Does 0.999... equal 1?

Yes, the limit is 0.
This means that f(x) cannot = 0. Ever. It can get infinitely close to 0, but can never actually equal it.

 

At no finite point it equals 0. But the limit means if x went for infinity it would be 0. When you subtract the sum of number n 9s from 1 you get 10^-n. Since the number of 9s goes on forever n would need to be infinity. Since you cannot directly substitute infinity into the equation you must use limits. And the limit of 10^-n as n approaches infinity is 0. Therefore the difference between .999... and 1 is 0. They are the same number.

 

He's right.


Although I wouldn't doubt what Sythe is saying either. He has a good point.
1/9 = .1111111 * 9 = .9999999999
1/9 * 9 = 9/9 = 1

 

I thought we were past this? .333... = 1/3. .333... is given as a representation of the decimal number obtained when you divide 1/3. The same goes for 2/3, except rather than getting .666..., you get .666...7. Then when you add them together, you get =1.

.999...=/=1 as it will always be a .000...1 off equaling 1.

Of course, if you roundup, then you do get 1.

 

You're going to argue A LOT with your calculus teacher.

Just saying.

 

The .000...1 is 0 since there are infinite number of 0s the 1 is never in the number.
At what point does dividing 2/3 give 7?

 

New invention called a calculator, try it some time. As I said, it is the number given to represent 2/3 as 1 can never be divided equally 3 times.

The .000...1 is a way of showing how small the number is =_="

 

.666... is exactly 2/3 because the 6es are repeated forever. If you stop writing the sixes at any finite point it would only be an approximation but since they go forever its exact. In the number .000...1, the .000... part means the zeros are repeated for infinity. How can you possibly have something past infinity? Doing so would mean that it would need to stop somewhere and since it goes forever there would never be a 1.

 

If you even say that 0.0..1 is showing how small a number is, you're doing it wrong. You can't represent an infinitesimal that way.

They work in strange ways. They're smaller than any number you can construct, but they're larger than zero.

 

This is obvious...NO! The numbers are different by .001, it might be a small difference but they are different numbers.

 

I thought about that as well but if you decide 1 By 3 it wont be 0.333 exactly.

 

1/3 is exactly .33333333(repeated forever)

 

no its because 1 cannot be devided by 3. Theres no complete answer to 1/3 ... it will always go on continuing.

 

You would be correct in saying there is no exact nonrepeating decimal representatino of 1/3. You said it your self it would keep continuing. That is what .333... means. The 3s go forever.

 

Welcome to the thread.

Note the "..." in 0.999...

 

once you hit grade 8 you will find out that they can...

 

thats why you cant get 1 because 0.333 forever would be 0.999 forever which just doesnt make one. and listen carefully. 1/3 isnt EXACTLY 0.333 forever. so therefore 3x 0.333 wont make 1!

 

Theoretically, .9 repetend, i.e. .99... is equal to 1.
This can be proved.

Let
x be .9... (1)
Therefore,
10x is equal to 9.9... (2)

Now, if you subtract equation (1) from (2), you get...
9x=9
Obviously, x is equal to 1. However, logically, this may not appear to be
true. As the number of digits in .999 tends to infinity, the number becomes 1.

 

no because 1 / 3 x 3 = 1 But 1 / 3 x 0.333 equals 0.999

 

**ALL rational numbers can be represented with infinitely repeated decimals.**

How isnt 1/3 exactly .333...?
x=.333...
10x=3.333...
10x-1x=3.333...-.333...
9x=3
x=3/9
x=1/3