1/3 = 0.333

phoenixthoth

easier way: 1/3 = 0.333... (three dots being key)

multiply by three to get
1 = 0.999...

that thread is closed but this is my two cents.

it's not that surprising, really, that every rational number (1 in this case) can be represented by a series (9/10 + 9/100 + 9/1000 + ...) because even fractions like 1/3 can be (3/10 + 3/100 + 3/1000 +...).

may your journey be graceful,
phoenix

HallsofIvy

Homework Helper
Oh, God, Not again!

Yes

easier way: 1/3 = 0.333... (three dots being key)

multiply by three to get
1 = 0.999...
This is "easier" because you are assuming all of the hard part. It works if you ASSUME 1/3= 0.333... (do you consider that to be simpler than 1= 0.999... itself?) and if you ASSUME that you can multiply a decimal number by multiply each digit. That's true but a good proof should show that.

it's not that surprising, really, that every rational number (1 in this case) can be represented by a series (9/10 + 9/100 + 9/1000 + ...) because even fractions like 1/3 can be (3/10 + 3/100 + 3/1000 +...).
Actually, any REAL number can be represented by such a series. That's what "decimal representation" means.

phoenixthoth

"do you consider that to be simpler than 1= 0.999... itself?"

no. it's all the same. but sometimes people are mystified by 1=0.999... but not by 1/3 = 0.333... . i argue that they are equally mystifying (and that neither of them are).

cheers,
phoenix

HallsofIvy

Homework Helper
If you agree that they are "equally mystifying", why do you argue that starting with 1/3= 0.333... is an "easier" way of proving that
1= 0.999...?

phoenixthoth

because to some, 1/3 = 0.333... is LESS mystifying, though not to me. that's why i start there. drawing on one's intuition on that equation, one can lead themselves to what is commonly held as more mystifying, that 1 = 0.999... it's just that 1, like many numbers, has two different decimal expansions. there is nothing mystifying about that except that our perceptions are often muddled by the "more" mysterious.

may your journey be graceful,
phoenix

Hurkyl

Staff Emeritus
Gold Member
It all depends on what you take as your starting point. If you're allowed to presume that decimals form a field, then the proof at the top of this thread is perfectly valid... and I can certainly imagine ways to present the decimals that allow you to show they form a field without needing to prove 0.999... = 1.

And, of course, most people learn (without proof) that the decimals form a field many years before they learn anything about limits, series, or sequences.

And on an amusing note, I have seen at least one use of the decimals where anything terminating in all zeroes was not considered a decimal, and at least one use of the decimals where anything terminating in all nines was not considered a decimal. And I have also seen the decimals presented as a formal sequence of digits, and 0.999... = 1.000... was a definition instead of something needed to be proved.

phoenixthoth

starting points

yes it all depends on your starting point. i was starting from a lower level version of decimals that is taught to grade-school kids. one can also start from the axioms of set theory and then define what the real numbers are from which you can prove that 1 = 0.999... of course, there are rigorous proofs of that equation out there just from first principles. this requires some effort, but it gives us insight into what a real number is as an equivalence class of the set of sequences of rational numbers moded out by the equivalence relation of being equivalent if two sequences get close to each other. field theory is very interesting indeed.

may your journey be graceful,
phoenix

Lyuokdea

let x = .999...
10x = 9.999...
10x - x = 9.999... - .999...
9x = 9
x = 1

phoenixthoth

correct but it's easier to start with 1/3 = 0.333... and multiply both sides by three to get 1 = 0.999...

may your journey be graceful,
phoenix

Integral

Staff Emeritus
Gold Member
As I stated in the other thread, both of the examples presented in this thread are DEMONSTRATIONS they do not constitute a proof.

phoenixthoth

no, they're not proofs, either the algebra or the multiplication of the expansion 0.333... but i don't think the algebra is needed for a demonstration of the fact. for a proof, you would have to argue from first principles and set theory.

cheers,
phoenix

HallsofIvy

Homework Helper
However, I will say again that I do not think that "1/3= 0.333... therefore 3*(1/3)= 1= 3(0.333...)= 0.999..." is EASIER than
"If x= 0.999... then 10x= 9.999... so 10x- x= 9x= 9.999...-0.999...= 9
and therefore 9x/9= x= 9/9= 1".
Your method introduces the new fact 1/3= 0.333... and you are simply assuming that people who object to 1= 0.999... will not object to 1/3= 0.333...

