Can you do it by subtracting x from the left side only and subtracting 0.999 from the right side only.
The question really is, can you do an operation on one side of the equation without doing the same thing to the other side.
2006-07-18 11:17:49 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The general answer is "NO".
Not clear what you are trying to do. It appears you are mixing the issue of how many repeating "9"s are needed to equal a 1.0 with questions about operations.
What you can do is divide both sides by 10 to give you x = 9.999+/10. Then assuming 9.9999 forever is the same as 10 you would simplify the 9.9999+/10 to equal 1. So that x = 1.
Once you have x=1 you can multiply both sides by 9 and get 9x = 9.
So you can conclude that 10x = 9.999999+ is equivalent to 9x = 9.
2006-07-18 11:21:41 · answer #1 · answered by Alan Turing 5 · 0⤊ 0⤋
No, you have to do the same thing to both sides of the equation. Also, since you don't know what "x" is until the problem is solved, you don't want to subtract x from one side and a number from the other.
Just to complicate things, if the ".999" is really ".9" with an infinite number of nines after it, then x=1 because .999... is equal to 1. Proof is as follows:
x = .999...
10x = 9.999...
10x - x = 9.999... - .999...
9x = 9
x = 1
I hope this helped!
2006-07-18 11:23:02 · answer #2 · answered by DAC 2 · 0⤊ 0⤋
If you're referencing a particular proof of .9999... = 1 (as I think you are), than in this context, yes. You left out one key detail: in this proof you start with let x = .9999... Thus, 10x = 9.9999..., and (10x - x) = 9x = 9.9999... - x = 9.9999... - .9999... = 9.
This proof seems flawed to me, though. On the left side of the above equation (10x - x = 9x) you're assuming that x =1 (or else 10x - x does not equal 9x), which is a very familiar and intuitive result, but contradicts the initial statement (let x = .9999...). Isn't this circular reasoning? You're using the answer to prove the answer.
The fact that .9999... = 1 can be proved in a variety of different ways, but the more rigorous proofs use limits or infinite series.
2006-07-18 11:26:35 · answer #3 · answered by Anonymous · 0⤊ 0⤋
No, because if 10x=9.999, then x=.9999, and .9999≠.999.
By the way, if you're wondering about the recent line of questions regarding whether .9999...=1, do keep in mind that 10*.9999... is 9.9999.... and not 9.9999...0. The reason for this is that to append the zero at the end of the number, there would have to be an end of the number to append it to, and this is not true of an infinite decimal expansion. Thus, it is valid to write x=.9999... → 10x=9.9999... → 10x-x=9 (because x IS equal to .9999...) → 9x=9 → x=1.
2006-07-18 11:26:16 · answer #4 · answered by Pascal 7 · 0⤊ 0⤋
No you have to do it on both sides unless your just changing the form of the fraction or something. For example: 50/10 is the same as 5.
2006-07-18 11:23:14 · answer #5 · answered by Anonymous · 0⤊ 0⤋
no. if you do something to one side you have to do it from the other. you would have to subtract x from both sides. the answer to this is simply 9.999 divided by 10 (divide each side by 10) which comes out to x = .9999
2006-07-18 11:23:10 · answer #6 · answered by coltsfan3874 4 · 0⤊ 0⤋
Yes, because x = 0.999, so you are subtracting equal quantities from bots sides.
2006-07-18 12:15:50 · answer #7 · answered by menezes_dean 2 · 0⤊ 0⤋
It might work this time, but It won't happen all the time. You should get in the habit of doing it the right way; and that is : Whatever you do to one side, you have to do to the other.
2006-07-18 11:22:39 · answer #8 · answered by I ♥ men in uniform 5 · 0⤊ 0⤋
only if the .999 is understood to be repeating.
2006-07-18 11:20:00 · answer #9 · answered by Anonymous · 0⤊ 0⤋
no. what's done on the one side has to be done on the other side.
2006-07-18 11:22:17 · answer #10 · answered by jade 3 · 0⤊ 0⤋