- #71
Isaac0427
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I thought it represented an infinitely small change of x. I could be wrong, however.jbriggs444 said:
I thought it represented an infinitely small change of x. I could be wrong, however.jbriggs444 said:dx is not a number. It is a notation.
Remember I said in previous posts that the length of the rectangles in a fixed Riemann sum is bounded below. You then consider what happens as this approaches 0 in the limit, but this, dealing with limits, is a separate issue.Isaac0427 said:dx is also infinitesimally small, but if it equaled zero, then any integral or derivative would equal zero or be undefined.
In the non-standard analysis, dx can be given meaning as an infinitesimal. In standard analysis, the dx in an integral is pure notation.Isaac0427 said:I thought it represented an infinitely small change of x. I could be wrong, however.
Isaac0427 said: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.
I know what a limit is. She explained how for practical purposes the limit .999... approaches 1 is considered .999... being equal to one.lavinia said:Ask your math teacher to explain to you what a limit is.
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.Isaac0427 said: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".Isaac0427 said:I know what a limit is. She explained how for practical purposes the limit .999... approaches 1 is considered .999... being equal to one.
HallsofIvy said:...
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]
...
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.Isaac0427 said: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.
Yes, that was a typo. It should have been [itex]ar^{n+1}[/itex] on both sides.gmax137 said: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?
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.Kegan said: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
Isaac0427 said:Instead of thinking of .999... as 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.micromass said:No. 0.99999... is a number.
MrAnchovy said: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.MrAnchovy said: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.
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.WWGD said: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.
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.MrAnchovy said: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.
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 ?Isaac0427 said: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 I Know i just forgot the...WWGD said:@Kegan It is not .999 , it is .9999... with an infinite string of 9s.
No, .999... doesn't approach 1 -- it is exactly equal to 1.Kegan said: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 ?
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 subjectMark44 said: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... 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.Kegan said: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 ?