chapotraphouse

if 0.333... = 1/3 then 0.999... = 3/3 = 1

The thing that really got 0.999...=1 to click on my mind is the fact that you can't find a number between 0.999... and 1. You might think "Just put something at the end of 0.999...," but there is no end to 0.999....

and wants to know why 0.999... = 1

\begin{align} 0.999.... &= 9\cdot(0.1+0.01+0.001+... ) \ &= 9\cdot( 0.1^1 + 0.1^2 + 0.1^3 + ... ) \ &= 9\cdot(\sum\limits_{k=1}^\infty ( \frac{1}{10^k} ) ) \ &= 9\cdot(\sum\limits_{k=0}^\infty ( \frac{1}{10^{(k+1)}}))\ &= 9\cdot(\sum\limits_{k=0}^\infty \frac{1}{10}*(\frac{1}{10^k})) \ &= \frac{9}{10}\cdot (\sum\limits_{k=0}^\infty (\frac{1}{10^k})) \ &= \frac{9}{10}\cdot \frac{1}{(1-(\frac{1}{10}))}\ &= \frac{9}{10}\cdot \frac{10}{9} = 1 \end{align}

The crux rests on a handy result on from calculus: the sum of an infinite geometric series looks likes s = 1/(1-r), when s = \sum\limits_k=0^inf r^k, and |r| < 1.

Sorry for the latex. When will hexbear render latex? This is a bit more readable:

(aesthetic edit for our big beautiful complex analysts)

0.999..=1 because it's funny that people get so mad about it

He's hungry

Feed him

I suppose I will post one myself, as I do not expect anybody else to have that one in mind.

The decimals '0.999...' and '1' refer to the real numbers that are equivalence classes of Cauchy sequences of rational numbers (0.9, 0.99, 0.999,...) and (1, 1, 1,...) with respect to the relation R: (aRb) <=> (lim(a_n-b_n) as n->inf, where a_n and b_n are the nth elements of sequences a and b, respectively).

For a = (1, 1, 1,...) and b = (0.9, 0.99, 0.999,...) we have lim(a_n-b_n) as n->inf = lim(1-sum(9/10^k) for k from 1 to n) as n->inf = lim(1/10^n) as n->inf = 0. That means that (1, 1, 1,...)R(0.9, 0.99, 0.999,...), i.e. that these sequences belong to the same equivalence class of Cauchy sequences of rational numbers with respect to R. In other words, the decimals '0.999...' and '1' refer to the same real number. QED.

I will be posting another another proof, one using the assumption that 0.999... is the sum of the series 9/10+9/100+9/1000+... (and that 1 is the sum of the series 1+0+0+0+...), unless somebody else comes here and feeds it to the little guy first.

(I haven't done real math since forever)

Part 1: Proving If x,y are distinct positive real numbers and x < y, there is a z in R such that x < z < y.

Let y = x + e, where e is in R. e is also positive because:

x < y
x < x + e
0 < e

x < z < y becomes x < z < x + e. We can then choose z = x + 0.5e so that

x < z < y
x < x + 0.5e < x + e
0 < 0.5e < e
0.5e - 0.5e < 0.5e + 0 < 0.5e + 0.5e
-0.5e < 0 < 0.5e

Since we've shown earlier that e > 0, -0.5e < 0 < 0.5e is true no matter what e.

Part 2: Decimal Notation

Decimal notation is written as the sequence a~m~a~m-1~...a~1~a~0~.b~1~b~2~...b~n~, where a,b are in the set {0,1,2,3,4,5,6,7,8,9}. The sequence a~m~a~m-1~...a~1~a~0~.b~1~b~2~...b~n~ itself is of the number a~m~10^m^ + a~m-1~10^m-1^ + ... + a~0~10^0^ + b~1~10^-1^ + b~2~10^-2^ + ... + b~n~10^-n^.

Part 3: The actual proof through proof by contradiction

Let x = 0.999... and y = 1. This means there exists a z that is between 0.999... and 1. So let's construct z = a~0~.b~1~b~2~...b~n~. a~0~ has to be either a 0 or 1 and since there is no number smaller than 1 with 1 as its first digit, a~0~=0. z = 0.b~1~b~2~...b~n~

0.999... = 9*10^-1^ + 9*10^-2^ + 9*10^-3^ + ... while z = b~1~10^-1^ + b~2~10^-2^ + ... + b~n~10^-n^. If 0.999... < z, then

0.999... - z < 0
9*10^-1^ + 9*10^-2^ + 9*10^-3^ + ... - (b~1~10^-1^ + b~2~10^-2^ + ... + b~n~10^-n^) < 0
(9*10^-1^ - b~1~10^-1^) + (9*10^-2^ - b~2~10^-2^) + (9*10^-3^ - b~3~10^-3^) + ... < 0
(9 - b~1~)*10^-1^ + (9 - b~2~)*10^-2^ + (9 - b~3~)*10^-3^ + ... < 0

But as we've established in Part 2, b~1~, b~2~, b~3~, etc have to be from the set {0,1,2,3,4,5,6,7,8,9}, meaning (9 - b~n~) > 0 for any n. Therefore, (9 - b~1~)*10^-1^ + (9 - b~2~)*10^-2^ + (9 - b~3~)*10^-3^ + ... cannot be less than 0.

The theorem requires for x and y to be distinct positive real numbers and x < y, and since 0.999... and 1 can be trivially shown to be positive real numbers, this means that 0.999... and 1 are not distinct. In other words, 0.999... = 1.

Alright, so, the other proof that I promised:

If we define 0.999... as the sum of the series 9/10+9/100+9/1000+..., then for every neighbourhood U(1) it is true that there exists a metric ball B_N = B(1, 1/10^N), where N is natural, such that B_N is a subset of U(1).

For all natural n > N it is true that d(sum(9/10^k) for k from 1 to n, 1) = |1 - sum(9/10^k) for k from 1 to n| = |1/10^n| = 1/10^n < 1/10^N, meaning that for all natural n > N it is true that sum(9/10^k) for k from 1 to n is in B_N, meaning that it is also in U(1).

However, sum(9/10^k) for k from 1 to n is the nth partial sum of the series 9/10+9/100+9/1000+..., which, together with the fact that every such sum is in U(1) for n > N, means that 1 is the limit of the sequence of the partial sums of the series 9/10+9/100+9/1000+..., meaning that 1 is the sum of that series. That means that 0.999... is 1 by definition.