The student's acceptance or rejection of 0.999 =1

[coolul007]

Did you know there are a number of people who think numbers aren't "exact"? That people get the idea that there is no such number as "one third", with reasoning that decimal can only provide an approximate value of such a thing? Non-terminating decimals really need some amount of coverage, to properly connect decimal notation to arithmetic and algebraic knowledge.

For the sake of this thread, I took it to mean the theory of decimal expansion of rational numbers. There are a multitude of theories and ways of describing numbers. I am not precluding any of that with my statements, just pointing out that 0.9999... is not the result of decimal expansion of a rational number in base 10.

 

[micromass]

0.9999... is not the result of decimal expansion of a rational number in base 10.

Sure it is. It is a decimal expansion of 1.

But first you need to say what you mean with "decimal expansion" in the first place?

 

[Mark44]

0.9999... is not the result of decimal expansion of a rational number in base 10.

Sure it is. It is a decimal expansion of 1.

Every rational number that has a terminating base-10 representation has an alternate nonterminating expansion.

Besides 1.0 being equal to 0.9999..., there are also
0.5 = 0.49999 ...
0.6 = 0.59999...

and on and on.

 

[coolul007]

Every rational number that has a terminating base-10 representation has an alternate nonterminating expansion.

Besides 1.0 being equal to 0.9999..., there are also
0.5 = 0.49999 ...
0.6 = 0.59999...

and on and on.

I would like to see your calculations, other than circular arguements that 1 = 0.999... and therefore 0.9999.. = 1. I refer you to the "Theory of Decimal Expansions" as outlined in chapter 13 of "Number Theory and its History" by Oystein Ore, a Dover paperback book.

 

[micromass]

I would like to see your calculations, other than circular arguements that 1 = 0.999... and therefore 0.9999.. = 1.

Again, how did you define decimal expansion??

I refer you to the "Theory of Decimal Expansions" as outlined in chapter 13 of "Number Theory and its History" by Oystein Ore, a Dover paperback book.

And specifically which part of the chapter do you think shows that 0.999... is not the decimal expansion of 1??

 

[Mark44]

I would like to see your calculations, other than circular arguements that 1 = 0.999... and therefore 0.9999.. = 1. .

You would agree that 1 - 0 = 1, right?

Consider the sequence {a_n} = $\{1 - 10^{-n}\}_{n = 1}^{\infty}$
a₁ = 0.9
a₂ = 0.99
a₃ = 0.999
.
.
.
a_n = 0.999...9 (terminates after the nth 9)

Each number in the sequence has a finite number of 9's following the decimal point, but it should be clear that the limit of the sequence is 1.

 

[coolul007]

You would agree that 1 - 0 = 1, right?

Consider the sequence {a_n} = $\{1 - 10^{-n}\}_{n = 1}^{\infty}$
a₁ = 0.9
a₂ = 0.99
a₃ = 0.999
.
.
.
a_n = 0.999...9 (terminates after the nth 9)

Each number in the sequence has a finite number of 9's following the decimal point, but it should be clear that the limit of the sequence is 1.

You get no argument from me about a limit of a sequence. One can create any sequence of digits we want. I specifically state that 0.9999... is not represented by an a/b to begin with. In fact since no rational example can be shown, I'm concluding it must be irrational. Because 0.9999...98 seems to have no rational representation either. my argument is based on the method we use to obtain decimal representation, division.

 

[hddd123456789]

From experience in another physics forum, this has the ring of a similar problem often encountered in physics: that of definitions. (Lay)People (like myself) often get confused about a topic by assigning it undue complexity/special stature when at its core it is a definition rather than a revelation.

If I'm understanding right, then this idea of 0.999... equaling 1 is a language definition, not a hint at some deeper property of real numbers. It's boring, but on the other hand, it saves one time from looking for meaning in something that was never defined to have any deeper meaning about numbers or infinitesimals.

 

[Hurkyl]

   .9999..
 +---------
1|1.0
   .9
 ----
   .10
   .09
  ----
   .010
   .009
   ----
   .0010
   .0009
   -----
   .0001
       :
       :

I like subtraction better, though. Because when you subtract 2 from 1, you get a 0 in the one's place because of the borrow:

$$\begin{matrix} \tiny{ 1} & . & \tiny{ 9} & \tiny{ 9} & \tiny{ 9} & \cdots \\ \not2 & . & \not 0 & \not 0 &\not 0 &\cdots \\ 1 & . & 0 & 0 & 0 & \cdots \\\hline 0 & . & 9 & 9 & 9 & \cdots \end{matrix}$$

Because 0.9999...98 seems to have no rational representation either.

Terminating decimals do have rational representations.

0.9999...98 = 9999...98 / 10000...00, where all of the ellipses represent the same number of digits.

 

[lurflurf]

You must be trolling 1=1/1=0.9999...

Decimal expansions are real numbers by definition. The sequence of digits is not equal, but they represent the same real number.
In binary we have
1=.111111111...
In trinary
1=.22222222222...

 

[HallsofIvy]

In my opinion, the simplest, most straightforward, proof that 0.9999... = 1 is in terms of "geometric series". 0.9999...= .9+ .09+ .009+ .0009+ ...= .9+ .9(1/10)+ .9(1/100)+ .9(1/1000)+ ...= .9(1+ 1/10+ 1/100+ 1/1000+ ...)= .9(1+ (1/10)+ (1/10)^2+ (1/10)^3+ ... That is, it is a geometric series $$\sum ar^n$$ with a= .9 and r= 1/10. The sum of that is $$\frac{.9}{1- 1/10}= \frac{.9}{.9}= 1$$.

