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

[jbriggs444]

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.

 

[Isaac0427]

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.

 

[Kegan]

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]

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

yeah I Know i just forgot the...

 

[Mark44]

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.

 

[Kegan]

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

 

[Isaac0427]

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.