1 + 1 + 1 = ?

[Hal.]

Is 1 + 1 + 1 = 3 ? and are there alternative answers ? like for instance 2.99999999999999999999999999999999999999999999999999999999999999999999991

[Cap'n Refsmmat]

2.99999999999999999999999999999999999999999999999999999999999999999999991 doesn't equal 1 + 1 + 1, but 2.999... with infinitely repeating 9s does.

 

[math]x = 2.999\ldots[/math]

 

[math]x-2=0.999\ldots[/math]

 

[math]10(x-2) = 9.999\ldots[/math]

 

[math]10(x-2)-(x-2)= 9[/math]

 

[math]9(x-2) = 9[/math]

 

[math]x-2 = 1[/math]

 

[math]x = 3[/math]

 

Hence [math]2.999\ldots = 3[/math], but only if the nines repeat infinitely.

[Hal.]

I see the math Cap'n and it's like a magic trick in front of my eyes . Don't confuse what I say with disagreement .

 

Ask yourself this Cap'n ,

 

Could you show us that ,

 

2.99999999999999999999999999999999999999999999999999999999999999999999999 recurring < 3 if you wanted to ?

Edited April 18, 2011 by hal_2011

[Cap'n Refsmmat]

No.

[insane_alien]

nope because it isn't <3

[Hal.]

Do you know somebody who can ?

[Cap'n Refsmmat]

No, because it's impossible.

[Hal.]

Could you try to disprove yourself as vigorously as possible ?

[insane_alien]

why? the disproof is trivial.

 

3!=<3

[Hal.]

Be careful 3!=6

[swansont]

Is 1 + 1 + 1 = 3 ?

 

Yes. Ref: Come Together by the Beatles.

[Fuzzwood]

It is 3 because you define 1 as an integer.

[mooeypoo]

No, because it's impossible.

 

What? Sorry, are we being cynical or is it really the claim that we can't show that 3-(smaaaal) is lower than 3?

 

I'm just checking here, is this some mathematical trick-of-words or are we just playing around the concept for fun?

[Hal.]

I assure you mooeypoo that my questions have a point . Try to prove and disprove as much as you can on your own methods .

[mooeypoo]

I don't understand the problem.

 

Not matter how many 9s you have in your 2.999, it's still, by definition, [math]3-\Delta[/math], where [math]\Delta[/math] is an arbitrarily small unit.

 

Which is also, quite clearly, less than 3.

 

What am I missing?

[DrRocket]

I don't understand the problem.

 

Not matter how many 9s you have in your 2.999, it's still, by definition, [math]3-\Delta[/math], where [math]\Delta[/math] is an arbitrarily small unit.

 

Which is also, quite clearly, less than 3.

 

What am I missing?

 

The only "arbitrarily small" non-zero number is 0.

 

 

.999999............ = 1

 

As can be seen:

 

x = .9999999.............

 

10x = 9.9999999..........

 

9x = 9.99999999..... - .99999999.... = 9

 

x =1

 

This is a perfectly legitimate mathematical proof.

 

If you desire gory detail here it is ;

 

 

[math]\displaystyle \sum_{n=0}^N x^n = 1 + x \displaystyle \sum_{n=o}^N x^n - x^{N+1} [/math]

 

[math](1-x)\displaystyle \sum_{n=0}^N x^n = 1-x^{N+1}[/math]

 

[math]\displaystyle \sum_{n=0}^N x^n = \dfrac {1-x^{N+1}}{1-x}[/math]

 

Similarly

 

[math]\displaystyle \sum_{n=1}^N x^n = \dfrac {1-x^{N+1}}{1-x} -1 [/math] [math] = \dfrac {x-x^{N+1}}{1-x} [/math]

 

So, if [math]|x|<1[/math]

 

[math]\displaystyle \sum_{n=0}^\infty x^n[/math] [math]=\displaystyle \lim_{N \to \infty} \displaystyle \sum_{n=0}^N x^n = \displaystyle \lim_{N \to \infty} \dfrac {1-x^{N+1}}{1-x}[/math] [math] = \dfrac {1}{1-x}[/math]

 

And

 

[math]\displaystyle \sum_{n=1}^\infty x^n[/math] [math]= \displaystyle \lim_{N \to \infty} \displaystyle \sum_{n=1}^N x^n = \displaystyle \lim_{N \to \infty} \dfrac {x-x^{N+1}}{1-x}[/math] [math] = \dfrac {x}{1-x}[/math]

 

 

[math]0.99999........ = \displaystyle \sum_{n=1}^\infty 9 (\dfrac{1}{10})^n[/math] [math] = 9 \displaystyle \sum_{n=1}^\infty (\dfrac{1}{10})^n[/math] [math] = 9 \dfrac {\frac {1}{10}}{1- \frac{1}{10}}[/math] [math] = 9 \dfrac {1}{9}[/math] [math] = 1[/math]

[Klaynos]

Ah, the joys of infinite decimal places in maths... it all becomes a little... odd...

