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

by thoughtweevil
Tags: acceptance, rejection, student
Mentor
P: 18,299
 Quote by coolul007 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?
Mentor
P: 21,295
 Quote by coolul007 0.9999.... is not the result of decimal expansion of a rational number in base 10.
 Quote by micromass 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.
P: 234
 Quote by 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.
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.
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P: 18,299
 Quote by coolul007 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??
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P: 21,295
 Quote by coolul007 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 {an} = ##\{1 - 10^{-n}\}_{n = 1}^{\infty}##
a1 = 0.9
a2 = 0.99
a3 = 0.999
.
.
.
an = 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.
P: 234
 Quote by Mark44 You would agree that 1 - 0 = 1, right? Consider the sequence {an} = ##\{1 - 10^{-n}\}_{n = 1}^{\infty}## a1 = 0.9 a2 = 0.99 a3 = 0.999 . . . an = 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.
 P: 82 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.
Emeritus
PF Gold
P: 16,091
   .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.
 HW Helper P: 2,264 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...
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,553 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.
 Emeritus Sci Advisor PF Gold P: 9,403 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.
P: 234
 Quote by 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.
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)
 P: 166 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.
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P: 21,295
 Quote by coolul007 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.
 Quote by coolul007 $\frac{10^{n}}{10^{n}} - \frac{1}{10^{n}}$ = 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)
Emeritus
PF Gold
P: 16,091
 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.
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P: 18,299
 Quote by coolul007 $\frac{10^{n} - 1}{10^{n}}$ = 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?
Emeritus
PF Gold
P: 9,403
 Quote by justsomeguy $\frac{10^n}{10^n} - \frac{1}{10^n} = 1$ only holds when n = 0.
When n=0, the left-hand side is 0.
P: 166
 Quote by Fredrik When n=0, the left-hand side is 0.
Curse you, subtraction!

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