# Why Do People Say That 1 and .999 Are Equal? - Comments

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

jbriggs444
Homework Helper
I know what a limit is. She explained how for practical purposes the limit .999... approaches 1 is considered .999... being equal to one.
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 limit is 1. The notation .999... denotes the limit. Hence .999... is 1.

lavinia
pwsnafu
I know what a limit is. She explained how for practical purposes the limit .999... approaches 1 is considered .999... being equal to one.
The part where I bolded indicates you don't know what a limit is. Limits don't move and they don't "approach".

lavinia
gmax137
Going back to post #2

...
Restore that by adding $ar^{n+1}$ to both sides:
$S_n- a+ ar^n= r(a+ ar+ \cdot\cdot\cdot+ ar^{n-1})+ ar^{n+ 1}$
...

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

FactChecker
Gold Member
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.
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.

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HallsofIvy
Homework Helper
Going back to post #2

Maybe (probably) I'm being dense, but didn't you just add $ar^n$ to the left and $ar^{n+ 1}$ to the right?
Yes, that was a typo. It should have been $ar^{n+1}$ on both sides.

i'm 11 i understand that 1 and .999 are two completely different numbers and in math you don't round numbers to get precise answers it just doesn't work like that

WWGD
WWGD
Gold Member
i'm 11 i understand that 1 and .999 are two completely different numbers and in math you don't round numbers to get precise answers it just doesn't work like that

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

i'm 11 i understand that 1 and .999 are two completely different numbers and in math you don't round numbers to get precise answers it just doesn't work like that
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?

Kegan
micromass
Staff Emeritus
Homework Helper
Instead of thinking of .999... as a number

No. 0.99999.... is a number.

pbuk
Gold Member
No. 0.99999.... is a number.
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.

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
Gold Member
The confusion seems largely to stem from the implicit assumption that a number can have only one representation. Instead, a number is an equivalence class; one sees this in daily life, e.g., today is Friday, and (ignoring modular issues) any date 7k days from now is also a Friday, so if the difference between the dates (again, re modularity) is a multiple of 7 , then both are the same day of the week. And then there is 2/2, 3/3 , etc.

pbuk
Gold Member
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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micromass
Staff Emeritus
Homework Helper
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.

pbuk
WWGD
Gold Member
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.
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.

jbriggs444
Homework Helper
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 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.

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

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.
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.

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 ?

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

yeah I Know i just forgot the....

Mark44
Mentor
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 ?
No, .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.

No, .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.
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 subject

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 ?
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.