1=.9999...

[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]

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]

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]

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]

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]

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]

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]

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.

[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]

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]

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]

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]

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

[<<<GUILLE>>>]

Have you heard of the Archimedean principle?

Wht is the archemidean principle?

[Zurtex]

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

[HallsofIvy]

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

[Zurtex]

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]

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.

[Zurtex]

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.
Hmm, well just checked my notes and that's what I have written down, guess that's another lecturer who gets their stuff mixed up, it annoys me greatly listening to so many misquotes and wrong names given from lecturers.

[Changbai LI]

I know 1/3=0.3…, then multiply by three to get 1=0.9….
Anyone knows it .
But it is easy to see that 1/3 was finished. In other words, it got the end. And 0.9… is going on on the contrary. In other words, it is not finished forever.
So 1/3 is not equal to 0.9…, I think.
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

[Zurtex]

I know 1/3=0.3…, then multiply by three to get 1=0.9….
Anyone knows it .
But it is easy to see that 1/3 was finished. In other words, it got the end. And 0.9… is going on on the contrary. In other words, it is not finished forever.
So 1/3 is not equal to 0.9…, I think.
You're right 1/3 does not equal 0.999..., 1/3 equals 0.333....

Would you please care to explain yourself a little better, what you wrote there makes very little sense.

[HallsofIvy]

I know 1/3=0.3…, then multiply by three to get 1=0.9….
Anyone knows it .
But those people who object to 1= 0.999... are exactly the people who object to 0.999....

But it is easy to see that 1/3 was finished. In other words, it got the end. And 0.9… is going on on the contrary. In other words, it is not finished forever.
I don't understand this. 1/3 contains only a finite number of symbols but 0.333... cetainly does "go on and on"- it's no different from 0.999... in that way.

So 1/3 is not equal to 0.9…, I think.
Surely you didn't mean that!

[myhbc]

1=0.999...
assume a= 0.999...
then 10 * a = 10 * 0.999...
10 * a = 9.999...
10 * a = 9 + 0.999...
or a = 0.999...
then
10 * a = 9 + a
10 * a - a =9
9 * a = 9
a = 1

[HallsofIvy]

Exactly what a thousand people have said before. And obvious IF you assume that 10*(0.99999...)= 1.00000... Have you proved that the simple arithmetic operations you apply every day to numbers with finite number of decimal places also apply to those with unterminating decimal expansions?

[nate808]

if the probability of something not happening, such as tossing a potentially infinite number of heads in a row is .99999 repeating, hiw could you say that it is 1 since there is still a chance of it happening

[HallsofIvy]

if the probability of something not happening, such as tossing a potentially infinite number of heads in a row is .99999 repeating, hiw could you say that it is 1 since there is still a chance of it happening


I don't understand what you mean by "a potentially infinite" number of heads. It's impossible to toss and infinite number of heads because you don't have infinite time in which to toss an infinite number of coins whatever they come up to!

However, there is a fundamental assumption here that is not true- it is only in "discrete" (finite) probability that "probability 1" means"certain to happen" or "probability 0" means "impossible". Take for example the experiment "choose a number between 0 and 1 inclusive" where we take the probability distribution to be the uniform distribution- every number is equally likely to be chosen. The probability of any specific number being chosen is 0- but clearly some number has to be chosen. The probability of choosing 1/2 or 1/3 or \pi or e is 0 but since this is a uniform distribution, they are as likely to be chosen as any other number. The probability that "any number except 1/2 will be chosen" is 1 but certainly it is possible that that will not happen- that 1/2 will be chosen.

[Zurtex]

if the probability of something not happening, such as tossing a potentially infinite number of heads in a row is .99999 repeating, hiw could you say that it is 1 since there is still a chance of it happening
If you toss an coin an infinite number of times, you will eventually get a tails, regardless if you get an infinite number of heads.

But that's really side tracking from the subject of the thread.

[Hurkyl]

If you toss an coin an infinite number of times, you will eventually get a tails, regardless if you get an infinite number of heads.

No, you will get a tails with probability 1.

Actually, there's a fundamental mathematical problem with this whole issue anyways: it's a nontrivial task to figure out how one can even apply the concept of probability to the set of all possible outcomes.

[Sydius]

Uhm… maybe someone already said this, but 0.333… comes infinitely close to 1/3rd, but is not 1/3rd. 0.999… repeating comes infinitely close to 1, but is not 1.

