# How exactly does 0.999~ = 1?

1. Jan 9, 2006

### QuantumTheory

Last edited by a moderator: Apr 21, 2017
2. Jan 9, 2006

### matt grime

It is true because of the way we have defined what decimal representations* of real numbers are, and exactly because of what limits are. 0.9recurring is the limit of a cauchy sequence of the partial sums of the obvious geometric progession

9/10+9/100+9/1000+...

which is 1.

End of story.

Shall I await the "but you never get to the end of an infinite sum so it never truly equals 1" arguments?

* I presume you have no problem with the fact that 1/2 and 2/4 (and an infinite number of other fractions) all represent the same rational number? If so why is it so surprising that *two* (and only 2) decimal representations are of the same real number?

Last edited: Jan 9, 2006
3. Jan 9, 2006

### QuantumTheory

Incredible. Well, I never fully finished algebra, geometry, or pre calc or calculus. It's like you konw bits and pieces and not the whole thing, which makes thigns like this very difficult.

I won't doubt you ,but I still dont understand. However, my math teacher did tell me that there is an infinite numbers between 1 and 2.

I thought the whole point of a limit was that you can get close and not reach it? Wait a second. Will 0.9 always repeat itself or eventually get to 1?

4. Jan 9, 2006

### QuantumTheory

Well it doesn't matter, but it could be 0.99999999999999999999999 and isn't that still 0.000000000001 close from 1?

5. Jan 9, 2006

### QuantumTheory

I guess it only equals one if ou let it go for infinity. But if you stop the 'process' then it does not equal one.

6. Jan 9, 2006

### matt grime

The reals are something that obeys a set of rules, one of those rules directly implies that in the decimal representations *we must identify those two numbers if they are to behave properly*. Decimals are just representations of the real numbers. If you want to understand those definitions then you need to do significantly more maths (analysis), but it is just a formal consequence of hte definitions. It is not mysterious.

Of course no finite truncation of the infinite sum is 1, but I don't see what that has to do with the infinite sum. You just appear to have some misconception about what a limit is. In general I'd say you're reading far too much into what after all are just strings of digits that obey certain rules. If you don't know what the rules of the game are you cannot play it. You're 'let it go to infinity' thing is one such indication. This isn't something that adds up things one at a time as if it were a machine, this isn't some process, you don't let things 'run forever'. It is just a symbol with certain properties.

One of which is it is the smallest number larger than any of the finite truncations.

7. Jan 9, 2006

### Hurkyl

Staff Emeritus
Right -- 0.999~ does not mean "a lot of 9's" -- it means the 9's keep going without ending.

More precisely, for every positive integer n, there is a 9 in 0.999~ located exactly n places to the right of the decimal point.

(There are infinitely many 9's in 0.999~ -- nothing is "going" to infinity)

8. Jan 9, 2006

### nazgjunk

1/3 is 0.333~, right?

3 * 1/3 is 1, right?

All you then need to accept is that 3 * 1/3 is "also" 0.999~.

To make it somewhat clearer, 2*2=4, 2 * 0.222~ is 0.444~. So if 3*3=9, 3*1/3=0.999~.

When I first stumbled across this, my main problem was accepting the seemingly illogical truth.

Last edited: Jan 9, 2006
9. Jan 9, 2006

### TD

I've seen this very often but what I don't see is why someone would accept 1/3 = 0.3~ and not accept 0.9~ = 1, yet this is often used as 'proof'.

10. Jan 9, 2006

### HallsofIvy

??Typo? I presume you meant "All you then need to accept is that 1/3 is "also" 0.333~".

11. Jan 9, 2006

### TD

I think he meant that people would already accept that 1/3 = 0.3~ (don't ask me why though), then they'd need to accept that 3*1/3 is not only 1 but also 3*1/3 = 3*0.3~ = 0.9~.

12. Jan 9, 2006

### Mindscrape

Yeah, I always thought this was strange.

13. Jan 9, 2006

### vaishakh

There is a lot of difference between something tending to 1 and something being 1. we approximate it to 1, but need not be that it is actually 1. The explanation given by Nagzun was really fantastic. Something tending to 1 being written as 1 when we don't know the actual value is a differant case. This is what happens in GP. We don't know the actual value and so the formula given is a/1-x wherew x<1. Here a Mathematician is sue that no one tommorow will give a unique perfect answer to this question about summation about GP to infinite values, So he approximated the formula. We write root-2 as 1.414, But 1.414 is a unique rational number. However root-2 is not really 1.414, it only tends to that value. We have no use of finding root-2 to hundred decimals to solve numericals. So we neglect the remaining. We can be sure about the fact that 1 is something very near to the number 0.999~. However we cannot tell, that both are equal.

14. Jan 9, 2006

### TD

There is a big difference between an approximation, which 1.414 is for sqrt(2), and another representation for exactly the same number, which 0.9~ is for 1. There is no "choice" for us, to say whether or not those two are equal. If we're working in the real numbers, there is no other possiblity than for those two to be exactly the same real number - it's just another way of writing it.

15. Jan 9, 2006

### matt grime

Excuse me, but such displays of ignorance tend to rile me. Who said 0.9.... tends to one? It is one, it is not tending anywhere; its partial sums tend to one. One is the smallest real number greater than any of the strictly increasing partial sums, hence is the limit **by the definition of the space in which the argument takes place**.

16. Jan 9, 2006

### Staff: Mentor

If nazgjunk's simple and elegant proof (minus the typo, of course )isn't enough to quench all doubt, perhaps a question:

If 0.999... is not equal to 1, then there must be a number between the two: what is it?

And don't say 0.000...1, because there is no such thing.

17. Jan 9, 2006

### Integral

Staff Emeritus
Last edited by a moderator: Apr 21, 2017
18. Jan 9, 2006

### Maxwell

Like a sticky with the topic:

"BEFORE YOU POST 0.999 = 1, READ THIS!!!!!!!!!" with a link to that post?

19. Jan 9, 2006

### Hurkyl

Staff Emeritus
I have been getting the feeling for a while now that each 0.999~ = 1 thread has been "better" than the previous one... I'm not sure what that really means, though.

20. Jan 9, 2006

### Sir_Deenicus

We simply happen to use a system where 1 and 0.999... are equal. On the hyperreals, R*, a *logically vaild* superset of our current system, 0.9999... and 1 are not equal.

Last edited: Jan 9, 2006