phoenixthoth

"Your method introduces the new fact 1/3= 0.333... and you are simply assuming that people who object to 1= 0.999... will not object to 1/3= 0.333..."

my method is easier for it uses no algebra. the x is like a contrived invention and doesn't really serve a purpose when you can just multiply both sides by 3. and, yes, i believe fewer people will object to 1/3 = 0.333... but my method can be taught at an earlier age before algebra is learned; they can wrap their brains around the fact that 1 = 0.999... and, in fact, every terminating decimal can be represented by a decimal that is "ending" in all nines. that some numbers (those with terminating expansions) can have this dual representation carries over to other bases besides ten.

may your journey be graceful,
phoenix

mathwonk

Homework Helper
I have taught arithmetic for elementary school teacher candidates, and they certainly did believe that 1/3 = .3333.... is true, but did not at all believe that 1 = .999..... Thus multiplying the first equation by 3, presented them with a major intellectual crisis, and I think a useful one.

In fact one graduate wrote me back later that she was trying to persuade her students and peers of this latter fact in her new job, and was trying hard to hold her ground in the face of major criticism and doubt.

Of course for more critical readers, the place to start is probably with a definition of an infinite decimal as a real number, in terms of least upper bounds. But for naive students the formal manipulations above pose genuine thought provoking conundrums.

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phoenixthoth

Since my last post on the subject, it occured to me that there is another way I hadn't thought of to prove 0.999...=1. I'm sure someone has already stated this.

Premise: 1. if |x|<e for all e>0, then x=0.

Conclusion: 0.999...=1.
proof: (I'm drunk right now so forgive me if this is wrong): let x be the difference between 0.999... and 1: x=1-0.999... . Pretty clear that x>=0 so it suffices to show that x<e for all e>0. Let e>0 be given. For sufficiently large n, 1/10^n is smaller than e. Then x < 1/10^n as the n+1st digit of x is 1/10^(n+1)<1/10^n.

Well, something like that. A bit briefer: ask them to show the difference between 0.999... and 1 in decimal form.

mathwonk

Homework Helper
this is of course the least upper bound approach. i.e. one defiens .999... as the smallest real number not smaller than any of its finite decimal approximations, and checks that this number cannot be smaller than 1.

Changbai LI

Prove:0.9…\=1

1. 1^n=1^10^n=1(even n->infinite) (1)
2. lim(1+1/n)^n=e(when n->infinite) (2)
3. lim(1-1/n)^n=1/e(when n->infinite) (3)
4. lim(1-1/10^n)=lim((10^n-1)/10^n)=(0.9…)^10^n=1/e (when n->infinite) (4)
5. It is the key. So write it more detail in series:
(1-1/10)^10=0.9^10=0.3486784401 (5)
(1-1/10^2)^100=… (6)
lim(1-1/10^n)^10^n =lim((10^n-1)/10^n)10^n=1/e (7)
1\=1/e of course.
So we get :
1\=0.9…. (If a^n\=b^n, then a\=b. ) End.
By the way, the limit of the series:
1.1,1.01,1.001,…
is like as the limit of the series 0.9,0.99,0.999,… in some way. In fact, the ︳1-0.9…|=
|1+1/10n-1︳=1/10n(n→∞). But perhaps no one said it is 1.

Changbai LI

Prove:0.9…\=1

phoenixthoth said:
easier way: 1/3 = 0.333... (three dots being key)

multiply by three to get
1 = 0.999...

that thread is closed but this is my two cents.

it's not that surprising, really, that every rational number (1 in this case) can be represented by a series (9/10 + 9/100 + 9/1000 + ...) because even fractions like 1/3 can be (3/10 + 3/100 + 3/1000 +...).