If you protest that you want an explanation that does not require infinite series, I would counter that you can't understand non-terminating decimal expanasions without using infinite series.

 

[Fredrik]

A "proof of 0.999...=1" only tells us that the series $\sum_{n=1}^\infty 9\cdot 10^{-n}$ is convergent and has the sum 1. This is interpreted as a proof of 0.999...=1, because the standard definition of 0.999... says that the string of text "0.999..." represents the sum of that series.

I think that a person who struggles with the equality 0.999...=1 would benefit as much or more from hearing an explanation of why that definition is standard, as from seeing a proof of $\sum_{n=1}^\infty 9\cdot 10^{-n}=1$. They probably think that this is an entirely different problem.

 

[coolul007]

In my opinion, the simplest, most straightforward, proof that 0.9999... = 1 is in terms of "geometric series". 0.9999...= .9+ .09+ .009+ .0009+ ...= .9+ .9(1/10)+ .9(1/100)+ .9(1/1000)+ ...= .9(1+ 1/10+ 1/100+ 1/1000+ ...)= .9(1+ (1/10)+ (1/10)^2+ (1/10)^3+ ... That is, it is a geometric series $$\sum ar^n$$ with a= .9 and r= 1/10. The sum of that is $$\frac{.9}{1- 1/10}= \frac{.9}{.9}= 1$$.

If you protest that you want an explanation that does not require infinite series, I would counter that you can't understand non-terminating decimal expanasions without using infinite series.

last word:

$\frac{10^{n} - 1}{10^{n}}$ = 0.999... = 1

$\frac{10^{n}}{10^{n}} - \frac{1}{10^{n}}$ = 1

$1 - \frac{1}{10^{n}}$ = 1

$1 - (1)- \frac{1}{10^{n}}$ = 1 -(1)

$- \frac{1}{10^{n}}$ = 0 Houston we have a problem!

That is why we cannot equate infinities to rational numbers without bending the rules.

(I know the next post is the limit of $- \frac{1}{10^{n}}$ = 0)

 

[justsomeguy]

Your step 2 needs some explaining. $\frac{10^n}{10^n} - \frac{1}{10^n} = 1$ only holds when n = 0. Unless you'd like to claim that $\frac{1}{1} - \frac{1}{10^5} = 1$ where n=5.

The limit expresses what happens as n approaches infinity. For any n < infinity, you don't have $.\overline{999}$, you have .999...90.

 

[Mark44]

last word:

$\frac{10^{n} - 1}{10^{n}}$ = 0.999... = 1

As already mentioned by justsomeguy, the expression in the left side is not equal to 0.999... In the limit, as n increases without bound, then yes, they're equal.

Since you don't include any indication that you are taking the limit, then none of the equations below follow.

$\frac{10^{n}}{10^{n}} - \frac{1}{10^{n}}$ = 1

$1 - \frac{1}{10^{n}}$ = 1

$1 - (1)- \frac{1}{10^{n}}$ = 1 -(1)

$- \frac{1}{10^{n}}$ = 0 Houston we have a problem!

That is why we cannot equate infinities to rational numbers without bending the rules.

(I know the next post is the limit of $- \frac{1}{10^{n}}$ = 0)

 

[Hurkyl]

If you protest that you want an explanation that does not require infinite series, I would counter that you can't understand non-terminating decimal expanasions without using infinite series.

I don't recall having trouble with infinite decimals before taking calculus.

One learns how to do arithmetic with terminating decimals early; if you learn the variations that work left to right, then those same elementary school algorithms for doing arithmetic can be used for non-terminating decimals too.

I've even seen people define the real numbers in this manner: numbers are strings of digits (modulo relations like 0.999... = 1.000...), and the arithmetic operations are given as explicit algorithms that operate on strings of digits.

The algorithms for addition and subtraction give yet another reason why 0.999... = 1.000...; for example, when subtracting 1 from 2, you can either have no borrows, or you can have a borrow in every place.

Of course, one could resolve the ambiguity by disallowing decimals ending in all 9's Interestingly, in the text I saw use this approach, I believe it did exactly the opposite: it disallowed terminating decimals! So "0.999..." was a legal decimal, but "1" was not.

 

[micromass]

$\frac{10^{n} - 1}{10^{n}}$ = 0.999...

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]

$\frac{10^n}{10^n} - \frac{1}{10^n} = 1$ only holds when n = 0.

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

 

[justsomeguy]

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

Curse you, subtraction!

 

[Fredrik]

A "proof of 0.999...=1" only tells us that the series $\sum_{n=1}^\infty 9\cdot 10^{-n}$ is convergent and has the sum 1. This is interpreted as a proof of 0.999...=1, because the standard definition of 0.999... says that the string of text "0.999..." represents the sum of that series.

I think that a person who struggles with the equality 0.999...=1 would benefit as much or more from hearing an explanation of why that definition is standard, as from seeing a proof of $\sum_{n=1}^\infty 9\cdot 10^{-n}=1$. They probably think that this is an entirely different problem.

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 $\frac{1}{3} * 3 = 1$, and likewise that $.\overline{333} * 3 = 1$ and thus that, finally, $.\overline{999} = 1$.

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.

$(\frac{1}{3})_{base-10} = (\frac{1}{10})_{base-3}$

$.\overline{333}_{base-10} = 0.1_{base-3}$

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.

$(\frac{1}{3})_{base-10} = (\frac{1}{3})_{base-6} = 0.2_{base-6}$

$0.2_{base-6} * 3_{base-6} = 1$

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]

The point that I try to make is that a limit is NOT equality.

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 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.

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]

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.

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

 

[Mark44]

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.

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]

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.

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.