[swansont]

If Zeno posts and says he's only halfway done, I'm locking the thread.

[Hal.]

Dr.Rocket , I like that as a proof , It looks like art that must be wondered about .

[ewmon]

Try looking at it this way.

 

Using regular subtraction,

 

0.99999..... ad infinitum = 1 – 0.00000..... ad infitinum .....1

 

The ad infinitum on the right side of the equation means you never get the opportunity to place the one because the zeros go on forever.

 

Therefore,

 

0.99999..... ad infinitum = 1

[DrRocket]

Try looking at it this way.

 

Using regular subtraction,

 

0.99999..... ad infinitum = 1 – 0.00000..... ad infitinum .....1

 

The ad infinitum on the right side of the equation means you never get the opportunity to place the one because the zeros go on forever.

 

Therefore,

 

0.99999..... ad infinitum = 1

 

 

argh !

[ewmon]

argh !

Uhhh..... Was that a "good" argh or a "bad" argh?

[Xerxes]

I think you may take it as a bad one. I assume it refers to fact that 0.000.......1, where the ellipses represent an infinite string, has absolutely no meaning whatever.

 

Moreover (though I doubt that DrRocket would subscribe to this), even if it did have meaning, your construction implies that 0.999..... both equals 1 and doesn't equal 1, a very strange state of affairs!

 

Seriously, this problem crops up over and again on boards like this, and, to use your expression, has been dispatched ad infinitum.

[mooeypoo]

I actually thought, at first, that you are all yanking my chain. This entire thing sounds ridiculous to me, honestly, but I must admit, mathematics has this tendency to have paradoxes that I simply don't get.

 

Useful link to anyone who's as confused as I was/am: http://en.wikipedia.org/wiki/0.999... (thanks swanstont)

 

 

So now that I know this is actually REAL, I have a few questions of my own:

 

First this seems to talk about infinite 9s after the decimal. Does this mean that the original post, with

2.99999999999999999999999999999999999999999999999999999999999999999999991

as in, there's a 1 at the end and it's not infinite 9s, is *not* the same as infinite 9s? The wiki has examples with limit goes to infinity, and this one doesn't...

 

Second, why is this different than the mathematical paradoxes out there, where I can get a nonsensical result out of mathematical manipulation? Is this not nonsensical? The definition of .999 is that it's not yet 1, isn't it?? so isn't this manipulation resulting in a nonsensical result?

 

I saw a math paradox where 1=0, simply by stating something like

1 = 1 + 0 + 0 + 0 + 0 + ...

and using rules of math, so I can change order of addition:

1 = 0 + 1 + 0 + 0 + 0 + ...

1 = 0 + 0 + 1 + 0 + 0 + ...

1 = 0 + 0 + 0 + 1 + 0 + ...

1 = 0 + 0 + 0 + 0 + 0 + ...

1=0

 

This isn't REAL. It's nonsensical, it's just abusing the laws of math to reach a nonsensical result, hence being called a PARADOX. the 0.999... thing is also a paradox, isn't it? (btw, it's under "math paradoxes" category in wikipedia, if it helps my point).

 

So.. what's the difference between producing nonsensical results that we laugh about and NOT treat seriously like the 1=0 one and this 0.999... one?

 

 

This makes no sense to me.

[Xittenn]

A paradox seems to have several definitions; I had to double check this because I recently had an English assignment that was developed around definition number two.

 

On the one hand a paradox is defined as:

 

1) A statement or proposition that, despite sound (or apparently sound) reasoning from acceptable premises, leads to a conclusion that seems senseless, logically unacceptable, or self-contradictory

 

On the other hand a paradox is also defined as:

 

2) A seemingly absurd or self-contradictory statement or proposition that when investigated or explained may prove to be well founded or true

 

so I think the use of the definition of a paradox as an argument with direction, especially under the context, is not paradoxical, but maybe misguided??? i.e. a paradox denotes absurdity ..