why does 1/3 not "really" equal .333... ? explain this NOW
>>16869408[math]0.4_{12}[/math]
>>16869408because it's missing the last little bit on the end, after the infinity
>>168694081/3 by definition is the least upper bound of the set {0.3,0.33,0.333,...}
trisect 1, fool!
>>16869413>>16869420OKAY SO .999... = 1 BUT .333... CANT EQUAL 1/3 COOL
because 3rds don't fit naturally into 10s, obviously
Start at 0 on the number line and move 3/10 to the right, then 3/100 to the right, then 3/1000 to the right, and so on. Some numbers will eventually be to your left if you take enough steps. Some numbers will never be to your left no matter how many steps you take. 0.333... is by definition the number at the boundary, and that number is 1/3.
>>16869695For those who want to know more:Consider the set of numbers {0.3, 0.33, 0.333, ...} that you can move to with enough steps.>Some numbers will never be to your left no matter how many steps you take.We call one of these numbers an upper bound on the set.>Some numbers will eventually be to your left if you take enough steps.These numbers fail to be upper bounds on the set.With some thought, you can see that the boundary between the upper bounds and the numbers which are not upper bounds must belong to the upper bounds. So this number is called the least upper bound, or supremum for short. We define 0.333... as the supremum of {0.3, 0.33, 0.333, ...}.
remember to stick with explanations within [math]\mathbb{Q}[/math] rather than attempting to use [math]\mathbb{R}[/math], since it's presence within [math]\mathbb{R}[/math] simply comes from it being in [math]\mathbb{Q}[/math]
>>16869408>why does 1/3 not "really" equal .333... ?It does.1/3 really does equal 0.333...
>>16869408>why does 1/3 not "really" equal .333... ? explain this NOWBecause "infinite sum" is an invalid concept and convergence is pure cope.
>>16870045Done. See: >>16867365
>>16870128cool!
>>16870128have these for posterity1/2
>>168701282/2
>>16869408It does.[math]9\cdot \frac{1}{9} = 1[/math][math]9\cdot \lim_{n\to \infty}\sum_{k=1}^{n}10^{-k} = 1[/math][math]\lim_{n\to \infty}\sum_{k=1}^{n}3\cdot 10^{-k} = \frac{1}{3}[/math]This argument works using limits in Q.
