/sci/ - Science & Math

>>15672380
Firstly,
0.999... = sum[k = 1 -> inf] 9/(10^-k)
Secondly,
if sum[k = 1 -> n] expr > x and expr > 0 for all n
then sum[k = 1 -> inf] expr > x
and lastly,
if sum[k = 1 -> inf] expr > x and expr > 0 for all n
then some n exists such that sum[k = 1 -> n] expr > x

If we have agreed so far:

subproof:
T(n) ≡ sum[k = 1 -> n] 9/(10^k) = 1 - 10^-k
base n=1:
0.9 = 1 - 0.1
step T(n) -> T(n+1)
sum[k = 1 -> n] 9/(10^k) = 1 - 10^-k /+ 9*10^-(k+1)
sum[k = 1 -> n+1] 9/(10^k) = 1 - 10^-k + 9 * 10^-(k+1) =
= 1 - 10 * 10^-(k+1) + 9 * 10^-(k+1) = 1 - (10 - 9) * 10^-(k+1) =
= 1 - 10^-(k+1)
By proof of induction we have shown that:
sum[k = 1 -> n] 9/(10^k) = 1 - 10^-k for all n element N.

Furthermore, let's assume that 0.999... < 1. Then there exists some x from [0, 1>
such that sum[k = 1 -> inf] 9/(10^-k) is not greater than x (so equal or less).
That means that for all n element N, sum[k = 1 -> n] 9/(10^k) < x.
Using the subproof, this implies that for this x, there is no n element N such that 1 - 10^-n >= x.
1 - 10^-n >= x /-x + 10^-n
1 - x >= 10^-n / log10
log10(1-x) >= -n

Since x is from [0, 1>, 1-x is from <0, 1], and for all such numbers log10 is defined, so its always possible
to find a natural number n that satisfies the inequation.
Therefore, sum[k = 1 -> inf] 9/(10^-k) >= 1.

Using the last premise from the top of the post,
it is trivial to show that sum[k = 1 -> inf] 9/(10^-k) <= 1
(there is no n such that the sum is greater than 1).

Therefore, sum[k = 1 -> inf] 9/(10^-k) = 1

>>15678342
> Let's assume a thought experiment in which you were given the gift of perfect deduction
> Since you know what the best decision is in every situation (in accordance to your morals and goals), there is no reason to do anything that is not that decision (or in the set of best decisions if there are multiple)
> Therefore, you are predictable

Determinism isn't really a prison. And since nothing inside the system can observe it as a whole, there are no practical effects of it either.