1/9 = .111... (1 repeating forever)
9 *.111...= .999... (9 repeating forever)
9 * 1/9 = 9/9 = 1
Therefore .999... = 1
Agree or disagree?
2006-09-06 04:51:44 · 19 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
agree. though not based on your proof. another example of mathematical proof that 0.999... = 1
10*0.999... = 9.999...
9.999... - 0.999... = 9
9/9 = 1
QED: 1 = 0.999...
the general form of this equation is as follows:
10 * X = 10X
10X - X = 9X
9X/9 = X
since X at the beginning is 0.999... and at the end is 1, they both must equal each other. this is observable in the math of physics as well. if 1 and 0.999... are not equal then there's alot of equations that don't work
2006-09-06 04:59:07 · answer #1 · answered by promethius9594 6 · 0⤊ 2⤋
Look, I found this as hard to believe when I heard it as all the posters who disagree. There is no real debate as to the truth of this, and those who do debate it debate the truth of the proofs of ALL infinite series. No reputable mathematician will claim that .999... != 1. However, you have to understand the nature of infinity to understand how .999999.... = 1. Infinity does not ever stop at a certain point. If it did, then all the people disagreeing would be correct, however, it does not.
This is an infinite geometric series with a = 9 and r = 1/10. By the theorems regarding infinite geometric series, if r < 1:
(SUM from k=0 to infinity of a*(r)^k) = (ar / 1 - r). In this case, the sum = [(9 * 1/10) / (1 - 1/10)] = [ (9/10) / (9/10) ] = 1.
You need to have taken the calculus of infinite series to understand this. If you need more convincing, carefully read the link below and the discussion tab. Again, before I took calculus, and even while I was taking it, I found this very hard to stomach, but the proof is rigorous and it will eventually make sense.
2006-09-06 14:32:36 · answer #2 · answered by TOB 3 · 2⤊ 1⤋
It may seem contrary to logical thought, but 0.999... (repeated infinitely) does equal _exactly_ 1. It doesn't "approach 1", or have a "limit of 1", it is exactly 1. Several methods have been shown above which prove this mathematically, in no uncertain terms.
They key here is that the decimal repeats infinitely. If it did stop or digress at some point, then it would be less than one. But an infinite series can, and in this case does, equal a specific non-infinite number.
Other examples:
0.333... (repeated infinitely) = 1/3. It does not approach 1/3 or have a limit of 1/3, it equals 1/3. So 0.333... * 3 = 0.999... = 1/3 * 3 = 1.
1/4 + 1/(4^2) + 1/(4^3) + 1/(4^4) + 1/(4^5) + ... = 1/3. Again, it does not approach 1/3 or have a limit of 1/3, it equals 1/3. So 9/10 + 9/(10^2) + 9/(10^3) + 9/(10^4) + 9/(10^5) + ... = 9/9 = 1.
Both of these are proofs that 0.999... = 1.
2006-09-06 06:10:52 · answer #3 · answered by stellarfirefly 3 · 4⤊ 2⤋
If .999 is equivalent to 1 I would like to know how because I can't believe how it could be. It doesn't matter how many 9s you put on the end of .999... it would still not equal 1. There can be no possible way so I think now you would have gathered that my answer is disagree.
2006-09-06 07:18:55 · answer #4 · answered by Elliot The Runescape Master 2 · 1⤊ 3⤋
the series that represent .999... has sum of 1, therefore I agree with it. In fact, any number like a.999... = a + 1
2006-09-06 05:09:42 · answer #5 · answered by Pedromdrp 2 · 1⤊ 2⤋
Disagree. Read following link for reasons. Someone has a PhD in physics and claimed that "a lot of equations just don't work if this is not true" [sic]
Well, I rest my case about academics - he/she is a liar and if indeed he/she has a PhD, then it just goes to show how stupid academics really are!
0.999... is STRICTLY LESS than 1.
The Dutchprof says that by definition an infinite sum is equal to its limit - this is untrue. There is no such thing as 'by definition'. Whenever academics are trapped, they try this trick as a last resort. Sorry, no 'by definitions' acceptable.
Look, mathematics works just fine without this falsehood. It is counter-intuitive and not required.
Beware of whatever you read on Wikipedia!!
2006-09-06 12:12:13 · answer #6 · answered by Anonymous · 1⤊ 2⤋
Agree.
By definition, a decimal fraction is a sum like
.9999 = 9*10^(-1) + 9*10^(-2) + 9*10^(-3) + 9*10^(-4)
An infinite decimal fraction is the infinite sum
.9999.... = SUM 9*10^(-i) i from 1 to infinity
which in turn by definition is equal to
LIMIT SUM 9*10^(-i) where n -> infinity and i from 1 to n.
Now SUM 9*10^(-i), with i from 1 to n, is equal to
1 - 10^(-n)
therefore
.9999.... = LIMIT (1 - 10^(-n)) = 1 - LIMIT 10^(-n) n to infinity
which is equal to 1 - 0 = 1.
2006-09-06 07:25:40 · answer #7 · answered by dutch_prof 4 · 3⤊ 2⤋
it's freakish.
another way to show that they're equal -
x = .9 repeating.
multiply both sides by 10: 10x = 9.9 repeating.
subtract x from both sides: 9x = 9.
divide both sides by 9: x = 1.
therefore: .9 repeating = 1.
2006-09-06 05:04:19 · answer #8 · answered by Moxie1313 5 · 4⤊ 3⤋
Disagree...... .999 approaches 1 but never actually equals 1.
2006-09-06 07:20:14 · answer #9 · answered by Tracy J 1 · 2⤊ 3⤋
Yes
2006-09-06 05:21:07 · answer #10 · answered by tjoyce71 1 · 2⤊ 1⤋