>>12061580

Also, it should be clear that '0.000...1' means a bunch of 0, and THEN a 1.[the sequence is of order type w+1] '0.999...' means a bunch [countably infinite] number of 9's, there is no 'after'. [sequence of order type w] Summing these in the only meaningful way would give 0.999... 1 [again, this is assuming 0.000...1 means anything which there is no system I am aware of that it does]. As you might have noticed, this does not produce 1.

...And to that, it should be noted that '0.000...1' has no meaningful definition. For a positive real number r, let a_0 be largest integer not 0 such that. r-a_0>0, then a_1 largest integer such that r-a_0-a_1/10>0, and so forth. This defines a sequence (a_1,a_2,...) called the decimal expansion. So, you have the 'decimal' (0.000...1), for what value of n in the sequence do you get this 1? If you were to actually carry this out for 1-0.999..., you would get (0,0,0,0,...), which, as you may have noticed, is just a string of 0's. And, the infinite sum a_0+a_1/10+a_2/100+... is equal to r, as the limit of the partial sums (the definition of infinite series) is r by the above algorithm. So the real number defines the decimal and vice versa-so since we do this for 1-0.999... and get decimal (0,0,0,...), what real number do you think we'll get? I am putting my money on... 0.

Also, 0.333... is 1/3 and it always has been and always will be. It's you whining against pretty much the most obvious concepts of mathematics. The left is infinite series with terms 3/10^n, which sums to... 1/3. What do you think it sums to, it is a geometric series so should converge to some number, if it isn't 1/3 done by basic plug and chug, what is it? Again, learn real analysis

Anyway you cut it, you are wrong from practically every angle imaginable