Math Is Fun Forum

that they repeat indefinitely?

Yes indefinite.

But not infinite.

These two are different.
How different? Back to my previous post, the former indefinite, the latter infinite.

I think there is misunderstanding that real numbers are already steady thing, dude. Actually, how many pi's decimals can be reached is refreshing but yet not arriving at infinite. As long as my method can peg with the decimals we reach so far for any real number, in a way it has count all the reals we know, or the knowable part of reals.

One easy way to generate a power set of a set of n elements is simply

list 0,1,...2^n-1

and then transfer them from mod 10 to mod 2
0,1,10,...,11(n)1

and then translate the binary sequence as turn on's for each element
0 is null, 1 is only the first element is in, 10 is the second one in, other excluded, ... the last is all in.

maxthon is pretty cool. You just need to drag a gesture on mouse while pressing right button to close a tab.

Thank you careless. That's kinda hard.

I encountered UBS test, with lots of tough graphic puzzles (similar to the puzzles at the end of a common iq test). It seems I passed although I answered less than half. Or maybe I didn't. They sent a rejection letter after one month. Is it because they find some better candidate through resume or they filtered me out by iq after 1 month?

But it seems illigal for a US company to test iq of its applicant. Then it makes sense why US lacks such books...

You know a lot ganesh

Yes luca, but that was previous me. Someone whose name begins with D reminded me of thinking all the decimals as a whole, altogether, and imagine it is reached infinite amount. Since then I adopted his idea and found when there is a whole, there is an infinitesimal or infinitesimals within it and it is at the right end and its backwards.

However, this is not only to you, luca. All set theorists and real number believers took one by one counting approach since Cantor named infinite set. And they are handicapped by counting method to discover the very logic flaw in infinite sets.

but we as humans can't get to it because we only live a finite amount of time.

Luca, you don't understand.
Because human cannot reach the end of 0.999... by counting one by one, then the end of 0.999... does not exist?
By the same reason, if humans cannot count 0.999... all over, does it mean 0.999... does not exist?
Because humans cannot finish the recursive division 1/9 under finite time, does it mean 0.111... does not exist?

Counting is only one method, and it is not sufficient at all even to rigorously show the very existence of infinitesimals. Since infinitesimals need imagination beyond empirical counting, it is very fair to examine it by logic beyond counting.

BTW, the "proof" of ∞=∞+1 is fake, or two proofs, one is mapping, the other is stating larger than finite.

for the former one
0.999...->0.0999... every 9 has already done one'o'one mapping
and +0.9 strictly adds in one more 9 to the latter
the only reason now to equate the 0.999... and the new 0.9999...
is by their similarity or unexaughstive counting, by simply looking at
0.9 0.99 0.999 ...
0.9 0.99 0.999 ...
and say since first second and third 9's are the same, and the examine process can go on beyond our human's ability to count all over
the 9's in the two are the same amount.

such finite "examination" is plausible because this method itself is defected since it is not done at all. (Have it examined "all" digits for either 0.999...?) You cannot force us to testify someone innocent by stressing s/he was not seen in the crime scene. And another piece of evidence might be sufficient to nail down the case.

The only exhaustive method is by wholistic mapping
when you do the division by 10 to 0.999...
every digit is already assumed moving rightwards by one digit
(or if you try to move the 9's one by one, you never can finish the division in your life
and you should not claim you can do the division in your life if you are rigorous),
I want to mention here not a single one more 9 is created or less 9 is diminished in this division
otherwise it is not defined as "every" or "all" (however it can be relaxed if Ricky you want)
then add in one more 9
yes just the one more 9 to compose 0.9+0.999.../10
and now we have already created ∞+1>∞
and then you can use my proof to just locate where the one more 9 is
yes the one more 9 is 0.9
but taking it at a different angle, suppose the 9's representing the same 9*10^-r play the same role and be treated the same in both 0.999...'s and from previous knowledge we know there is one 9's difference,  one not and one has, the former can be filled with 0.
And this pair of 0 and 9 will not have a succeeding 9 'n' 9 pair to compose the paradox
...09
...99
Logically it follows this pair is on the right end. A 9 has no 9's on its right is clearly the right end of a bunch of 9's

Simple,huh? And I did not use countive mapping to try to conclude not enough evidence for the last 9. Additional evidence has already been provided to pin it down.

appendix
"I want to mention here not a single one more 9 is created or less 9 is diminished in this division
otherwise it is not defined as "every" or "all" (however it can be relaxed if Ricky you want)"
Now I relax it, but could anyone tell me which 9 can be diminished? The 9 at the right end? or the 9 without another 9 on its right? Still, the answer leads directly to the 9 at the end.

The reason why is that the proof is in that essay I submitted.