# Rational Numbers

What rational number does the repeating decimal 0.9999...equal?

### Re: Rational Numbers

hmmm... "no gutta", since we are dealing with an ifinite (decimal)sequence. Me "gutta" the finite (partial) sum
x_n= 9.\sum__{k=1}^n 10^{-k},
or
x_n= 9.10^{-1} + 9.\sum__{k=2}^n 10^{-k} (1)
and then we take the limit
x= \lim_{n\to\infty} x_n.
Thus, we multiply (1) by 10^{-1} (as usual in geometric progressions),
10^{-1}.x_n = 9.\sum_{k=1}^n 10^{-(k+1)}=9.\sum_{k=2}^{n+1} 10^{-k},
or
(1/10).x_n = 9.\sum_{k=2}^n 10^{-k} + 9.10^{-{n+1)}. (2)
By subtracting (1)-(2), both sums cancel and we have
(9/10).x_n = (9/10) - 9.10^{-{n+1)}. (3)
Take the above limit and the right-last term in (3) yields to zero. (and x_n yields to x)
So x = 1.

### Re: Rational Numbers

I think you prove it in geometric progressions .That's OK.You are right.And my proof is right too,I think.

### Re: Rational Numbers

Equals 1.Prove it below:
Supposed x=0.9999...
then 10x=9.999...
subtract the first equation from the second equation
(10-1)x=9.999...-0.999...
9x=9
therefore x=1