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BY : Byron A Jeff email (remove spam text)

On Sun, Jun 03, 2001 at 12:22:50PM -0300, Alexandre Domingos F. Souza wrote:

> >Earlier in this thread, the most simple (IMO)) proof that

> >0.999(Repeating) = 1 was shown using no magic, mirrors or divides by 0.

> > There are 3 thirds in a whole (ie 3*(1/3) = 1)

> > and, 1/3 = 0.33333333(repeating forever) (divide it out and see!)

> > Multiply that by 3 and get 0.99999999(repeating forever) (try it!)

> > Therefore 1 = 0.99999999(repeating forever)

> >It may be uncomfortable, but it is true!

> >Show me any flaw, trick, or error in this example.

>

> It seems to be something like "angel's gender" but let's revise it.

>

> When you part one something MATERIAL (numbers aren't material, they just TRY to represent it), you get three thirds. One equals another. That's ok.

>

> When you divide the number one, by three, you get ALMOST three equal parts. ALMOST because a calculator cannot represent the part who tend to infinity. If you divide something physical, you can get all three parts with the same volume or weight. But a calculator cannot represent this.

>

But this is exactly the point. When you divide the number 1 by 3 you get

exactly (note EXACTLY!!!!) 3 equal parts. However the representation of that

exactly is .3 (repeating).

Everyone is getting confused by the repeating. "There is a difference at

every step therefore there must be a difference at the 'infinite' step." is

close to the quote I've read in this thread.

There is no infinite step. The threes stop ending at the same place that

you find the biggest integer! Now would anyone care to discuss what the

biggest integer is?

The repeating series is the only way to represent the value exactly. Any

attempt to cut off the repeating and it's no longer an exact representation.

You have to take the baby with the bathwater. 1/3 is exactly .3 (repeating) and

three times each value is 1 and .9 (repeating) which are exactly (note

EXACTLY!!!) the same.

And no finite string of 3's or 9's, no matter how long the string is is

exactly equal to the infinitely repeating string. And that's why this works.

Any attempt to make the string finite gives a value less than the exact value.

However the infinite string is the exact value.

Let's consider a couple of cases. Consider two numbers x1 and x2. x1 is

exactly equal to 9 (repeating). x2 is equal to 2^1024 9's after the decimal

point (a lot of 9's to be sure). Now we should be able to agree on a couple

of things.

1) That x1 > x2. x1 has more 9's than x2, so it's bigger.

2) That 10*x2 - x2 is less than 9. Consider:

10 * 0.9999 (2^1024 of them) = 9.99999 (2^1024 - 1 9's after decimal point)

(-x2) -0.99999 (2^1024 9's after decimal point)

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

8.99999 (2^1024 -1 9's and a 1 on the end)

No problem right? Now do the same with x1 and you see the difference.

10 * .9 (repeating) = 9.9 (repeating)

-0.9 (repeating)

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

9.0 (repeating)

Now the difference is that since in both cases the .9 (repeating) are exactly

the same. That's the difference with infinity. Arithmetic operations on

infinity result with exactly equal infinite values. That's the catch. With

any finite string, multiplying by 10 and subtracting has a difference however

small. But the same operation on a infinite string gives no, zero, 0 , nada

difference. Not an infintessimally small difference. NO DIFFERENCE!!! It's

like trying to define the smallest real number larger than 0. That number is

0. Sounds strange, but it's true because the limit of an infinite series

approching 0 is 0.

Everyone who's arguing the difference between the sum of an infinite series

and the limit of the series isn't getting that when you deal with infinity

there is no end, therefore there is no difference.

Just remember that to justify your argument you must, absolutely must, be

able to define the largest integer. Once you can do that, then everything

else falls into place.

Good luck. ;-)

> When you bring these three parts togheter (ops, where is my english lexican?) it becomes 1 again. But in a calculator, that ROUND NUMBERS PER SE, it will not bring you the same result. It's valid for any calcule that brings the "dizima periodica" (anyone who speaks portuguese, can tell me how we say this expression in english? It means the infinite .33333333 of the 1/3), because a calculator has a limit of HOW FAR can it store numbers.

>

> So, I think it's nice to speak that 0.9999999999999 TENDS to 1. But it can never be 1.

>

> And of course, In my HP48 I trust :o) And it gives me 0.99999999999999 when I multiply 0.33333333333 x 3 :o)

>

Calculators don't help. They're finite. Infinity has different rules than

finite. I'm sure your calculator also has a largest integer or real that it

can store. So does that mean that the next larger integer or real (which

clearly does exist) does not actually exist because your calculator says so.

So to summarize. 1/3 = .3 (repeating). 3*(1/3) = 3 * .3 (repeating). Therefore

1 = .9 (repeating). End of story.

And by the way if you fell compelled to refute this argument, please bring

along the largest possible integer with you. You'll need it.

BAJ

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In reply to: <200106031222500660.007A31DB@smtp.vix.terra.com.br>; from taito@TERRA.COM.BR on Sun, Jun 03, 2001 at 12:22:50PM -0300

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