was algebraic. Xchanged for a HP48G and it is RPN. Completely crazy, but once you got used to, you start to search for the "enter" key on other calculators :o)

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2001\06\01@201618 by Paul Hutchinson

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> To tie things in with previous discussions, I suspect that this sort of
> thing is one of the differences between "degreed" engineers and "self
> taught" engineers.  While repeating decimals are theoretically high-school
> or earlier math, the sort of "gut feel" for non-intuitive values of limits
> and such might be something that one gets drummed into them during the two
> years of calculus and physics one endures as "perparation" to using Smith
> charts and phasors for EE (not to mention that numerical analysis class.)
>
> There'd be exceptions, of course.

I guess, I'm an exception. I don't personally know any other self taught
engineers so I can't say how it stacks up but, my gut instinct is that you
are right Bill. I do know two degreed engineers who don't have a good feel
for mathematics, both close personal friends, and they both got out of
design engineering (one does engineering project management and the other
cranks out asp web code).

Paul

=========================================
Paul Hutchinson
Chief Engineer
Maximum Inc., 30 Samuel Barnet Blvd.
New Bedford, MA 02745
spamBeGonephutchinson@spam@spamimtra.com
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=========================================

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2001\06\01@225753 by Kyle Stemen

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One thing I don't like about the round to even method is that you lose
precision. Say you are measuring a value from 0 to 9. You have one sig
fig, and 10 different possiblities of what that number can be. Now lets
say that you had to do a calculation such as add .5. You would have to
round back to one digit. Here are the different possiblities with round
to even:
0.5 -> 0
1.5 -> 2
2.5 -> 2
3.5 -> 4
4.5 -> 4
5.5 -> 6
6.5 -> 6
7.5 -> 8
8.5 -> 8
9.5 -> 10

with always round up you have

0.5 -> 1
1.5 -> 2
2.5 -> 3
3.5 -> 4
4.5 -> 5
5.5 -> 6
6.5 -> 7
7.5 -> 8
8.5 -> 9
9.5 -> 10

With the round to even you can only have an even number so there are 6
possibilites of what it can be, and you started out with 10 so you have
lost precision. With always round up there are still 10 possibilies so
you have lost no precision, but they do not average out the same so you
have lost accuracy.

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2001\06\02@053746 by Roman Black

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Paul Hutchinson wrote:

{Quote hidden}

Thanks Paul, you are absolutely right and I stand
humbled. I should indeed have a thorough understanding
of this.

Unfortunately the last time I sat in a classroom
doing calculus was over 15 years ago, and the amount
of times over the years I have actually needed formula
more intense than the usual freq/reactance etc stuff
has been pretty minimal.

Thanks for the nice links, I will check them out
this weekend. :o)
-Roman

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2001\06\02@055045 by Roman Black

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Bob Barr wrote:

> A similar story is told of a mathematician and an engineer presented with a
> problem concerning a beautiful woman located 10 feet away. Each is told that
> he can halve the distance to her every 10 minutes and asked how long it will
> be until he reaches her.
>
> The mathematician states that he will never be able to reach her.
>
> The engineer figures that, within the hour, it'll be close enough for him.
>:=)


Ha ha ha!! <cheeky> Hey, for some of us, 30 minutes
would be close enough... Think about it. ;o)
</cheeky>
-Roman

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2001\06\02@074137 by David W. Gulley

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Kyle Stemen wrote:
{Quote hidden}

Actually, you lose no precision or accuracy, if your premise is that the
x.5 values are +/- 0.5 in accuracy to start with.
 To demonstrate this take the average of the groups above:
  0.5 + 1.5 + ... 9.5 = 50/10 = 5   (the list of numbers)
  0 + 2 + 2 + ... 10  = 50/10 = 5   (Rounded to Even)
  1 + 2 + 3 + ... 10  = 55/10 = 6   (Rounded Up)

Rounding by adding 0.5 to x.5 results in a number that is different from
the original by the same amount that subtracting 0.5 from x.5 would
yield.

