Yes. The fact that the 9's go on for ever means that the two are exactly identical.
Really, they are! None of this "approaches 1" stuff -- that's limit talk, and a number is not a series that has a limit.
The thing that causes problems for most people is the sensation that the number "almost equals" 1 but somehow doesn't quite "get there." But the number isn't "getting anywhere" -- it's not moving or changing its value in any way. Those 9's aren't being added on, one by one... they're all there, right now. The whole infinite string of 9's.
Look at this way: if the numbers aren't the same, then there must be some finite difference between them. (That's called the "density of the real numbers" -- if two numbers are not the same, there are other numbers between them.) So, how much is that difference? You can try to subtract them -- 1.000... - 0.999... = 0.000? but there's no place to write down that 1 you have the urge to put down, because there's no "last 9."
Once you think of it that way, and get away from the mental concept of "adding 9's one by one," you can begin to accept that they're the same.
Here are two pseudo-proofs that might help:
1/3 = 0.3333....
Multiply both sides by 3:
3 * 1/3 = 3 * 0.3333...
Simplifying,
1 = 0.9999...
Here's a better one, with a little algebra:
Let x = 0.9999...
Multiplying both sides by 10, all we need to do on the right side is move the decimal point by one place:
10x = 9.9999...
Now, subtract the first equation from the second:
10x - x = 9.9999... - 0.9999...
9x = 9.0000...
And, dividing both sides by 9,
x = 1.0000...
Hope that helps!
2006-07-25 06:44:19
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answer #1
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answered by Jay H 5
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This question comes up every few days. It always inspires a host of people saying that 'it is approximately 1' or 'it is not quite 1'.
I WISH THESE PEOPLE WOULD NOT ANSWER QUESTIONS IF THEY DON'T KNOW THE CORRECT ANSWER!
The infinite decimal .999..(repeating) is *exactly* equal to 1. It is all a matter of what the decimal expression means. It means the number that is the limit of the sequence of partial decimals:
.9, .99, .999, .9999, .....
The limit of that sequence is 1. That means the decimal expression equals 1. Period. No debate.
The point is that any finite decimal expression represents a number that also has an infinite decimal expansion with an infinite number of 9's. This is just one of the quirks of decimal expansions.
2006-07-25 07:02:48
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answer #2
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answered by mathematician 7
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YES!!!
if you believe that 9/9 (nine divided by nine) = 1.
follow the pattern:
1/9=0.11111111111111111111...
2/9= 0.2222222222222222222...
3/9=0.3333333333333333333...
4/9=0.4444444444444444444...
5/9=0.5555555555555555555...
6/9=0.6666666666666666666...
7/9=0.7777777777777777777...
8/9=0.8888888888888888888...
9/9=0.9999999999999999999...
if the nines go to infinity...then it has to equal 1.. the (infinity+1) digit is rounded to 10, carry the 1 it comes all the way back to 1.0
2006-07-25 06:49:20
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answer #3
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answered by kmclean48 3
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Yes, yes, a thousand times yes. There is no rounding involved, 0.9999... is exactly equal to 1. People seem to be confusing the decimal representation of a number with the number itself, and think something along the lines of "oh, these numbers have different decimal representations, therefore they must be different numbers." This is NOT true, any number with a terminating decimal expansion has an infinite number of different decimal representations, all of which are equivalent (for insance, 3 = 3.0 = 3.00 = 3.000 = 3.0000 etc., as well as 3 = 3.000... (repeating) and 2.999... (repeating)). Please consider the following:
A decimal number 0.abcd... (where a,b,c,d, etc. are decimal digits) is defined as a/10+b/100+c/1000+d/10000 and so forth. Now when there are an infinite number of digits in the decimal expansion, you have an infinite series. The sum of an infinite series is defined as the limit of the partial sums of the first n numbers as n approaches infinity. This is not something to quibble with, it is a matter of definition. And it is obvious that the limit of 9/10+9/100+9/1000+9/10000... is in fact 1.
Possible objection: "but the number .9999... only approaches 1, it doesn't actually get there"
Answer: .9999... is a single number, with a single value. It doesn't approach anything. Now, it is true that the partial sums themselves never equal 1, but so what? The number is not the same as any of its partial sums unless (ironically) you start rounding. We are not rounding, therefore the number is the result of the entire infinite sum, which is defined as the limit of the partial sums, and clearly these partial sums have only one limit, and that limit does not approach 1, it IS 1.
