2007-07-06 05:21:48 · 6 answers · asked by dwadefranchizemia03 5 in Science & Mathematics ➔ Mathematics
You just proved that .999... = 1.
1 = 1/3 + 1/3 + 1/3 = .33333.... + .33333..... + .33333.... = .999999.....
2007-07-06 05:24:40 · answer #1 · answered by triplea 3 · 0⤊ 1⤋
This has been asked in different ways.
Briefly, .9999...(continuous) = 1
You'll get a lot of differing opinions, but the key is that .999... does not have an ending point. As soon as you stop with a 9 an do not continue, the equality with 1 is no longer true.
The straightforward proof is let x = .9999....
Then 10 x = 9.999....
9x = 10 x - x = 9.999... - .9999... = 9
so 9x = 9, and x =1
2007-07-06 12:28:41 · answer #2 · answered by John V 6 · 1⤊ 0⤋
Even though I'm only 13 I think as fractions it depends on how many of that certain fraction you have before you can tell:
1/3 x 3 = 3/3
3/3 x 3 = 9/3 ( improper fraction ) = 3 ( wholes )
9/9 div. 3/3 = 1/3 .... I think ......
2007-07-06 12:33:27 · answer #3 · answered by Baseball_Girl 3 · 1⤊ 1⤋
Hi. You are dealing with an infinite series. When carried to infinity they difference becomes infinity small. This is the same as 0 as far as math is concerned. So, both.
2007-07-06 12:26:53 · answer #4 · answered by Cirric 7 · 1⤊ 0⤋
What are all those periods? Isn't 3 * (1/3) = 1? then why isn't your sum equal to 3?
Ron.
2007-07-06 12:26:09 · answer #5 · answered by Anonymous · 0⤊ 2⤋
1/3=.33 .33x3 + .33x3 + .33x3 =
2007-07-06 12:34:18 · answer #6 · answered by cinammon l 2 · 0⤊ 0⤋