In other words, rounding 0.5 to 0 adds no more imprecision than rounding
to 1.

Having now stated that Round to Even does not increase the imprecision,
I must also state that neither does it does it increase the precision.
What Round to Even accomplishes is most valuable when accumulating a set
of values, since it tends to center the accumulated result in the range
of accuracy, while the Round Up method approaches the upper bound of the
accuracy range, and performing a truncate forces the result to the lower
bound of the accuracy range.

The original question in this thread was (essentially):
 "How do you round a number?"

When I encounter a task that requires rounding, I first consider what
the appropriate method might be and then implement the most reasonable
solution.
 For some jobs, I might just truncate; for others add 0.5 and truncate
(which is Round Up); but, if the rounded values are being accumulated I
(may) use the Round to Even methodology.

But, if you ask me "how do you round a number," my reply is round up if
greater than 0.5, round down if less than 0.5 and if exactly x.5, round
up if previous digit odd, round down if previous digit even.


David W. Gulley
Destiny Designs

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2001\06\02@101114 by michael brown

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> I guess, I'm an exception. I don't personally know any other self taught
> engineers so I can't say how it stacks up but, my gut instinct is that you
> are right Bill. I do know two degreed engineers who don't have a good feel
> for mathematics, both close personal friends, and they both got out of
> design engineering (one does engineering project management and the other
> cranks out asp web code).
>
> Paul
You know me, sort of anyway!  Although, I wouldn't rise to call my self an
"engineer", just a guy that loves to play with electricty, computers and
math.  I like to learn everything the "hard way", hopefully it will stick
better that way.  The guy cranking out the asp code is making some bucks,
isn't he?  I still think you should round at the end and not all along the
way. ;-D

michael

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2001\06\02@121914 by Thomas McGahee

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Sigh.... I did not say .99999 I said .99999 REPEATING.
As in **forever**.

Do not get confused between the VALUE of a number and its
representation. Whether you like it or not, mathematically the mumber
1 exactly equals .9 repeating. This is not a matter of personal
opinion. It is simple mathematical fact.

Fr. Tom McGahee

{Original Message removed}

2001\06\02@145026 by Chris Carr

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> Do not get confused between the VALUE of a number and its
> representation. Whether you like it or not, mathematically the mumber
> 1 exactly equals .9 repeating. This is not a matter of personal
> opinion. It is simple mathematical fact.
>
Now if I remember my maths correctly 0.9 repeating to infinity will
equal 1........but..........you can never reach infinity, you can only
"Approach Infinity".  So whilst 0.9 repeating will get so mind boggling
close to 1
that an amoeba couldn't push his reproductive organ in the gap it will
never actually reach the point where it will equal 1.

Chris Carr

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2001\06\02@152744 by Thomas McGahee

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I really do not want to keep this discussion about .9 repeating
being exactly equal to 1 going and going and going, but as
an educator I really want to make sure that at least some of the
PIC list members "get it".

9/9 is a RATIONAL number. It can be **represented** in decimal
notation in (at least) two ways that LOOK different but have the same
exact VALUE. These are 1.000 (with the zero repeating forever),
and .999 with the 9 repeating forever. The mathematical difference
between the two is .000 with the zero repeating forever.

This concept is difficult for some people to grasp at first,
because .9999 (9 repeating) just LOOKS smaller than 1.0 (0 repeating).

Looks can be VERY deceiving.
777/777 may **look** bigger than .001/.001 to some
people, but believe me the VALUE is the same. 1.

Fr. Tom McGahee

{Original Message removed}

2001\06\02@184812 by Bill Westfield

face picon face
> Now if I remember my maths correctly 0.9 repeating to infinity will equal
> 1........but..........you can never reach infinity, you can only "Approach
> Infinity".  So whilst 0.9 repeating will get so mind boggling close to 1
> that an amoeba couldn't push his reproductive organ in the gap it will
> never actually reach the point where it will equal 1.

