The student's acceptance or rejection of 0.999...=1

micromass

This equality is simply not true!! You do know that there is an infinite number of 9's in 0.999... right?? Your left hand side makes it seem like a finite number.

What is n anyway?

Fredrik

When n=0, the left-hand side is 0.

justsomeguy

Curse you, subtraction!

Fredrik

A few more thoughts along these lines. I think almost everyone will agree that 0.999... should be defined as the sum of that series (if such a sum exists and is unique). And I think almost everyone who's still with us after that will agree that the sum of that series should be defined as the limit of the sequence 0.9, 0.99, 0.999,... (if such a limit exists and is unique). This brings us to the definition of the limit of a sequence. I like to state it in the following way:

A real number L is said to be a limit of a sequence S if every open interval that contains L contains all but a finite number of terms of S.

Since this definition takes a while to understand if you haven't seen it before, I can imagine that some students won't agree that this definition is a good idea. So let's return to that in a while. First, here's a proof that 1 is a limit of that sequence, and that no other real number is.

Let M be any real number other than 1. Define t=|M-1|/2 and consider the intervals (1-t,1+t). and (M-t,M+t). Note that they are disjoint. The former contains all but a finite number of terms of the sequence, so the latter contains at most a finite number of terms of the sequence. This means that 1 is a limit, and M is not.

Now all that remains is to explain why we're choosing that particular definition of "limit". There are of course many reasons why we're doing it, but the one that's the most relevant here is that we want the sequence 0.9, 0.99, 0.999,... to have 1 as its unique limit.

This brings us to the point that I think isn't emphasized enough in these discussions. The real reason why 0.999...=1 is that we want this equality to hold. There was never any chance that it wouldn't hold, because if our first choice of definitions would have given us another result, we wouldn't have accepted the result. We would have changed the definitions.

justsomeguy

I don't really understand why this causes such consternation for some. When I went to school, we were exposed to the idea of infinite repeating decimals at a very young age; the first time 1/3 came up when learning to convert back and forth between fractions and decimals. It's easy to show that [itex]\frac{1}{3} * 3 = 1[/itex], and likewise that [itex].\overline{333} * 3 = 1[/itex] and thus that, finally, [itex].\overline{999} = 1[/itex].

It wasn't until a year or two later that I was first introduced to the idea of different bases, but I still contend that is a much easier (and less contentious) way to make the entire problem disappear in a puff of smoke. Not all fractions in a given base can be represented as decimals in that base. If you switch to a base that has all of the prime factors of the divisor, then the problem disappears entirely.

[itex]
(\frac{1}{3})_{base-10} = (\frac{1}{10})_{base-3}
[/itex]

[itex]
.\overline{333}_{base-10} = 0.1_{base-3}
[/itex]

If you switch to a base that is at least 2x the product of all the prime factors of the divisor, then you can do it without even manipulating the divisor.

[itex]
(\frac{1}{3})_{base-10} = (\frac{1}{3})_{base-6} = 0.2_{base-6}
[/itex]

[itex]
0.2_{base-6} * 3_{base-6} = 1
[/itex]

In other words, it's a matter of representation of values -- not of arbitrary definitions or conventions. I encountered all of this long before limits or any greek symbols.

justsomeguy

For what it's worth, the idea of primes and their relationship to bases also helped me "accept" the idea of irrational numbers like pi. When I saw the Leibniz formula for pi I knew immediately that no decimal or fractional representation was possible. Because there are an infinite number of primes, a base that could present pi as a fraction or rational number would have an infinite number of factors.

coolul007

The point that I try to make is that a limit is NOT equality. A limit is a number that will not be reached and will not go beyond, because one can always add an iteration to the sequence. The fact that limits have been used so much to achieve success, doesn't mean that they are EQUAL. We also treat pi and e as if they were not limits, but numbers, nice habit, but not accurate.

Fredrik

This is wrong. For example, the limit of the sequence 0.9, 0.99, 0.999 is equal to 1. To understand this, you need to study the definition of "limit".

Number Nine

The limit of a convergent sequence of real numbers is a real number. This is a fairly elementary result in real analysis (it's generally considered to be an axiom of the real numbers, actually).

pwsnafu

Completely and utterly wrong. Each sentence is wrong. coolul007 you need to go back and study limits again, from the very beginning.

Mark44

I think that you have a fundamental misunderstanding of what a limit actually means, based on what you are saying above, particularly the part about "that will not be reached".

There is a big difference between saying a_n = L and ## \lim_{n \to \infty} a_n = L##, and you seem to not be getting that difference.

micromass

There is little point in continuing this discussion. I do understand that you think that 0.999... is not equal to 1. I even understand why you think that. The problem is that you seem to be missing quite some basic knowledge about real numbers. Without this knowledge, I think it would be impossible to fully grasp the 1=0.9999... situation.

So, if you're truly interested in understanding the equality 1=0.9999...., then I can only suggest you to start studying limits and real numbers. Any good analysis book should cover these things very well. So please, do yourself a favor and try to study these things from the very beginning.

I'll keep this thread open to see if we can get further discussion. But if people keep commenting without taking the effort of studying limits and real numbers, then we will be forced to lock.