Is There a Shorter Proof for 0.999... = 1?

[mathwonk]

gee, what do you think the definition of the limit of a series is anyway?

the limit of a series of positive terms is most naturally and simply defined to be the lub of the sequence of partial sums. that is exactly the definition i have given only in simpler terms.


i.e. look: the limit of a series is by definition the limi tof the sequence of partial sums.
now if the series has only positive terms, as in this case, then the seuence of partial sums is increasing.

so we only need the definition of a limit of an increasing sequence.

there is a trivial theorem that the limit of an increasing sequence is the lub of that sequence,

so a simpler definition of the limit of an increasing sequence, and hence of a series of positive terms, is simply as the lub of the sequence, or of the sequence of partial sums in the first place.

anyone who understands anything about convergence would immediately realize this.

the general definition of limit of a sequence, or series is more difficult for young students, and is not needed in this trivial case.

thats why i explain limits in easy stages in my class. at least i am not surprized that someone doesn't want to take time to read my notes. it takes effort to learn something.

by the way, how do you KNOW the sum of the geometric series 1+r + r^2 +... is 1/(1-r)? [when |r| < 1).

duhh, you prove it by showing that 1/(1-r) is the lub of the sequence of partial sums!

or you could just memorize it if you do not want to understand anything.

 

[Curious3141]

How would you use limits to define recurring numbers such as 0.1234512345r?

0.\overline{12345} = \lim_{n\to\infty}\sum_{k=1}^{n} \frac{12345}{10^{5k}} = \frac{12345}{99999}

 

[Icebreaker]

0.\overline{12345} = \lim_{n\to\infty}\sum_{k=1}^{n} \frac{12345}{10^{5k}} = \frac{12345}{99999}

Touche INCREASING THE LENGTH OF THE MESSAGE TO MEET OBLIGATORY LENGTH

 

[calculus1967]

or
\sum^{+ \infty}_{n=1} \frac{12345}{100000} (\frac{1}{100000})^{n-1} = \frac{12345}{100000} + \frac{12345}{100000}\frac{1}{100000} + \frac{12345}{100000}(\frac{1}{100000})^2+ ... + \frac{12345}{100000}(\frac{1}{100000})^{n-1} + ... = \frac{\frac{12345}{100000}}{ 1 - \frac{1}{100000}} = \frac{4115}{33333}

 

[mathwonk]

come on. these are trivial facts.

 

[Night Owl]

gee, what do you think the definition of the limit of a series is anyway?

...

by the way, how do you KNOW the sum of the geometric series 1+r + r^2 +... is 1/(1-r)? [when |r| < 1).

duhh, you prove it by showing that 1/(1-r) is the lub of the sequence of partial sums!

or you could just memorize it if you do not want to understand anything.

I'll assume that was directed at me.

Why so hostile? I apologize for not reading your intimidatingly long block of text, but frankly, I figured it would be a better use of my time to simply indicate that I hadn't and then include my own short reply. If you find this offensive, then ignore my post. You have to understand, it's finals week here, and I really have no desire to read another person's class notes when I have my own that I'm reading (right now, in fact).

Additionally, what gives you the idea that I simply memorized the formula for geometric series without understanding anything? Is it because I didn't include it in my previous post? I'm sure you're aware of the fact that if you include every proof for every single thing that needs to be proved in every statement, you'll have an unnecessarily long explanation. You don't need to re-invent the wheel. The topic wasn't "How do you evaluate a geometric series?"; it was "Why does 0.9999 recurring equal 1?". I was not interested in teaching people what a geometric series is or how to evaluate one, and judging by your previous posts, neither were you. You still haven't posted a proof of why a geometric series equals 1/(1-r). If you want to, go ahead, but I won't read it. That's what my class notes are for.

Just because I chose not to read your post is no reason to get so hostile. You never once said that your notes directly addressed the issue we were discussing, and come on, that's a lot to read for something that requires a relatively simple explanation. If you think my explanation is flawed in some way, then please, by all means, correct it.

Now if you'll excuse me, I have things to study for.

 

[Integral]

I believe that can be written as:
12345 \Sigma_ {n=1} ^{\infty} 10^{-5n}

 

[mathwonk]

never mind, night owl, everyone takes whatever instruction they are ready to absorb.

 

[toocool_sashi]

we cannot prove, or rather justify that 3*0.33333... = 0.999999... isn't it?? in arithmetic multiplication, we start multiplying from the rightmost and here...we can't reach the righmost. we are dealing with infinity here(infinite number of digits) and arithmetic fails with infinity. Maybe my thinking is a bit too "simple" for u all...but i don't know how u can say 3*.3333... = .9999... arithmetic cannot be used as a proof for a theorem or a result ... best to use the word "justification" ;) cheers!

 

[lurflurf]

we cannot prove, or rather justify that 3*0.33333... = 0.999999...

This means
3(\frac{1}{3})=1
It is often better to think of decimal expansions as ways to write numbers, not as strange and mysterious objects.
.999999999... is just one possible way to write 1 here are a few others
\frac{151}{151}
\frac{d}{d(x^2)}x^2
\lim_{n\rightarrow\infty}\frac{n^2+1}{n^2-1}
\int_0^11dx

 

[EnumaElish]

...
1/3 = 0.333...
1.3 * 3 = 0.333... * 3
1 = 0.999... also works.
...

Doesn't this merely shift the punchline of the proof, so to speak? Doesn't one have to prove 1/3 = 0.333... ? How's that easier than proving 1 = 0.999... directly?

Either I am the smartest one in this bunch or I missed the point of a lot of the postings here.

 

[TenaliRaman]

Doesn't this merely shift the punchline of the proof, so to speak? Doesn't one have to prove 1/3 = 0.333... ? How's that easier than proving 1 = 0.999... directly?