You don't have to "reach" infinity.  You SAID that it repeats to infinity,
so it DOES.  It doesn't matter that you can't write it down...

BTW, the "half way" problem is equivilent to 0.1111 repeating in BINARY,
and that's 1.0 too.  The mathematician shouldn't have had a problem.  (The
Philosipher, on the other hand...)

And amoebas don't have reproductive organs...

BillW

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2001\06\02@204500 by michael brown

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---- Original Message -----
From: "Thomas McGahee" <RemoveMEtom_mcgaheeEraseMEspamKILLspamSIGMAIS.COM>
To: <spamBeGonePICLISTspam_OUTspamRemoveMEMITVMA.MIT.EDU>
Sent: Saturday, June 02, 2001 11:19 AM
Subject: Re: [OT]: Brain Burp Rounding??


> Sigh.... I did not say .99999 I said .99999 REPEATING.
> As in **forever**.
>
> Do not get confused between the VALUE of a number and its
> representation. Whether you like it or not, mathematically the mumber
> 1 exactly equals .9 repeating. This is not a matter of personal
> opinion. It is simple mathematical fact.
>
> Fr. Tom McGahee

Yeah, thats the ticket.......  And infinity < 2*infinity.........


>
> {Original Message removed}

2001\06\02@212246 by michael brown

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----- Original Message -----
From: "Chris Carr" <.....nyedspamRemoveMEBTINTERNET.COM>
To: <PICLISTspam@spam@MITVMA.MIT.EDU>
Sent: Saturday, June 02, 2001 1:47 PM
Subject: Re: [OT]: Brain Burp Rounding??


{Quote hidden}

Precisely! The fact that the number is "repeating" implies infinity as a
piece of the formula.

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2001\06\02@212716 by michael brown

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----- Original Message -----
From: "Thomas McGahee" <EraseMEtom_mcgaheeRemoveMEspamSTOPspamSIGMAIS.COM>
To: <RemoveMEPICLISTKILLspamspamTakeThisOuTMITVMA.MIT.EDU>
Sent: Saturday, June 02, 2001 2:30 PM
Subject: Re: [OT]: Brain Burp Rounding??


{Quote hidden}

Don't forget the 1 at position 'infinity + 1'

>
> This concept is difficult for some people to grasp at first,
> because .9999 (9 repeating) just LOOKS smaller than 1.0 (0 repeating).

It is.

>
> Looks can be VERY deceiving.
> 777/777 may **look** bigger than .001/.001 to some
> people, but believe me the VALUE is the same. 1.

In this I have no problem.

{Quote hidden}

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2001\06\02@222346 by David VanHorn

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At 07:39 PM 6/2/01 -0500, michael brown wrote:
>---- Original Message -----
>From: "Thomas McGahee" <tom_mcgaheespamspamSIGMAIS.COM>
>To: <spam_OUTPICLISTspam_OUTspamspam_OUTMITVMA.MIT.EDU>
>Sent: Saturday, June 02, 2001 11:19 AM
>Subject: Re: [OT]: Brain Burp Rounding??
>
>
> > Sigh.... I did not say .99999 I said .99999 REPEATING.
> > As in **forever**.
> >
> > Do not get confused between the VALUE of a number and its
> > representation. Whether you like it or not, mathematically the mumber
> > 1 exactly equals .9 repeating.

hold on jack..

That's as nonsensical as saying that 2=3 for large values of 2.

I have a problem with that.

0.99(followed by any finite or infinite number of nines) is by definition,
not equal to 1.0

The value of 0.99(inf..) is less than 1.0 by an infinitely small (but
non-zero) amount.




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2001\06\02@223902 by Bob Ammerman

picon face
Sorry Dave,

He's right and you are wrong.

The value .9 repeating is the same as the infinite series:

9/10 + 9/100 + 9/1000 + 9/10000 + ...

