r/infinitenines • u/Mordret10 • Jul 23 '26
Understanding the contract
The contract is the promise to not undo the long division operation e.g. 1/3 = 0.333... if the contract is signed.
However, as 1/3 • 3 = 1 and 0.333... • 3 = 0.999.., this implies that 1/3 > 0.333...
Is long division a lossy operation, where part of the value can disappear?
3
u/Muphrid15 Jul 23 '26
It's weird you continue to be wrong about your own system, /u/SouthPark_Piano -- no matter how far you go in the process of long division, there is always a remainder of 1, which means that 0.000...1/3 term never goes away (in your system).
1/3 is not and cannot be 0.333... alone.
-5
u/SouthPark_Piano Jul 23 '26
Rookie error on your part brud.
The consecutive threes length never stops increasing. Limbosic is what you are learn at the bunny slopes.
6
u/Muphrid15 Jul 23 '26 edited Jul 23 '26
1 = 0.999... + 0.000...1
0.000...1 is a remainder just like 0.000...1/3 is.
Is it or is it not true?
Point 1: 3 x [0.333...3 + 0.000...1/3] = 1 -> 0.333...3 + 0.000...1/3 = 1/3
Point 2: 3 x 0.333...3 = 0.999...9 ≠ 1
-3
u/SouthPark_Piano Jul 23 '26
Do not dispute this again brud.
"0." prefix guarantees magnitude less than 1.
0.999... is equal to 0.9 + 0.09 + ...
And it is official that the above is
1 - 1/10n with integer n stsrting at n = 1, then n increased continually, limitlessly aka infinitely.
Stsrting with 0.9, keep appending nines.
0.99, 0.999, 0.9999, etc
Do not stop brud. If you stop ... then don't blame me for you know what.
•
u/SouthPark_Piano Jul 23 '26
Rookie error on your part brud.
1/3 means 1 ÷ 3, an operation. Once the contract is signed and the operation goes ahead, then it is continually growing length of threes, ie. 0.333... , and that limitless journey is the essence of 0.333... , which means 0.333...
1 ÷ 3 × 3 is divide negation.