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.
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.
The ellipses mean infinity. Anything less than infinity and the equation is false. It looks unintuitive because infinity is unintuitive
The ellipses mean infinity
No, it does not. The ellipsis notation is used to denote a repeating pattern. It is not used to denote points that are named ‘infinity’ in spaces like Aleksandrov compactification and the extended space of real numbers.
I am sorry, but your reply lacks rigour.
Is this a bit? I meant that the ellipses, in this context, mean that the nines go on forever. Obviously there are other symbols used to represent similar ideas, like ∞
I’m being serious. The word ‘infinity’ refers to some points from spaces like Aleksandrov compactification. It does not refer to any sort of repetition of any pattern. You cannot, for example, write ‘0.999∞’ or ‘0.999infinity’ to mean the same thing as ‘0.999…’ or ‘0.(9)’.
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
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.
