Re: devreden

loofangyenn wrote:

Here's a nice little proof

assume that
a=0.9999...
then multiplying by 10
10a=9.999.....
subtracting a
10a-a = 9.9999... - 0.99999
that is
9a = 9
but then
a=1

That's the method I was talking about when I wrote:

"There was a whole section in my high school algebra 1
text about converting repeating decimals to fractions.
It appeared in the chapter on solving linear equations
(e.g. equations such as 4x = 12, 3x+2 = 20, 5x-1 = 8, etc.),
which was near the beginning of our text."

This method can be generalized beyond repeating decimals
(where you multiply by 10^n, where n is a multiple of the
period length) to infinite radicals, to continued fractions,
to x^x^x^... = 2, etc. Of course, you need to know the limit
exists to apply this method, or you might find yourself
getting 1 + 2 + 4 + 8 + 16 + ... = -1. [If S is this sum,
then 2S = 2 + 4 + 8 + 16 + ... = S - 1.] Even just in the
case of how it works for periodic decimals, you can get
the formula for the sum of a convergent infinite geometric
series. [If S = a + ar + ar^2 + ..., then S - rS = a,
and hence S = a/(1-r).]

Dave L. Renfro

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