Either I am the smartest one in this bunch or I missed the point of a lot of the postings here.

I am not going to justify the post made there, but many people find it easier to comprehend the fact that 1/3 = 0.3333... than 1 = 0.9999...
Why??
Divide 1 by 3 using long division and it seems apparent (to them) however there is no such process to get 0.99999... from 1, so obviously they think its an error. However naive this argument may sound, i believe it appeals to their thought process and hence 1=0.99999... becomes a much talked abt issue in many forums.

No wonder, matt wants it to be immortalised as an FAQ

-- AI

 

[EnumaElish]

... however there is no such process to get 0.99999... from 1, so obviously they think its an error. ... -- AI

The closest expression I can come up with is: 1 = 0.999... * 1.000...

Or 1/1.000... = 0.999...

It's sort of like saying "1 is the geometric average of 0.999... and 1.000..."

Example:

Suppose a population grew -0.000... in the first year and +0.000... in the second year; what is the cumulative growth rate (CGR) in two years?

1 + CGR = (1-0.000...)(1+0.000...)
1 + CGR = 0.999... * 1.000...

Intuitively there was zero growth, so CGR = 0.

1 = 0.999... * 1.000... There.

If one is prepared to accept the CGR = 0 intuition then it becomes possible to show 1 = 0.999... directly:

1 + CGR = (1-0.000...)(1+0.000...) = 1² - (0.000...)²

Intuitively CGR = 0 so,

1 = 1 - (0.000...)(0.000...) = 1 - 0.000... = 0.999... So there!

 

[boteet]

Hello enumaelish, if cgr=0 intuitivley, you will not be able to get
1 + CGR = 0.999... * 1.000...

 

[EnumaElish]

Hello enumaelish, if cgr=0 intuitivley, you will not be able to get
1 + CGR = 0.999... * 1.000...

Can you elaborate, even it's along the lines of "obviously you haven't had your 3rd cup of coffee this AM, because ..."?

 

[rachmaninoff]

Why is this a five page thread? What's wrong with the three-line epsilon proof:

\begin{align*}<br /> &amp;\mbox{For } \epsilon &gt; 0, \exists n &gt; \log ( \epsilon^{-1} ) &gt; 0 \mbox{ such that} \\<br /> &amp; |1 - \sum_{k=1}^{n}9*10^{-k} | = 10^{-n} &lt; \epsilon \\<br /> &amp; \mbox{Therefore: } 0.99\overline{9}\equiv \lim_{n \rightarrow \infty} \sum_{k=1}^{n}9*10^{-k} = 1 \end{align}

Is anyone not happy?

 

[EnumaElish]

Why is this a five page thread? What's wrong with the three-line epsilon proof:
...
Is anyone not happy?

Not creative enough!

 

[Icebreaker]

Is anyone not happy?

Anyone who's math-literate enough to read that wouldn't need it in the first place.

 

[toocool_sashi]

This means
3(\frac{1}{3})=1
.999999999... is just one possible way to write 1 here are a few others

this is exactly what we have 2 prove isn't it!? i liked the epsilon proof...it involed no arithmetic that's why lol ;) but see...if u can convince urself that 3*0.3333333... = 0.99999999... then u mite as well convince urself that 1-0.9999999... = 0.00000000.. this wud become a shorter *proof*

 

[TenaliRaman]

1-0.9999999... = 0.00000000.. this wud become a shorter *proof*

Oh! But there will be a 1 at the end

-- AI

 

[Zurtex]

Oh! But there will be a 1 at the end

-- AI

Yes at the end of an infinite number of 0s

 

[TheGinkgoNut]

how about just this simple one with arguably less demands on the idea of "infinity":

<br /> \begin{align*}<br /> x &amp;=&amp; 0.999\overline{9} \\<br /> 10x &amp;=&amp; 9.999\overline{9} \\<br /> 10x-x &amp;=&amp; 9 \\<br /> 9x &amp;=&amp; 9 \\<br /> x &amp;=&amp; 1 \\<br /> \end{align}<br />

 

[matt grime]

and you've proven that arithmetic is well defined on strings of recurring (or any infinitely long) decimals have you?

 

[TheGinkgoNut]

i've opened a pandora's box haven't i...

 

[EnumaElish]

i've opened a pandora's box haven't i...

More like a Russian doll, IMO.

 

[mathwonk]

gingkonut, that was the arguemnt i was shown in 8th grade and i always remembered it and liked it. that's the right level for it, kids who don't need rigor, but enjoy being amazed.

 

[matt grime]

There is no harm in using white lies when explaining maths, often it is the best way of getting across information. And the multiplying by ten thing is a good example: we know what *ought* to happen when we do it, and as mathwonk says that is often good enough. the problem comes from the die hards who refuse to accept this *motivational* reason and then dismiss the full on proof with all the details and carefully laid out definitions (I would suggest because they do not understand the formality of mathematics). hence in this case (and this is only my opinion now because of the attitudes displayed repeatedly in this forum where the audience is of all levels) i tend to immediately go for the overkill approach since there is a good chance the reader is not a kid to be amazed.

 

[TOKAMAK]

Does anyone have any links to the original discussion from battle.net? Thanks in advance.

 

[bao_ho]

x = 0.9999
9x = 9
11x = 10.9999
2x = 1.9999?

 

[HallsofIvy]

x = 0.9999
9x = 9
11x = 10.9999
2x = 1.9999?

Starting from x= 0.9999, how do you get 9x= 9? According to my calculator 9x= 9(0.9999)= 8.9991! (And 11(0.9999)= 10.9989, not 10.9999.)

Anyway, why are you talking about 0.9999? Everyone else is talking about 0.99999... (which is easily shown to be equal to 1.)