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**Smartiebell****Member**- Registered: 2012-09-16
- Posts: 16

My personal view- but I do not like the wording.

I (thought) it's like the "proof" of 1=2. There has to be a hidden mistake.

The problem( copied and pasted )

~~~~~~~~~~~~~~~~~

Let X = 0.999...

Then 10X = 9.999...

Subtract X from each side to give us:

9X = 9.999... - X

but we know that X is 0.999..., so:

9X = 9.999... - 0.999...

or: 9X = 9

Divide both sides by 9:

X = 1

But hang on a moment I thought we said X was equal to 0.999... ?

Yes, it does, but from our calculations X is also equal to 1. So:

X = 0.999... = 1

Therefore 0.999... = 1

~~~~~~~~~~~~~~~~~~

Common Sense: The integer 1 has a 1 in the ones place, and 0.999... has a zero.

But, (oh crack, I just got confused!) 1/3 = 0.333... and 1/3 3 =3/3=1, and 0.333... 3 =0.999...

so 0.999 would equal (1)! ( put parenthesis so not to be mistaken for 1!)

I need some diffrent opinions...

EDIT: I say yes now, thnx bobby!

*Last edited by Smartiebell (2012-09-16 09:48:14)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hi;

Who says that there is only only representation of the integer 1?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Smartiebell****Member**- Registered: 2012-09-16
- Posts: 16

There IS

1.0

01

1 1

2-1

x^0

etc...

I AM going both ways on this subject.

I'm just a little kid, don't confuse me too much!

*Last edited by Smartiebell (2012-09-16 08:55:42)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hi;

How little?

In your post you admit or agree that

1 / 3 = .3333333...

So you are admitting that 1 / 3 has two representations, 1 / 3 and .33333333...

So why not 1?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Smartiebell****Member**- Registered: 2012-09-16
- Posts: 16

Agree.

Not 3rd grade little, but not College big.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hmmm, there is alot of room in there.

To continue, no one argues that 3 pieces of a pie that is cut in thirds does not equal the whole pie do they?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Smartiebell****Member**- Registered: 2012-09-16
- Posts: 16

Yeah, I have never disagreed with that.

I know that 1 can be representated more than once, but what I am trying to figure out is if 0.999... written in standard form is the same as 1 written in standard form. Now I am 0.6 leaning toward Yes.But every so often my brain goes back to no.

And talking about brainpower, I am higher then 5th grade.

*Last edited by Smartiebell (2012-09-16 09:21:38)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Very good!

If 1 / 3 = .3333333... as we agree then we know that 3 of those 1 / 3's equals the whole thing or 1. Since

3 * .333333.... = .999999....

or

3 * .333333.... = 1

therefore you must agree that .999999.... = 1, it is just written differently. It is actually much easier to accept this one even if you can not follow it.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Smartiebell****Member**- Registered: 2012-09-16
- Posts: 16

Thats what I wrote, right?!?!

And that means that 0.999... has the same properties as 1.

Like 50.999... = 5.

So we're sure about that, right?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hi;

This is not correct.

Like 50.999... = 5.

You have two decimal points in there.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Smartiebell****Member**- Registered: 2012-09-16
- Posts: 16

What do you mean?

the period after the 5 is a sentence stopper period.

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**Smartiebell****Member**- Registered: 2012-09-16
- Posts: 16

Like (5 times 1= 5) and (5 times 0.999... = 5)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hi;

Not the sentence stopper, this

50.999... we have an ellipsis on the end and two decimal points.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Smartiebell****Member**- Registered: 2012-09-16
- Posts: 16

ok.

i thought meant multiply.

So all that I had the problem with was the wording of the problem on the site. I say yes.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Hi;

Confusion gone? Are you okay with the idea?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Smartiebell****Member**- Registered: 2012-09-16
- Posts: 16

Yes. I just got a glimpse of (cue mystical music) The world of MAATH!

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**Smartiebell****Member**- Registered: 2012-09-16
- Posts: 16

Oh cool, I just turned into a Member!

sorry, off topic.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Congratulations!

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**Smartiebell****Member**- Registered: 2012-09-16
- Posts: 16

Soo... let the people who have a problem with the idea cometh here! Cause I just figured it out, and I like to hear new ideas.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,364

Very few who come in here with that problem can ever come around to any new ideas.

*Last edited by bobbym (2013-02-16 01:42:35)*

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**john_gabriel****Member**- Registered: 2013-01-29
- Posts: 7

All the phony proofs in favour of 0.999... being equal to 1 are easily refuted:

thenewcalculus.weebly.com/uploads/5/6/7/4/5674177/proof_that_0.999_not_equal_1.pdf

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**n872yt3r****Member**- Registered: 2013-01-21
- Posts: 392

I think that 0.999... = 1.

- n872yt3r

Math Is Fun Rocks!

By the power of the exponent, I square and cube you!

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**muxdemux****Member**- Registered: 2012-12-23
- Posts: 80

john_gabriel wrote:

All the phony proofs in favour of 0.999... being equal to 1 are easily refuted:

thenewcalculus.weebly.com/uploads/5/6/7/4/5674177/proof_that_0.999_not_equal_1.pdf

Interesting paper, John. Though I'll admit I'm skeptical.

In particular, on page 6 you prove that

by induction -- which *does* make sense to me. Where I start to have my doubts is where you show that

is absurd. Didn't you only prove

for the natural numbers?Offline

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,380

hi john_gabriel

Welcome to the forum.

I read the article. You wrote it?

One of the forum rules is:

"No Personal Attacks or Put-Downs. "

Hopefully the writer will understand that before contributing here.

Bob

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Also a good point is that we should "ban" 0.999..., and similar numbers (that end with 9 recurring), because we would then have one representation in decimal notation for every number.

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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