1. Dec 1, 2003

### Justin Horne

Hi. I'm a freshman in High school, and my algebra teacher gave me a problem. She said that .9 repeating = 1 Now apart from all the ways this didn't make sense to me, I thought of a way that might make it not work. If you were to put something like .0 [infinity of 0's] 1, wouldn't that be able to be added to .9 repeating, making 1? Correct me if I am wrong here, please. Thanks,
Justin.

2. Dec 1, 2003

### chroot

Staff Emeritus
Last edited by a moderator: May 1, 2017
3. Dec 1, 2003

### sridhar_n

There is a limit to how much we can measure. In fixing this limit we also fix our approximations. When there are infinites of nine we say that it is very close to one and hence approximate it to 1. We could approximate 0.999999999999999999999999999999999999999999999... to 0.9, but 0.9999999999999999999999999999999999999999.... is more closer to 1 than 0.9. Similarly, 0.0000000000000000000000000000000000000000001 is more closer to 0 than 1. Thus it is basically because there is a limit to our measurements and approximations that we approximate to te closest digit possible.

Sridhar

4. Dec 1, 2003

### chroot

Staff Emeritus
Sorry, this is not correct. $0.\overline{9}$ is not approximately 1, it is absolutely 1. Remember, I'm not talking about a lot of nines, I'm talking about an infinite number of nines. That's a whole different bucket o' spaghetti.

- Warren

5. Dec 1, 2003

### HallsofIvy

I know it is hard to accept but even teachers are right ocassionally! The problem with you idea is that you can't have an "infinite number of 0s" and then put a 1 on the end. With an infinite number of 0s there is NO end.

A non-terminating decimal such as 0.99999... is DEFINED as the limit of the sequence .9, .99, .999, .9999, etc. and that limit IS 1.0.

6. Dec 1, 2003

### HallsofIvy

I really have to object to this. When we say that 0.999.... is 1, we are not talking about "measurement" and we are not talking about approximations.

0.99999.... is not close to 1 it is 1: exactly equal to 1. As I said in another post, 0.99999... is defined as the limit of the sequence 0.9, 0.99, 0.999, ... and it is easy to show that that limit is 1.0.

7. Dec 2, 2003

### Organic

When 1.000... and 0.999... are two representations of the same number then:

1.00... = 0.999...

0.100... = 0.0999...

0.0100... = 0.00999...

0.00100... = 0.000999...

0.000100... = 0.0000999...

0.0000100... = 0.00000999...

Therefore we can write:

0.100... + 0.0100... = 0.0999... + 0.00999...

0.0100... + 0.00100... = 0.00999... + 0.000999...

But this is not true because:

0.1100... not= 0.0999... + 0.00999... = 0.10999...8

0.01100... not= 0.00999... + 0.000999... = 0.010999...8

and so on ...

Digit 8 is not the last digit but the limit digit or the unreachable digit of 0.999...

Therefore 1.000... is not the limit of 0.999...

Last edited: Dec 2, 2003
8. Dec 2, 2003

### HallsofIvy

I sincerly hope that that was your idea of a joke.

9. Dec 2, 2003

### suyver

Check out the other thread with the identical post to convince yourself that it wasn't a joke...

10. Dec 2, 2003

### chroot

Staff Emeritus
[moderator hat]

This thread should be locked, since it just parallels the other one.

[/moderator hat]

- Warren