This little guy craves the light of knowledge and wants to know why 0.999… = 1. He wants rigour, but he does accept proofs starting with any sort of premise.
Enlighten him.
(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 amam-1…a1a0.b1b2…bn, where a,b are in the set {0,1,2,3,4,5,6,7,8,9}. The sequence amam-1…a1a0.b1b2…bn itself is of the number am10m + am-110m-1 + … + a0100 + b110-1 + b210-2 + … + bn10-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 = a0.b1b2…bn. a0 has to be either a 0 or 1 and since there is no number smaller than 1 with 1 as its first digit, a0=0. z = 0.b1b2…bn
0.999… = 9*10-1 + 9*10-2 + 9*10-3 + … while z = b110-1 + b210-2 + … + bn10-n. If 0.999… < z, then
0.999… - z < 0
9*10-1 + 9*10-2 + 9*10-3 + … - (b110-1 + b210-2 + … + bn10-n) < 0
(9*10-1 - b110-1) + (9*10-2 - b210-2) + (9*10-3 - b310-3) + … < 0
(9 - b1)*10-1 + (9 - b2)*10-2 + (9 - b3)*10-3 + … < 0
But as we’ve established in Part 2, b1, b2, b3, etc have to be from the set {0,1,2,3,4,5,6,7,8,9}, meaning (9 - bn) > 0 for any n. Therefore, (9 - b1)*10-1 + (9 - b2)*10-2 + (9 - b3)*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.
so, start with:
x = 0.999…
now multiply each side by 10
10x = 9.9999…
now we subtract x from the left side, and 0.999… from the right, which is fine because they are equal:
9x = 9
and from there it should be fairly obvious that x is also equal to 1, which means 0.999… is also equal to 1
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…
Yeah that’s something that people have to get used to in maths, if the limit of a sequence exists we can just pretend to have “reached infinity” and work with like any number
I’m not sure what you are trying to say here, and I have a background in math. I think this is just going to confuse lay people.
Im trying to make the distinction between a function that approaches a value as it’s input grows, for instance a sequence seen as a function on the domain of the natural numbers, and the value itself.
I have seen a lot of people view 0.999… as a number that “approaches one”, so formally speaking as the sequence (0.9, 0.99, 0.999, …) and not the number itself which that sequence approaches which they would agree is 1.
The “its the largest number which is less than 1” type of thinking.
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)
No. 0.99 is 0.9+0.09. The proof I gave shows that 0.99999999999999999999999999999999999(…) is equal to 1.