Which sums to 1.

Remember --- infinite is infinite. If you carry the series out infinitely
there will be _no_ 'residual'.

We have the same thing in binary where

.1 repeating (binary) is the same as the infinite series:

1/2 + 1/4 + 1/8 + 1/16 + ...

Which again, after summing an infinite number of terms, comes to exactly 1.

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

{Original Message removed}

2001\06\02@224305 by Jinx

face picon face
> .1 repeating (binary) is the same as the infinite series:
>
> 1/2 + 1/4 + 1/8 + 1/16 + ...
>
> Which again, after summing an infinite number of terms, comes to
> exactly 1.

Wouldn't it come infinitely close to 1 ?

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2001\06\02@225948 by David VanHorn

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At 10:35 PM 6/2/01 -0400, Bob Ammerman wrote:
>Sorry Dave,
>
>He's right and you are wrong.
>
>The value .9 repeating is the same as the infinite series:
>
>9/10 + 9/100 + 9/1000 + 9/10000 + ...
>
>Which sums to 1.

It sums to almost one. The degree of "almost" is a function of your patience.
As you expand the sequence, the difference becomes small, but it is never zero.

You claim 1 = 0.9...
Is it then true that 1(inf) = 0.9...(inf)?

There may be a rule which says so (Which I suspect exists to avoid ugly
problems in calculation), but you're not going to convince me that a number
which is "less than 1 by an infinitely small amount" is equal to 1.


Next, you'll want to sell me 5 pounds of potatoes multiplied by the Sqrt of
-1 :)


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2001\06\03@003649 by James Newton. Admin 3

face picon face
"...by an infinitely small (but non-zero) amount...."

To paraphrase a popular space opera "Never underestimate the power of the
infinity." Infinity is a mind-numbingly large (or small) thing. A totally
incomprehensible value. Anyone who can actually understand it's magnitude,
would be insane. It is one of the many things in life that one must not try
to completely investigate but rather to, at some point, simply accept.

I always enjoyed a story that a teacher of mine told about a class of
juvenal delinquents that he once enjoyed and tried to inspire. He told them
that they could not walk from one end of a class room to the other. He felt
that in a class of "normal" kids, the students would have sat, smiling,
waiting for the teacher to go on, but in this class one of the kids popped
up, walked to the back wall, then to the front then said "shows what you
know, sucker" and sat down to the applause of his classmates. The teacher
then went on to explain that in order to walk from one end to the other, you
first must walk half way. But then to complete the task, you have to walk
half the remaining distance, and so on (here it comes) * infinitely * so
that you never actually reach the other end, because you always have half
way left to go. The kids thought about it and then one said, "that's bull
shit teach, I know can still walk all the way" and  another said "no man,
he's right, you can get that close, but you actually never get there!" And
before the shanks came out, the teacher said "Does it really make a
difference? Can you get close enough that it doesn't matter? If you just
keep walking WILL YOU STILL BREAK YOUR NOSE!!!???" and the kids were willing
to accept that it probably isn't a good thing to understand that you can't
actually get from one wall to another.

Anyway, the point is, if you can believe that you can walk from one end of a
room to the other, then you can believe that infinitely small IS zero and
that 4.999 repeating is 5 and that you need to round even to avoid bias.

In fact, this is the main difference between engineers and technicians. The
difference between things making sense and things working.

James Newton, PICList Admin #3
jamesnewtonspam_OUTspampiclist.com
1-619-652-0593 phone
http://www.piclist.com


{Original Message removed}

2001\06\03@003717 by Bob Ammerman

picon face
----- Original Message -----
From: "Jinx" <RemoveMEjoecolquittKILLspamspam@spam@CLEAR.NET.NZ>
To: <PICLISTspamBeGonespam.....MITVMA.MIT.EDU>
Sent: Saturday, June 02, 2001 10:42 PM
Subject: Re: [OT]: Brain Burp Rounding??


