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Originally posted by Sublime620
You start with reviewing limits. Then you have to master derivatives, a pretty large equation involving limits. Once you get that equation down forwards and backwards, the class gets a bit easy and you can start learning reverse derivatives.
Can you please post your proof that .333... + .333... + .333... is not equal to .999...?
Originally posted by politHowever, .99999999 repeating does _not_ equal 1. 1/3 + 1/3 + 1/3 = 1/1. Whoever said .333repeating + .333repeating + .333repeating equals .999999 repeating did not add up all of the decimal places!
But I can see how it is easily confused.
Originally posted by MrDead
.999999 is so close to 1 that it's really not important right? You'd go on for infinity and be 1 digit away. Same as .333333333333333 etc.
It's such a tiny amount that you could never reach, and nothing is perfect anyway.
Originally posted by OXmanK
And your first problem has a mistake.
10 * .999 = 9.99
10x - x = 8.991 not 9
Originally posted by American Mad Man
Wow - this actually has my attention now - and I usually HATE math
OK so taking 10x - x = 9x, how did you get to 9x = 9?
As I recall, you can manipulate both sides, so long as you do the same to both (although I do recall there being exceptions to this rule).
If you were to devide by x, that would give you:
(10x-x)/x = 9x/x
this would then become 10x/x - x/x = 9x/x, each "x" cancels the other out (x/x = 1), thus:
10 - 1 = 9
so I don't think that does anything.
Originally posted by Crysstaafur
10x=9.999 Multiplied by 10 on both sides.
10x - x = 9x Opps.... While the statement is true it is a non-connected system not in relation to the former operation...
alternate yet legal manipulation:
10x -9.999 = 9.999 - 9.999 (subtract by -9.999~ on both sides)
10x - 9.999 = 0 (simplified above operation)
(9x + x) - 9.999 = 0 (disunioned 10x into (9x+x))
9x - 9.999 = -x (subtracted both sides by x)
(9x-9.999)/-1 = (-x)/-1 (divided both sides by -1)
-9x+9.999 = x (simplified above operation) (result sign change)
(-9x+9.999)/9 = x/9 (divided by 9)
(-x+1.111) = x/9 (simplified above operation)
(-x+1.111)+x = x/9 + x (added x to both sides)
1.111 = x/9 + x (simplified above operation)
(1.111) * 9 = (x/9 + x) * 9 (multiplied both sides by 9)
9.999 = x + 9x (simplified above operation)
10x = 9.999 (re-added 9x + x which would be 10x)
(10x)/10 = (9.999) /10 (divided by 10)
x = 0.999 (final result matches *inits*)
still comes back in one peice.. the gotcha was in the above (marked opps) operation...
[edit on 24-3-2005 by Crysstaafur]