/sci/ - Science & Math

>>9388943
The irony being you probably actually do take meds. I haven't been sick in 7 years. Meanwhile you're festering in your own death waiting for the inevitable.

Keep trying 56%.

1/3 = 0.333[3
2/3 = 0.666[6
3/3 = 0.999[9
3/3 = 1
Therefore
1 = 0.999[9

this works when referencing the thirds up to the whole. Referencing thirds beyond the whole is different.

4/3 = 1.333[2

By motive of proving 0.999[9 = 1, adding 0.333[3 to 0.999[9 results in 1.333[2

1.333[2 - 0.999[9 should give us 0.333[3, which it does, and 0.999[9 × 2 = 1.999[8, but what is 1.999[9 then?

One note of providing acceptance for 0.999[9 = 1 is how close it is to 1, but 1.999[8 = 2 is less accurate. It would be more reasonable to say 1.999[9 = 2,
but 1.999[9 is not the result of 0.333[3 × 6. So if there is a number greater than the infinite sequence that is more accurate, this creates a problem of how to interpret the infinite sequence.

The infinite sequence is more than just infinitely repeating. It could be any number and will be any number by nature of calculating each successive digit. Even when it repeats a single digit, the potential for the next digit to be any other number remains.

by nature of truncating an infinite amount of decimal 3's from the fraction 1/3 as something like 0.333[3, we invoke this special property of infinity that increments an imaginary value to satisfy not only 0.333[3 × 3 = 0.999[9 = 1, but also 0.333[3 × 9 = 2.999[ㅌ = 3.
Disregarding the special property would make ㅌ = 8, yet defining the end limit as 8 would leave room for the larger number ㅌ = 9, meaning 2.999[8 is not actually as equal or more equal to 3 than 2.999[9 would be. This is where the special property of infinity can be utilized to fill in the gap between 2.999[8 and 3, giving us 2.999[9 towards 3 instead. Realistically, it would fill in the gap from the time of its first invocation at 0.333[3 where every increment of 1/3 also increments the value of the property defining the gap, thus 0.333[3 × 3 is actually understood as (0.333[3 × 3) + (ㅌ × 3)

0.333[3 × 3 = 0.999[9
(0.333[3 × 3) + (ㅌ × 3) = 1
1 - (ㅌ × 3) = 0.999[9
0.999[9 =/= 1
ㅌ × 3 =/= 0

>>7257158
whoever wrote that got the ordering wrong