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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,666

It is the same as .3333333... which you already accepted!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,523

Hm, but, you see, I do not allow for repeating 9's after the decimal point. I find that every number has a unique decimal representation that way!

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,666

But you allow for .333333... and .6666666... why stop at .9999999...?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,523

Because there is no other way to write the first two. The third is needless.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,666

Not really it also represents 3 x .3333333... and 9 x .1111111..., all of which you agree exist but suddenly when we multiply they disappear like magic?

*Last edited by bobbym (2013-03-06 04:22:42)*

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,523

Yes, but those are not decimal representation of that number.

And, they don't disappear. Both of those are 1.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,666

They are also digit by digit multiplication. We are allowed to multiply are we not?

.999999999... exits because as you say .33333333... exists.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,523

We are allowed to multiply. 3*0.3333...=1.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,666

How do you know that 3 * .3333333... = 1?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,523

Because 3*0.333...=3*(1/3)=1.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,666

Hmmm, that is because you know and accept that 1 / 3 = .3333333. What would you do for this one?

What fraction is that?

*Last edited by bobbym (2013-03-06 04:56:51)*

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,523

It is 1/2071.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,666

How do you know that?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,523

It was easy finding it.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,666

No it was not. It was easy for your package to find it.

Supposing I gave you one that your package could not get. Let us assume it is that one. Prove that 2701 x that number is one?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,523

Well, if I had a package, I would simply divide 1 by 2071 and check to see if it is that number.

*Last edited by anonimnystefy (2013-03-06 05:51:27)*

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,666

Did you try that?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,523

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,666

Hohohohohohohohohoh! That is a good one. Very good!

Ever wonder how wolfram did that? Supposing you had to do it without alpha, could you prove that 2701 x that decimal expansion is 1?

For that matter you have not proved that 3 * .3333333... = 1. What if you did not know that 1 / 3 = .33333333...?

*Last edited by bobbym (2013-03-06 06:08:31)*

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,523

Well, I do know that 1/3=0.333... Supposing that I do not is very unrealistic.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,666

I am asking for the proof without that knowledge because for the bigger fraction you would only have the repeating decimal not the fraction. Do you think that mathematica knows that fraction offhand?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,393

hi bobbym and Stefy

bobbym wrote:

Supposing I gave you one that your package could not get. Let us assume it is that one. Prove that 2701 x that number is one?

Well it's only a long division. So given the following I could do it without a computer of any kind:

(i) large sheet of paper (A0 or bigger)

(ii) A pencil and sharpener

(iii) A sufficient incentive to maqke the effort ($1000000 should do it)

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,666

Hi;

Where I am going with this is simple. Multiplying .333333333333333... by 3 is never going to produce that magical 1. It is going to produce a long string of nines. anonimnystefy gets around that by calling it 1 / 3 and doing rational arithmetic. But what happens when you do not know the rational equivalent of the decimal? He is now forced to do the multiplication

Always the string of nines.

*Last edited by bobbym (2013-03-06 07:14:51)*

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,523

That is because you do not have enough digits of precision.

Here lies the reader who will never open this book. He is forever dead.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,393

But my point is, you can achieve the same without doing a multiplication. You do a division instead. By long D you get all those digits and then the recurring clicks in and you can stop. No precision problem at all.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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