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I know what a limit is. She explained how for practical purposes the limit .999... approaches 1 is considered .999... being equal to one.Ask your math teacher to explain to you what a limit is.

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- #76

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I know what a limit is. She explained how for practical purposes the limit .999... approaches 1 is considered .999... being equal to one.Ask your math teacher to explain to you what a limit is.

- #77

jbriggs444

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It is not "for practical purposes". It is exact. The limit of the sequence .9, .99, .999, ... does not approach 1. Successive terms in the sequence approach 1. The limitI know what a limit is. She explained how for practical purposes the limit .999... approaches 1 is considered .999... being equal to one.

- #78

pwsnafu

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The part where I bolded indicates youI know what a limit is. She explained how for practical purposesthe limit .999... approaches 1is considered .999... being equal to one.

- #79

gmax137

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...

Restore that by adding [itex]ar^{n+1}[/itex] to both sides:

[itex]S_n- a+ ar^n= r(a+ ar+ \cdot\cdot\cdot+ ar^{n-1})+ ar^{n+ 1}[/itex]

...

Maybe (probably) I'm being dense, but didn't you just add [itex]ar^n[/itex] to the left and [itex]ar^{n+ 1}[/itex] to the right?

- #80

FactChecker

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You can prove that there is no difference at all because the 9's go forever. This proof may be your first encounter with a mathematical "proof by contradiction". Suppose you assume that there is any difference between .999999... and 1. Say it is over 0.00001 (a formal mathematical proof would use an arbitrarily small ε > 0). Now use enough 9's (0.999999999999) to show that there is less difference than that and that the difference will only decrease as you add more 9's. That contradicts to your original assumption that the difference is greater than 0.00001 (or ε>0). It doesn't matter how small your assumed difference is; you can add enough 9's to get closer to 1 and contradict that assumption. So it proves that there can be no difference between 0.99999... and 1.Ok, I just talked to my math teacher. She explained it to me in person which helped. Thanks guys it really does help. I mostly understand it now, that as something gets so close to 1, for all practical purposes, it is equal to 1.

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- #81

HallsofIvy

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Yes, that was a typo. It should have been [itex]ar^{n+1}[/itex] on both sides.Going back to post #2

Maybe (probably) I'm being dense, but didn't you just add [itex]ar^n[/itex] to the left and [itex]ar^{n+ 1}[/itex] to the right?

- #82

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- #83

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Instead of thinking of .999... as a number, think of it as the limit of the sequence (.9,.99,.999,.9999,...) as n approaches infinity. That is what the 3 dots mean. The limit of that sequence as n approaches infinity is 1.

Here's another example: (1/0!)+(1/1!)+(1/2!)...=e, however if you stop the sequence at any value of n, the answer will not be e. Does this make sense?

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- #86

pbuk

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To be fair, Isaac didn't say it was not a number. He just suggested that it might be easier to see that 0.9999... is the limit of the sequence 0.9, 0.99, 0.999, ..., and that the limit of that sequence is also equal to 1, than it is to make the jump from the recurring decimal to 1 directly.No. 0.99999....isa number.

This illustration is not IMHO any less rigorous than the 9.9999... - 0.9999... illustration of the identical equality to 1. Whether it is easier to comprehend or not is a matter of personal taste.

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WWGD

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pbuk

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I don't think days of the week, which are analogous to integers mod 7 is a helpful analogy to the reals. Edit: the rest of this is rubbish ~~And 2/2 is an expression which is equal to 1, it is not a valid representation of a rational number: the numerator and denominator must have no common factor.~~

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- #89

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I don't think days of the week, which are analogous to integers mod 7 is a helpful analogy to the reals. And 2/2 is an expression which is equal to 1, it is not a valid representation of a rational number: the numerator and denominator must have no common factor.

2/2 is a perfect representation of a rational number.

- #90

WWGD

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Yes, maybe it is not the best example, but the point I wanted to make is that two things don't need to be strictly equal in order to be considered the same. It is more a " to the effects of what we are doing, these two expressions are equal" EDIT: Maybe non-trivial, i.e., non-identity isomorphisms would be a better example.I don't think days of the week, which are analogous to integers mod 7 is a helpful analogy to the reals. And 2/2 is an expression which is equal to 1, it is not a valid representation of a rational number: the numerator and denominator must have no common factor.

- #91

jbriggs444

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I was going to write that one could consider the distinction between numerals or formulas on the one hand and numbers on the other. It seems that the distinction you want to make is between exemplars of an equivalence class and the class itself.Yes, maybe it is not the best example, but the point I wanted to make is that two things don't need to be strictly equal in order to be considered the same. It is more a " to the effects of what we are doing, these two expressions are equal" EDIT: Maybe non-trivial, i.e., non-identity isomorphisms would be a better example.

6 of one, ##\frac{dozen}{2}## of the other.

- #92

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That was the point. I understand it is a number, but the value of the number can be thought of the limit of that sequence. IMO, this is easier to comprehend, but as you said, it's a mater of opinion.To be fair, Isaac didn't say it was not a number. He just suggested that it might be easier to see that 0.9999... is the limit of the sequence 0.9, 0.99, 0.999, ..., and that the limit of that sequence is also equal to 1, than it is to make the jump from the recurring decimal to 1 directly.

- #93

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Instead of thinking of .999... as a number, think of it as the limit of the sequence (.9,.99,.999,.9999,...) as n approaches infinity. That is what the 3 dots mean. The limit of that sequence as n approaches infinity is 1.

Here's another example: (1/0!)+(1/1!)+(1/2!)...=e, however if you stop the sequence at any value of n, the answer will not be e. Does this make sense?

yeah that does make more sense so what you saying is that .999... to infinity approaches the number 1 so close it is consideribly the same number is that what you are saying ?

- #94

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@Kegan It is not .999 , it is .9999..... with an infinite string of 9s.

yeah I Know i just forgot the....

- #95

Mark44

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No, .999... doesn't approach 1 -- it is exactly equal to 1.yeah that does make more sense so what you saying is that .999... to infinity approaches the number 1 so close it is consideribly the same number is that what you are saying ?

The limit of the sequence {.9, .99, .999, ...} is 1, which means the farther you go in the sequence, the closer a number in the sequence is to 1.

BTW, it's redundant to write "to infinity" after .999... The dots (an ellipsis) already means that the 9 digits repeat endlessly.

- #96

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Ok that makes sense the more 9's you have the closer it is to the number 1 ,and i'm only eleven and language arts is not my best subjectNo, .999... doesn't approach 1 -- it is exactly equal to 1.

The limit of the sequence {.9, .99, .999, ...} is 1, which means the farther you go in the sequence, the closer a number in the sequence is to 1.

BTW, it's redundant to write "to infinity" after .999... The dots (an ellipsis) already means that the 9 digits repeat endlessly.

- #97

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No, .999... is NOT a sequence. It is a number. The ... means that the value of the number is equal to the limit of the sequence (.9, .99, .999, ...). If it helps, you can think of it as the term of the sequence in which n=infinity. The value of this term, by definition, will be the limit of the sequence as n approaches infinity. If you do not understand this, look up limits.yeah that does make more sense so what you saying is that .999... to infinity approaches the number 1 so close it is consideribly the same number is that what you are saying ?

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