Quote (Hammer_Hdin @ Jul 9 2010 08:43pm)
I still think that 0.999... = 1, but there is one thing which I am uncertain of.
If you add 0.999... + 0.333... you get 1.333... with a 2 on the end, but this 2 does not exist because it is after an infinite amount of 3's. If you added it up a thousand, a million or a billion times even, the number at the end would never get near infinity. But let's just say that you have this equation: 1 - 0.999... = 0.000... 1 = 0
The concept of infinity states that any number after an infinite series does not affect the number, so the 0.000...1 is actually 0.
But let's say we had (1-0.999...)+(1-0.999...)+(1-0.999...)+(1-0.999...)+(1-0.999...) = 0.000... 5, but once again the 5 is ignored because infinity - 1 digit = infinity.
But what if it was (1-0.999...)+(1-0.999...)+(1-0.999...)... meaning you add the numbers in brackets infinity times.
Would the "error" on the end of the infinite series eventually end up so large (infinity digits) that it would replace the numbers that are before the infinite series, and the series itself?
I suppose it sort of revolves around the concept of infinity - infinity = undefined.
Please explain.
(1-0.999...)+(1-0.999...)+(1-0.999...) + ... equals 0
You add an infinite amount of 0, which equals 0.
0,9999... is the same thing as a limit
(x->1)x
And (1-0,999) or 1 - limit
(x->1)x which equals 0.
And 1 - 0,999... is equal to limit
(x->0)x, which equals 0.
Your equation does equal 0 and you can do it with integrals.
Sygma (x=0) to (x = infinite) of limit (x->infinite) 1/x
You transform it into an integral (which I don't want to do), and it will equal 0.
Btw, limit (x->infinite) 1/x is the same thing as 1 -0,999;...
But the addition of an infinite amount of numbers that is not 0 can equal to a number that is not infinite. This is where maths starts to be counterintuitive.
This post was edited by Harmonium on Jul 9 2010 08:11pm