Alternatively, consider what would happen if.9999... were less than 1. That means that the difference (1-.9999...) would be nonzero. Call this difference D. Now, clearly, if there were any natural number N such that N*D>1, then .9999... + D > 1, and so there can be no such natural number. Let us call a nonnegative number with the property that there is no N such that N*D > 1 infinitesimal. Now, clearly the set of infinitesimals has an upper bound, such as 1. The real numbers are defined such that any set of umbers with an upper bound in the real numbers has a least upper bound in the real numbers (this is what distinguishes the reals from the rationals, which have upper bounds for things like "the set of all numbers x such that x^2 < 2, but there is no least upper bound (since the square root of 2 is not rational)). Clearly, then, the infinitesimals have a least upper bound, which may or may not itself be an infinitesimal. Call this upper bound B. If B is greater than zero, then B/2 < B, and therefore must be in the set of infinitesimals. However, for all N, N*4 is a natural number, and so if B/2*N is < 1 for all N, so too is B/2*N*4. But this means that 2B*N < 1 for all N, which means that 2B is infinitesimal, yet it is greater than the least upper bound B, which is a contradiction. Therefore, there are no nonzero infinitesimals. And since D is infinitesimal, D=0, which means that .9999.... = 1. Q.E.D.
Alternatively, note that we can perform arithmetcal operations on infinite decimal expansions, and thereby use algebra to find the number we are looking for. Let X=0.9999... Multiply both sides by 10 to get 10X=9.9999... Then subtract X from both sides to get 9X=9.9999... - 0.9999... = 9. 9X=9, so divide both sides by 9 to get X=9/9=1. Q.E.D. (Note: this method can be readily generalized to prove that all repeating decimal expansions correspond to rational numbers)
Objection: "but 0.9999... != 9.999..., because the last digit gets changed to a zero"
Answer: What last digit? This is an infinite decimal expansion. That means it doesn't end. There is no last digit to change to a zero.
Objection: "You can't perform arithmetical operations on infinite decimal expansions because subtraction is performed right to left, and there's no rightmost digit to start on"
Answer: As long as you keep in mind that some digits of your answer are provisional (because you might need to borrow or carry later), you can perform arithmeitcal operations just as sensibly from left to right as from right to left.
Objection: "You can't perform arithmetical operations on infinite decimal expansions because you can never get to the end"
Answer: It is no more necessary for me to complete an infinite number of operations to perform an arithmetical operation on an infinite decimal expansion as it is for me to make 100 additions to know that the sum of all natural numbers from 1-100 is 5050. If there is a regular pattern in the results, and I can show rigorously that this pattern must hold, I can use the pattern in the place of brute force and complete the operation in finite time.
Objection: "But that can't be! 1 is a rational number, and there are no pairs of integers whose quotient is .9999..."
Answer: Funny, I just got through showing that 9/9 is in fact .9999... (which may also be seen by multiplying the decimal expansion of 1/9 by 9 or 1/3 by 3). The misconception here is that every real number must have a unique decimal expansion, and that runningthrough the division algorithm by brute force produces that expansion. This is wrong: running through the division algorithm by brute force produces A decimal expansion of the number, but there is no guarantee that it be unique, and as I have shown, it is not. Ultimately, this is the real problem behind most of the "objections" - people have confused the decimal expansion of a number with the number itself, and think that because the decimal expansions are different that the number they represent must also be different. Once you realize that that is not the case, the mathematically well-established result that .9999... =1 will seem not only sensible, but even obvious.
2006-07-25 07:30:36
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answer #4
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answered by Pascal 7
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isn't that assertion, a million.00000....a million = a million only like asserting 10 = 13? I recommend, equivalent means precisely an identical. even although they are very very close at the same time, they are nevertheless 2 distinctive numbers. No evidence ought to ever substitute that fact. Math is stable logic subsequently. additionally on your evidence on the coolest, the line 10x = a million.000000....a million is fake 10x = a million.00000... a million (that being one decimal to the left) in actuality, a million.00000... a million isn't a real type. you elect a definitive like a million x 10^-one thousand multiply it by using 10 and you get 10 x 10^-999 it may well be much less complicated to work out in case you suggested 9x = 9.0000... 9 the completed equation you made relies upon on that mistake, with the numbers being wisely expanded you get a distinctive effect.
2016-12-10 15:23:25
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answer #5
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answered by Anonymous
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Yes. It is approximately 1 but always less than 1.
You will however, find proofs that use the completeness principle (incorrectly) to show that 0.999... = 1. In actual fact, the limit of the 'number' 0.999... = 1. There is an article on wikipedia with a huge debate on this. Why don't you go and read it?
2006-07-25 06:34:08
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answer #6
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answered by Anonymous
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yes it is 9 * 1/9. just like 0.333333333333333333333.... equals 1/3
2006-07-25 06:31:35
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answer #7
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answered by gjmb1960 7
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No, its .1 away from equaling one.
2006-07-25 06:32:15
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answer #8
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answered by Drew H 1
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Yes.
If you accept that a repeating decimal like .9999... represents a real number, then it has to be 1.
Proof: Let x = .9999.... Since it is a real number, we can do real arithmetic with it
10x = 9.99999999
9x = 10x - x = 9.9999... - .99999... = 9
9x = 9, so x = 1.
2006-07-25 06:31:37
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answer #9
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answered by rt11guru 6
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NO, but you round it off to 1.
2006-07-25 06:31:17
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answer #10
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answered by JAM123 7
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