Infinitely close, for all practical purposes, might as well be “equal”, which is why they say “1/3 = 0.333…” instead of “1/3 (squiggly equal sign) 0.333…” – it would just confuse students and make middle and high school math teachers get stressed out trying to explain something, that, for all practical purposes, does not matter. Regardless, though, being infinitely close to something does not make it that something and therefore saying “1/3 = 0.333…” is infinitely close to the truth, but not the truth. :smile:

[TD]

Maybe you didn't read this thread properly.

0.3333 \approx 1/3 and 0.333333333 \approx 1/3 is even better but 0.3333... = 1/3 is exactly the same, at least assuming that you mean an infinite expansion of decimal 3's.

[Sydius]

Maybe I am not reading it properly, but 0.3333… (With an infinite expansion of decimal 3’s) is NOT = to 1/3, it is just infinitely close.

Anytime you get a repeating decimal, you actually have an answer which is infinitely close to the true answer, but not the true answer. The repeating 3 is just a way to express “this is infinitely close to 1/3rd” – not that it *IS* 1/3rd…

Just like 0.9999… is infinitely close to 1, but not 1.

0.333… is infinitely close to 1/3, but not 1/3.

Maybe that is what you are saying, and I am just missing it. :-|

[TD]

No, that's what you're saying, but it's not correct.
I suggest you read the first 2 pages of this thread first :smile:

[Sydius]

It is true. Maybe we are just arguing about the applicability of the “equal” sign, but in the strictest term of the word, ONE divided by THREE does *NOT* EQUAL 0.3333 (Repeating)

Just because a number is infinitely close to another number, does not make them the same.

If you wrote a function that simply kept adding a “3” to the decimal, the LIMIT of that function would be 1/3rd, but it would NEVER reach 1/3rd, even if it went on for infinity.

Infinitely close to x != x

[Gokul43201]

...but in the strictest term of the word, ONE divided by THREE does *NOT* EQUAL 0.3333 (Repeating)Yes, it DOES.

Just because a number is infinitely close to another number, does not make them the same.Yes, it DOES.

[TD]

I'm affraid this is useless, but would you want to comment here beginning with "Uhm… maybe someone already said this," if you clearly didn't read the previous replies yet.

[Sydius]

*cries* I don’t get it!

So, if any function comes infinitely close to a number, than that function(infinity) = that number?

[HallsofIvy]

Maybe I am not reading it properly, but 0.3333… (With an infinite expansion of decimal 3’s) is NOT = to 1/3, it is just infinitely close.

Anytime you get a repeating decimal, you actually have an answer which is infinitely close to the true answer, but not the true answer. The repeating 3 is just a way to express “this is infinitely close to 1/3rd” – not that it *IS* 1/3rd…

Just like 0.9999… is infinitely close to 1, but not 1.

0.333… is infinitely close to 1/3, but not 1/3.

Maybe that is what you are saying, and I am just missing it. :-|

No, you are just completely wrong. 1/3 means (by definition of a "place ten numeration system") the sum of the infinite geometric series
\frac{3}{10}+ \frac{3}{100}+ \frac{3}{1000}+ ... and that, as any good high school graduate should know is
\frac{\frac{3}{10}}{1-\frac{1}{10}}=\frac{\frac{3}{10}}{\frac{9}{10}}= \frac{3}{9}= \frac{1}{3}.

0.333.... is exactly equal to 1/3.

Similarly, 0.999.... means the sum of the infinite geometric series
\frac{9}{10}+ \frac{9}{100}+ \frac{9}{100}+ ...
which is
\frac{\frac{9}{10}}{1-\frac{1}{10}}= \frac{\frac{9}{10}}{\frac{9}{10}}= 1

0.999... is exactly equal to 1.0

[Sydius]

I'm affraid this is useless, but would you want to comment here beginning with "Uhm… maybe someone already said this," if you clearly didn't read the previous replies yet.
I read them, but probably did not understand them. (I guess that is obvious now)

[TD]

Ok, that's no problem then :smile:

The things is, you shouldn't see this 'geometrically' where the 0.9, 0.99, 0.999 etc gets closer and closer to 1 but never there. Purely analytically, this is a limit (infinite number of 9's) and that limit doesn't "approach" 1, it's "equal" to 1.

[Sydius]

Okay, okay, sorry – lack of comprehension on my part. That, and math teacher telling me I was right when I was not (sigh)… I apologize.

I get the math, but still do not understand it in my mind, how a number infinitely close to another number is that number…

[Sydius]

Ok, that's no problem then :smile:

The things is, you shouldn't see this 'geometrically' where the 0.9, 0.99, 0.999 etc gets closer and closer to 1 but never there. Purely analytically, this is a limit (infinite number of 9's) and that limit doesn't "approach" 1, it's "equal" to 1.

