Re: A Basic Theorem
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Fri, 27 Jan 2006 14:36:26 +0000 (UTC)
In article <1138366550.139225.54070@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
TheNakedOne <Gooberglob@xxxxxxxxx> 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.
What makes you think it is incorrect?
A series is an expression of the form
sum_{i=0 to oo} a_i
where a_i are real numbers. By definition, the series converges to a
real number S if and only if the sequence of partial sums converges to
S, meaning that
lim_{n->oo} (a_0+...+a_n) = S.
The nonterminating decimal expansion .33333.... represents the value
of the series
3/10 + 3/10^2 + 3/10^3 + 3/10^4 + ... + 3/10^n + ...
This is a geometric series of the form
a + ak + ak^2 + ak^3 + ak^4 + ... + ak^n + ...
(with a=3/10 and k=1/10).
The series converges if and only if |k|<1, in which case it is
straightforward to verify that it converges to a/(1-k).
In this case, we have a=3/10, k=1/10, so a/(1-k) = (3/10)/(9/10) =
3/9 = 1/3.
It is also easy to verify that if you have a converging series
a_0 + a_1 + ... + a_n + ...
that converges to s, and r is any real number, then the series
r*a_1 + r*a_1 + ... + r*a_n + ...
converges to r*s. That is, symbolically,
sum_{i=1 to oo}(r*a_i) = lim_{n->oo} sum_{i=1 to n}(r*a_i)
= r* lim_{n->oo}(sum_{i=1 to n}(a_i)) = r*(sum_{i=1 to oo} a_i).
That's if the original series converges.
Since the series (3/10 + 3/10^2 + ... + 3/10^n + ...) converges to
1/3, it follows that the series
3*(3/10 + 3/10^2 + ... + 3/10^n + ... )
= (9/10 + 9/10^2 + ... +9/10^n + ... )
also converges, and coverges to 3(1/3) = 1.
Since the latter series is represented in decimal expansion by
..9999999....
that means that indeed .99999999... is equal to 1. You can also verify
it through the formula I give above: now you have the geometric series
with a = 9/10 and k=1/10, so it converges to
(9/10)(1-(1/10))=(9/10)/(9/10)=1.
So the manipulations are, in fact, correct, when you suitably
interpret the decimal expansion.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx
.
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