Re: A Basic Theorem
- From: "Peter Allen" <peteronusenet@xxxxxxxxxxx>
- Date: Sat, 28 Jan 2006 00:40:07 -0000
TheNakedOne wrote:
> .3 repeated = 1/3
> 3(.3 repeated) = 1
> 3(.333 repeated)=1
> .999 repeated=1
> .9 repeated=1
> Q.E.D.
> Hey everyone, I'm new to sci.math, and this is a proof I gave to many
> teachers at my school, yet none of them could give me a clear enough
> reason as to why this was incorrect. My personal belief is that
> because the repeated (sorry, I couldn't find a superline key) sign
> means the task is being done an infinite number of times, then it
> actually reaches the number which can not be reached in decimal form.
> I wanted to know your thoughts.
This is sort-of correct. The problem is you need to know exactly what you
mean by '.3 repeated' to say it equals 1/3. What you mean is the limit of
the sequence whose first few terms are 0.3, 0.33, 0.333, 0.3333, 0.33333,
.....
which you can prove is equal to 1/3 (this is _not_ completely trivial).
Thereafter your manipulations are essentially straightforwards and fine.
IOW, your first assumption that '.3 repeated' means something and is equal
to 1/3 (note 2 assertions here, prove both!) is the main issue. Once you can
prove that - and you haven't, here - you can repeat the rest of your logic
to prove 0.9999... = 1.
I suggest the easiest approach to the first bit is to write an epsilon-delta
proof that the 0.3, 0;33, ... sequence does tend to 1/3.
That said, if you can write such a proof, you might as well write
immediately a proof that 0.9, 0.99, 0.999, ... tends to 1, so 0.9999...=1
directly.
Peter
.
- References:
- A Basic Theorem
- From: TheNakedOne
- A Basic Theorem
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