> > .1 repeating (binary) is the same as the infinite series:
> >
> > 1/2 + 1/4 + 1/8 + 1/16 + ...
> >
> > Which again, after summing an infinite number of terms, comes to
> > exactly 1.
>
> Wouldn't it come infinitely close to 1 ?

Not after an _infinite_ number of terms.

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

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2001\06\03@003725 by Bob Ammerman

picon face
Dave,

The point is that it does _not_ take patience.

.9 repeating is already an _infinite_ series. Not just a "sorta infinite as
long as you want to go on" series.

Its value is exactly 1.

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

{Original Message removed}

2001\06\03@005515 by David VanHorn

flavicon
face
>
>Anyway, the point is, if you can believe that you can walk from one end of a
>room to the other, then you can believe that infinitely small IS zero and
>that 4.999 repeating is 5 and that you need to round even to avoid bias.

That's Zeno's Paradox, and the problem is that they kept decreasing the
time intervals over which the motion was made. (Half the distance, in half
the time) The paradox points out only that you can't really make those
measurements, not that you can't get there.
Indeed, you'll MUCH sooner cross the quantum uncertanity limit, where you
can't say where you are, and how fast you're moving.
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2001\06\03@010945 by David VanHorn

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At 12:18 AM 6/3/01 -0400, Bob Ammerman wrote:
>Dave,
>
>The point is that it does _not_ take patience.
>
>.9 repeating is already an _infinite_ series. Not just a "sorta infinite as
>long as you want to go on" series.

At each point, it evaluates to <1. Repeating it in smaller steps forever
doesn't change that.
Show me the point at which it changes, and evaluates to 1.
I know, it's infinitely far away, and it can never be reached, but if you
wish real hard...

>Its value is exactly 1.

Can you show any physical thing that requires this to be true?


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2001\06\03@015012 by David VanHorn

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>
>Its value is exactly 1.

So I did a little digging, and found this link.
http://www.maths.abdn.ac.uk/~igc/tch/ma1002/appl/node57.html

However, all that is said about a converging series is that you can
determine a value that it approaches.  I see no requirement or statement
that it REACHES that value.

Here also,
http://forum.swarthmore.edu/dr.math/problems/may7.8.98.html

S = 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/2^n + ...

This series is described as convergent (obvious) and approaching 1, but not
described as being  equal to 1

Here:
http://www.misd.wednet.edu/~kim_schjelderup/Integrated%203/Pages/Seq&Series/4.7l%20Inifinite%20Series%20(WP).pdf

Finally the statement:
If the sequence of partial sums of an infinite series has a limit, then
that limit is the sum of the series.

Looks to me like we are defining "sum of the series" as something special,
and we are not saying that the series is equal to the limit, in a manner
similar to the way that "spin" is used in quantum mechanics.

It's certainly useful in calculation, because it causes those awkward
infinities to dissapear, by ignoring the infinitely tiny difference between
the actual result, and the defined result.

Inverting Zeno's paradox, the fallacy is that a finite distance (or number)
does not become infinite, simply because it can be divided into an infinte
number of smaller distances (or numbers)




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2001\06\03@024811 by michael brown

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----- Original Message -----
From: "David VanHorn" <KILLspamdvanhornspam.....CEDAR.NET>
To: <spam_OUTPICLISTspamKILLspamMITVMA.MIT.EDU>
Sent: Saturday, June 02, 2001 9:57 PM
Subject: Re: [OT]: Brain Burp Rounding??


