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samuel bird
Regular poster

Post Number: 69
 Posted on Thursday, 29 May, 2014 - 06:58 pm:

I have just got 'The Mathematical Olympiad Handbook' by Gardiner and am confused about one thing. In the second last exercise of the first part (numbers), you are asked to prove that 0.99... = 1 and 0.49... = 1/2. I understand what it is asking but am not sure what sort of proof it is asking for and what would be appropriate. Previously in the chapter, he just stated a few things such as the convergence and limit and the fact that the sequence tends to x.

Basically, I am not sure what sort of answer you are meant to give. A rigorous proof? ...or a brief explanation? What type of answer would / did you give? Thanks
Jeremy Barker
Prolific poster

Post Number: 398
 Posted on Thursday, 29 May, 2014 - 07:20 pm:

I don't have that book, but has the author covered the convergence of GPs with an infinite number of terms?

That is how I would tackle these two results.
samuel bird
Regular poster

Post Number: 70
 Posted on Thursday, 29 May, 2014 - 07:30 pm:

He mentions it, yes, and explains that this means it will constantly converge to 'x'. This is what I did for my answer as well.
Dominic Yeo
Veteran poster

Post Number: 433
 Posted on Friday, 30 May, 2014 - 09:06 am:

How to define 0.999... is a sensible question to start with. You certainly can't prove things about a 'new' notation without defining it first.

(You could just define it to be equal to 1, but that means you then have to define 0.499... to be equal to 1/2 and so on. Adding an infinite number of definitions is not very practical really.)
lebensborg
Regular poster

Post Number: 44
 Posted on Friday, 30 May, 2014 - 09:10 am:

If you set n=0.99999999...
then what is 10n?

If you carry on messing around with n in about two more steps you will get n=1
and therefore 0.999...=1 *

Once you have done this, for the second part simply divide both parts of *by 10.

Then 0.4999...=0.4+0.0999...=0.4+0.1=0.5.
samuel bird
Regular poster

Post Number: 71
 Posted on Friday, 30 May, 2014 - 09:30 am:

Great, that clarifies the process. I will try to apply a similar depth of answer for other (*) problems.
Arkan Megraoui
Prolific poster

Post Number: 248
 Posted on Sunday, 01 June, 2014 - 11:46 am:

Here's the limiting approach.

Let $N=0.999...$, and consider the number with $k$ 9's in the decimal, i.e. $0.\underbrace{999...}_{k\ \textrm{times}}$. This may be written $\left(\underbrace{999...}_{k\ \textrm{times}}\right)/\left(1\underbrace{000...}_{k\ \textrm{times}}\right)$, or $(10^k-1)/(10^k)=1-\frac{1}{10^k}$. Thus, $N=\underbrace{\lim}_{k\rightarrow \infty}\left(1-\frac{1}{10^k}\right)=1$ since $\underbrace{\lim}_{k\rightarrow \infty}\left(\frac{1}{10^k}\right)=0$.

Or you can be like $0.111...=\frac{1}{9}$ so $0.999...=\frac{9}{9}=1$.
lebensborg
Regular poster

Post Number: 53
 Posted on Monday, 02 June, 2014 - 05:51 pm:

The only problem with '0.111...=1/9 so 0.999...=9/9=1' is that it is assuming that 0.111...=1/9, which is not necessarily obvious, so probably requires a proof.
lebesgue
Veteran poster

Post Number: 3412
 Posted on Monday, 02 June, 2014 - 10:15 pm:

What do you obtain by long division 1:9; I believe that 0.111.. Alternatively, 0.111... is the geometric series 1/10+1/10^2+1/10^3+..., whose sum is 1/9.
Jeremy Barker
Prolific poster

Post Number: 400
 Posted on Tuesday, 03 June, 2014 - 08:41 am:

There is a slight subtlety here that in going from 0.111... = 1/9 to 0.999... = 9/9 = 1 you're assuming that you can manipulate an infinite series the same as a finite one, i.e. that if a series with general term a(n) converges to S, then the series with general term ka(n) converges to kS, but this is nothing that can't be patched up with a basic result from Analysis