if: x=.999....
then
10x(9.999...)-x(.999...) equals 9x
so
x=1
so then 1=.999... what's the deal?
2006-09-26 18:14:19 · 9 answers · asked by john s 1 in Science & Mathematics ➔ Mathematics
KKK- I should have noted that the parenthese were meant to be grammatical, not mathematical. sorry.
2006-09-28 17:53:29 · update #1
.999... DOES equal 1. In math, if you have any two numbers, you can always find another number between them. Well, if that's true, then if you have two numbers and there is NOTHING between them, then the two numbers must be the same thing.
That's the case with .9999... Try to find a number closer to one then .999.... You can't. Since there is no number between .999... and one, they must be the same thing.
2006-09-26 18:21:10 · answer #1 · answered by garyhorne55 1 · 1⤊ 0⤋
You should use calculus, although many will argue .9 recurring equals 1 in reality. In the example you displayed, it is the LIMIT, not the actual value for which x = 1.
To prove this point, suppose you have a piecewise function
_______1, x < 1
f(x) = 2, x >= 1
when x=.9 recurring, f(x) = 1 because 1, and only 1, is not included in the first step of the piecewise. However, the limit as x approaches 1 is 1, therefore the limit .9 recurring equals 1. This is not possible algebraically, yet it is possible using limits.
2006-09-27 01:27:15 · answer #2 · answered by Ben 2 · 1⤊ 0⤋
The equation in the question is wrong, follow the below steps:
10x(9.999...)-x(0.999...)= 9x
Since x=0.999..., replacing 0.999... with x we get,
10x(10x)-x(x) = 9x
(10x)^2 - x^2 =9x
(10x-x)(10x+x)=9x
(9x)(11x)=9x
dividing each side with 9x, we get
11x = 1
dividing each side with 11, we get
x = 0.090909...
which is against the initial equation of x =0.999...
Therefore, the equation to start with is incorrect.
The use of limits does not apply in this case.
2006-09-27 01:40:17 · answer #3 · answered by Kanti Kiran Kare 1 · 0⤊ 0⤋
0.9 recurring equals 1.
When people first learn about recurring numbers, they think that there is some infinitesimally small difference between 0.9 recurring and 1, but there isn't.
2006-09-27 01:17:31 · answer #4 · answered by tgypoi 5 · 1⤊ 0⤋
tgypoi is correct.
Limit of 0.9 recurring is 1
2006-09-27 01:21:49 · answer #5 · answered by astrokid 4 · 0⤊ 0⤋
Best demonstration of the concept I've seen. (maybe I was sleeping in algebra class that day)
2006-09-27 01:26:09 · answer #6 · answered by Helmut 7 · 0⤊ 0⤋
0.999.... really is equal to 1. you can't find a number between 1 and 0.999...... and if 0.3333...=3/9=1/3, hte 0.999...=9/9=1.
2006-09-27 03:02:10 · answer #7 · answered by woof! 2 · 0⤊ 0⤋
im doing similar stuff and i reely dont get it
2006-09-27 01:22:07 · answer #8 · answered by Anonymous · 0⤊ 0⤋
This is true
Ana
2006-09-27 01:17:24 · answer #9 · answered by MathTutor 6 · 0⤊ 0⤋