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

Hurkyl

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.

coolul007

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

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

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

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

Again, how did you define decimal expansion??

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

Mark44

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 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

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:

[tex]
\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}
[/tex]





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 [tex]\sum ar^n[/tex] with a= .9 and r= 1/10. The sum of that is [tex]\frac{.9}{1- 1/10}= \frac{.9}{.9}= 1[/tex].

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

last word:

[itex]\frac{10^{n} - 1}{10^{n}}[/itex] = 0.999... = 1

[itex]\frac{10^{n}}{10^{n}} - \frac{1}{10^{n}}[/itex] = 1

[itex]1 - \frac{1}{10^{n}}[/itex] = 1

[itex]1 - (1)- \frac{1}{10^{n}}[/itex] = 1 -(1)

[itex]- \frac{1}{10^{n}}[/itex] = 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 [itex]- \frac{1}{10^{n}}[/itex] = 0)

justsomeguy

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

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

Mark44

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.

Hurkyl

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.