{Quote hidden}

patience.
> As you expand the sequence, the difference becomes small, but it is never
zero.
>
> You claim 1 = 0.9...
> Is it then true that 1(inf) = 0.9...(inf)?
>
> There may be a rule which says so (Which I suspect exists to avoid ugly
> problems in calculation), but you're not going to convince me that a
number
> which is "less than 1 by an infinitely small amount" is equal to 1.
>
>
> Next, you'll want to sell me 5 pounds of potatoes multiplied by the Sqrt
of
> -1 :)
I'm with Dave on this one.  Sqrt of -1 = i    "i" - that amazingly
ridiculous, non-existant value that allows mathematicians to continue on
creating even more ridiculous mathematics coming up with answers that don't
(can't) exist.  Sort of like "e" the fascinating irational number that
somehow ocupies an exact position on a number line.  Where as pi is a
supposedly "irrational" number easily represented as a fraction (which by
definition makes it rational) as in pi = C/D  where C is some number
representing circumfrence and D representing diameter of a circle.  You
cannot use infinity in an equation and expect any sanity in the result.
However, we are supposed to accept these things and then also believe that
you cannot divide by 0 (answer is undefined even though the real answer
should be infinity).  Since division is nothing but repetive subtraction,
you can easily prove (using a pic ;-D ) that dividing by 0 is infinite and
not "undefined".  Look Dave its true 1=2 ;-)

    a = x            [true for some a's and x's]
  a+a = a+x          [add a to both sides]
   2a = a+x          [a+a = 2a]
2a-2x = a+x-2x       [subtract 2x from both sides]
2(a-x) = a+x-2x       [Factor left side]
2(a-x) = a-x          [Simplify right side]
    2 = 1            [divide both sides by a-x]

Of course we can't divide by zero, which is precisely what happens in the
last step.  Back to the .999 rpt.=1 arithmetic with infinity is NOT allowed,
because infinity is not a number. And, just like our little puzzle, we get
answers that make no sense.

>

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2001\06\03@035319 by Chris Carr

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>
> And amoebas don't have reproductive organs...
>
Are you sure ?

According to the arguments going on here they must have.
Because nothing is the same as infinitely small so....
if they have no reproductive organ they must have
a reproductive organ that is infinitely small.

All this was explained in the service manual for Starship Titanic's
Infinite Improbability Drive. Unfortunately no longer available since
the Starship suffered a Spontaneous Massive Existence Failure.

8-)

Chris Carr

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2001\06\03@043345 by Roman Black

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David VanHorn wrote:

> > > Sigh.... I did not say .99999 I said .99999 REPEATING.
> > > As in **forever**.
> > >
> > > Do not get confused between the VALUE of a number and its
> > > representation. Whether you like it or not, mathematically the mumber
> > > 1 exactly equals .9 repeating.
>

> hold on jack..
> The value of 0.99(inf..) is less than 1.0 by an infinitely small (but
> non-zero) amount.


I'm with you David, if you are going to believe in
the mystical concept of infinity then it has to be
credited both ways. 0.99 (inf) is different to 1.0
by an infinitely small difference. If it was equal
to 1.0 then it would be written as such...

Seems the argument is "does an infinitely small
thing exist?" and I think by the very concept of
infinity the answer has to be yes.
-Roman

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2001\06\03@052228 by Roman Black

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michael brown wrote:

>      a = x            [true for some a's and x's]
>    a+a = a+x          [add a to both sides]
>     2a = a+x          [a+a = 2a]
>  2a-2x = a+x-2x       [subtract 2x from both sides]
> 2(a-x) = a+x-2x       [Factor left side]
> 2(a-x) = a-x          [Simplify right side]
>      2 = 1            [divide both sides by a-x]
>
> Of course we can't divide by zero, which is precisely what happens in the
> last step.


Actually I think I proved this with one of
my early PC programming experiments. I programmed
it to "divide N by zero", and yes, the PC sat there
for an *infinitely* long time.... I think other PC
programmers may have duplicated my test results,
Microsoft have done some astounding research in
this field... ;o)
-Roman

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2001\06\03@090310 by Bob Ammerman

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----- Original Message -----
From: "David VanHorn" <RemoveMEdvanhornRemoveMEspamEraseMECEDAR.NET>
To: <KILLspamPICLISTspamspamBeGoneMITVMA.MIT.EDU>
Sent: Sunday, June 03, 2001 12:58 AM
Subject: Re: [OT]: Brain Burp Rounding??