may your journey be graceful,
phoenix
1. 1^n=1^10^n=1(even n->infinite) (1)
2. lim(1+1/n)^n=e(when n->infinite) (2)
3. lim(1-1/n)^n=1/e(when n->infinite) (3)
4. lim(1-1/10^n)=lim((10^n-1)/10^n)=(0.9…)^10^n=1/e (when n->infinite) (4)
5. It is the key. So write it more detail in series:
(1-1/10)^10=0.9^10=0.3486784401 (5)
(1-1/10^2)^100=… (6)
lim(1-1/10^n)^10^n =lim((10^n-1)/10^n)10^n=1/e (7)
1\=1/e of course.
So we get :
1\=0.9…. (If a^n\=b^n, then a\=b. ) End.
By the way, the limit of the series:
1.1,1.01,1.001,…
is like as the limit of the series 0.9,0.99,0.999,… in some way. In fact, the ︳1-0.9…|=
|1+1/10n-1︳=1/10n(n→∞). But perhaps no one said it is 1.

Changbai LI

Prove:0.9…\=1

HallsofIvy said:
Oh, God, Not again!

Yes

This is "easier" because you are assuming all of the hard part. It works if you ASSUME 1/3= 0.333... (do you consider that to be simpler than 1= 0.999... itself?) and if you ASSUME that you can multiply a decimal number by multiply each digit. That's true but a good proof should show that.

Actually, any REAL number can be represented by such a series. That's what "decimal representation" means.
1. 1^n=1^10^n=1(even n->infinite) (1)
2. lim(1+1/n)^n=e(when n->infinite) (2)
3. lim(1-1/n)^n=1/e(when n->infinite) (3)
4. lim(1-1/10^n)=lim((10^n-1)/10^n)=(0.9…)^10^n=1/e (when n->infinite) (4)
5. It is the key. So write it more detail in series:
(1-1/10)^10=0.9^10=0.3486784401 (5)
(1-1/10^2)^100=… (6)
lim(1-1/10^n)^10^n =lim((10^n-1)/10^n)10^n=1/e (7)
1\=1/e of course.
So we get :
1\=0.9…. (If a^n\=b^n, then a\=b. ) End.
By the way, the limit of the series:
1.1,1.01,1.001,…
is like as the limit of the series 0.9,0.99,0.999,… in some way. In fact, the ︳1-0.9…|=
|1+1/10n-1︳=1/10n(n→∞). But perhaps no one said it is 1.

Zurtex

Homework Helper
Changbai LI said:
|1+1/10n-1︳=1/10n(n→∞). But perhaps no one said it is 1.
Have you heard of the Archimedean principle?

<<<GUILLE>>>

Zurtex said:
Have you heard of the Archimedean principle?
Wht is the archemidean principle?

Zurtex

Homework Helper
<<<GUILLE>>> said:
Wht is the archemidean principle?
There always exists a natrual number $n \in \mathbb{N}$ such that:

$$\frac{1}{n} < \epsilon \quad \epsilon \in \mathbb{R}^+$$

Applying this "Archimedean principle" it's not hard to get:

$$\lim_{n \rightarrow \infty} \frac{1}{n} = 0$$

And therefore if you can show a sequence Sn:

$$S_n \leq \left\{ \frac{1}{n} \right\}_{n \in \mathbb{N}} \quad \forall n\geq k \quad \text{for some} \quad k \in \mathbb{N} \quad \text{and} \quad S_n \geq 0\quad \text{for} \quad n\geq k$$

Then it stands that:

$$\lim_{n \rightarrow \infty} S_n = 0$$

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HallsofIvy

Homework Helper
An equivalent statement of the Archimedean Principle is that, given any integer n,
there exist a real number larger than n.

Zurtex

Homework Helper
HallsofIvy said:
An equivalent statement of the Archimedean Principle is that, given any integer n,
there exist a real number larger than n.
Yes I think we called it the law of large numbers or something along those lines.

HallsofIvy

Homework Helper
Actually, I stated it backwards: The Archimedian principle is that, given any real number x, there exist an integer n such that n> x.

It's true both ways but the way I originally stated it, it is easy to prove (given any n, take x= n+ 1/2). The true Archmedian Principle is more subtle than that.

No, that is NOT the "law of large numbers". The law of large numbers is a probability law, essentially saying that the larger a sample you take from a given probability distribution, the more likely the mean of the sample is to be close to the mean of the distribution.

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