Ohhhhhhhh! I see now. That makes sense :) Thanks...

[Sydius]

Just to be sure, though…

When you see x = some-repeating-decimal, it actually means x = limit(some-function-that-would-generate-that-repeating-decimal)?

[HallsofIvy]

When you see "x= some decimal, repeating or not" it actually means that x is the sum of the infinite series implied by that decimal:

if x= 0.abc.... then x= (a/10)+ (b/100)+ (c/1000)+ ...

[Hurkyl]

When you see x = some-repeating-decimal, it actually means x = limit(some-function-that-would-generate-that-repeating-decimal)?

Yes, that is one way to define the decimals.


Incidentally, the decimals can instead be defined without resorting to calculus simply by specifying how to do the arithmetic with infinite strings of digits. (which includes things like 0.999~ = 1)

[HallsofIvy]

Then you are doing better than I could! I couldn't figure out what was meant by "some-function-that-would-generate-that-repeating-decimal"!

[Hurkyl]

I presume he means something like the sequence of partial sums of that series.

I.E. f(1) = 0.9, f(2) = 0.99, f(3) = 0.999, ...

[Sydius]

Yup, that is what I meant. :biggrin: I get it now. Thanks.

[moose]

I guess I will never believe .999 repeating = 1

I mean, if something asymptotes at 1, it gets infinitely close to 1, but it never actually reaches one... you can have .999 repeating but not 1

.99999 repeating is infinitely close to one, but how can it be one? I have read all these proofs, but for example, when you take the sum of the infinite series, it's always defined as approaching that number but not actually being that number soooo, i dont know I will just always find there is something wrong in my head...

[Hurkyl]

In the real numbers, the only infinitessimal number is zero. Therefore, if x is infinitely close to y, then x - y = 0, and therefore x = y.

but how can it be one?
The same way the fraction 2/2 can be one -- their numerical values are equal.

the sum of the infinite series, it's always defined as approaching that number
No, it's not. The sum of an infinite series is nothing more than a number. A number cannot be "approaching" some other number.

It's the sequence of partial sums that is approaching something. It is true that no partial sum of the series \sum_{i=1}^{\infty} 9 * 10^{-i} will ever be equal to one. It is true that this sequence of partial sums becomes arbitrarily close to one. (Meaning that if you pick a positive number ε, then there is a partial sum whose distance to one is less than ε)

But the sum of the series \sum_{i=1}^{\infty} 9 * 10^{-i} is one.

[polyb]

Having randomly bopped into this one I think I may have a small point to make of this issue. You guys have done some pretty interesting twists and turns with your logic but you have seem to missed something very simple. So bear with me:

As stipulated, if:

1/3=0.33333...

and

2/3=0.66666...

then by a conclusion we say,

1/3+2/3=0.99999...

Which looks like

1=0.99999...

Which plenty of arguments were given to make this so, but here is my issue:

If

1=0.99999...

then

1-0.99999...=0 ; which is obviuously not true. So what are we missing? Perhaps the notion of an infinitesimal is applicable. If we say this:

1=0.99999...+dx,where dx is an infinitesmal number, then the equation holds true and does not leave us with the earlier contardiction.

But if this considered true, then might it better to say,

1/3=0.33333...+dx?

I dont know if you call this a definition but I did find this interesting enough to pass along.

Thoughts?

[arildno]

1-0.99999...=0 ; which is obviuously not true.
Wrong; it is true.

[HallsofIvy]

1-0.99999...=0 ; which is obviuously not true. So what are we missing?

Perhaps you meant to type "which obviously IS true"? If not then what YOU are missing (I don't know about anyone else) is that 1- 0.999...= 0 is completely true. "1" and "0.999..." are just different ways of expressing exactly the same thing. Their difference is exactly 0.

[arildno]

Consider the following, polyb:
1/2, 2/4 and 3/6 are merely different ways of expressing the same number. You don't have any problems with that, do you?

[russ_watters]

1-0.99999...=0 ; which is obviuously not true. So what are we missing? Perhaps the notion of an infinitesimal is applicable. If we say this:

1=0.99999...+dx,where dx is an infinitesmal number, then the equation holds true and does not leave us with the earlier contardiction.

But if this considered true, then might it better to say,

1/3=0.33333...+dx?

I dont know if you call this a definition but I did find this interesting enough to pass along.