> At 12:18 AM 6/3/01 -0400, Bob Ammerman wrote:
> >Dave,
> >
> >The point is that it does _not_ take patience.
> >
> >.9 repeating is already an _infinite_ series. Not just a "sorta infinite
as
> >long as you want to go on" series.
>
> At each point, it evaluates to <1. Repeating it in smaller steps forever
> doesn't change that.

> Show me the point at which it changes, and evaluates to 1.

It doesn't change at any given step, yet, _after_ infinite steps (if one can
comprehend that), in the _limit_ that is its value.

> I know, it's infinitely far away, and it can never be reached, but if you
> wish real hard...
> >Its value is exactly 1.

Show me where infinity is and I'll show you the 'point' you are looking for.

>
> Can you show any physical thing that requires this to be true?
>

Obviously not, physical reality is limited by Heisenburg, quantum effects,
etc.

>
> --
> Dave's Engineering Page: http://www.dvanhorn.org

Bob Ammerman
RAm Systems
(contract development of high performance, high function, low-level
software)

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2001\06\03@095955 by Alexandre Domingos F. Souza

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>In fact, this is the main difference between engineers and technicians. The
>difference between things making sense and things working.

       HUAHUHAUHAUHAUHUAHUHAUHA, I think this one brings an end to this thread :oD Congratulations James!


---8<---Corte aqui---8<----

Alexandre Souza
taitospamspamterra.com.br
http://planeta.terra.com.br/lazer/pinball/

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2001\06\03@130049 by David VanHorn

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>
>I'm with Dave on this one.  Sqrt of -1 = i    "i" - that amazingly
>ridiculous, non-existant value that allows mathematicians to continue on
>creating even more ridiculous mathematics coming up with answers that don't
>(can't) exist.

Hold on there. Sqrt -1 is a valuable tool, but it is an imaginary number.


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2001\06\03@135116 by Byron A Jeff

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On Sat, Jun 02, 2001 at 08:03:04PM -0500, David VanHorn wrote:
{Quote hidden}

Dave,

You need some infinity training. Let's try a few examples:

1) What's the largest integer?
2) What's the value of 1/0?
2A) What's the value of 1/infinity?
3) What's the smallest real number larger than 0? (BTW that's the answer to
your question above)
4) Pick any two real numbers. How many real numbers are between the two?
5) What's the last digit of PI?

It's a simple mathematical fact that once you start dealing with infinity
the rules change. You find your difference above in the same place as the
largest integer and the smallest number bigger than 0. The answer is
infinity and 0. The last one sounds incongruous, but true. That's the nature
of infinity. 1/infinity doesn't approch zero. It doesn't approximate zero. It
isn't infinitely close to zero. IT IS ZERO! EXACTLY ZERO!

There is no difference between 1 and .9 (repeating). Not in infantisimal
difference. No difference. They are the same number. Of course any attempt
to make an infinite value finite make it lose it's infiniteness.

All of this fuss is over trying to attribute infinity to an approximation.
We use approximations to gain understanding in our little finite minds!
But you have to cross over and realize that once something is defined as
infinite, it is no longer an approximation of anything. It's exact.
And the math that works on infinity fails once you express finiteness on it.

The best treatment on the subject I ever read was Issac Asimov's "Asimov on
Numbers". He has several essay's on the nature of infinity.

Just remember that whenever you're having an infinity discussion, that you must
bring the largest integer to the table. Once you know that integer, the rest
is easy.

BAJ

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2001\06\04@022854 by Russell McMahon

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> 5) What's the last digit of PI?


The last digit of Pi is 4.5, on average


           RM :-)

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2001\06\04@044703 by uter van ooijen & floortje hanneman

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> Next, you'll want to sell me 5 pounds of potatoes multiplied by the Sqrt
of
> -1 :)

Why not, if he pays you with imaginary money?

Wouter

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