Thoughts? The part you are missing is that the "..." covers that "+dx" term that you are trying to add.

[polyb]

arildno,

No I do not have a problem with that. The fractional or ratio representation of those numbers in terms of a decimal number does not carry the same continued aspects as such numbers like 1/3 does.

My logic was more keen on trying to resolve the discrepencies between fractioanl or ratio representation and the decimal form. The hope was that the representations were more correct.

Ironically, this has a long history to it! The old greeks didn't really like those irrational numbers and they could not trisect an angle. Of course, neither can we!

russ,

I see what you are saying but

1-0.99999...=0 still does not hold true and hence why I introduced the infinitesimal to resolve the discrepency. It is an interesting curiosity, isn't it? This special case where a ratio is not exactly the same as it's continued decimal form.

Halls,

I have to disagree. Perhaps this is a best way of representing the equation:

1.00000... \neq 0.99999... because

1.00000...-0.99999... \neq 0

This is the tautology I'm batting around.

Remember these are just some small thoughts of a random poster!:biggrin:

Please excuse any misuse of language, it's Saturns' day!

[russ_watters]

russ,

I see what you are saying but

1-0.99999...=0 still does not hold true and hence why I introduced the infinitesimal to resolve the discrepency. It is an interesting curiosity, isn't it? This special case where a ratio is not exactly the same as it's continued decimal form. No, it isn't curious at all, because 1-0.9...=0 does hold! If it didn't, then there'd be a number other than 0 that 1-.9... equaled. Since there isn't, it must equal 0.

This thread will probably need to be closed: polyb, you are arguing against definitions here. There's just nowhere to go with an argument like that.

[Integral]

Polyb,
I have seen and heard your same argument so many times that it is losing its humor.

Here you are, in a forum, arguing basic architecture of the Real number system with mathematicians. An analogy of this is if you were to argue the height of a building with the man holding the blue prints.

A correct approach would be to attempt to ask questions about and learn why what seems obvious to you is wrong.

My history of this is that it does no good to present you with mathematical proofs, because you do not have the mathematical sophistication to understand them. We can talk till we are blue in the face, nothing changes.

So the word is, change your approach or this thread will be locked.

[polyb]

OK Integral,

Explain why

1.00000...=0.99999...

Please!

My reasoning tells me that for all pratical purposes it is true, in reality that is usually well within the tolerances of observation. But when I look at that equation I see the discrepency of an infinitesimal.


My history of this is that it does no good to present you with mathematical proofs, because you do not have the mathematical sophistication to understand them. We can talk till we are blue in the face, nothing changes.

Please enlighten me if you are so inclined, otherwise how am I to posit that you have an explanation?

Why would this line of reasoning kill the thread? In essence this boils down to the concept of an infinitesimal, which by no means a simple idea.

Does this have more to do with mixing the idea of real numbers with that of integers?

Physical consideration:

Theoretically you can approach the speed of light but only asymptotically, you never reach c. So in the physical concept c does not eaqual 0.999....c. Perhaps this is where I am drawing some of this reasoning and why I am posting the thought.

[krab]

It has nothing to do with an infinitesimal. It has to do with limits; something you don't appear to understand.

[cdm1a23]

If only we used a base-9 system... there would be no arguments

1/3 = .3 :)

[polyb]

It has nothing to do with an infinitesimal. It has to do with limits; something you don't appear to understand.

elaborate and enlighten please!:biggrin:

If only we used a base-9 system... there would be no arguments

1/3 = .3 :)

Too funny!

Hey, what else is there to do on a staurday night!?!?!:surprised

[cdm1a23]

My friends and I argued this forever, and came to the conclusion that

1/3 = .333... mathematically simply based on definition, but that there is no such thing as .333... in anything except math, so really it must be right because the only context in which it is possible to have .333... contains the definition that .333... = 1/3

[Gokul43201]

Explain why

1.00000...=0.99999...'Cause it's defined that way for the reals. There are no infinitesimally small numbers in the reals, so the two representations point to the same real number, the multiplicative identity.

If you won't take anyone's word here, in addition to the dozens of demonstrations, please read the following completely. It's wriiten by the 1998 Field's Medal winner, Tim Gowers.

http://www.dpmms.cam.ac.uk/~wtg10/decimals.html

[polyb]

I can easily see hours being dwindled away on this topic.

Can be fun though!

I'm guessing that perhaps the thing here is the fact that the ratio of two whole numbers or integers(?) being defined in terms of a real number. Perhaps this is where I need to exercise some more discrimination.

As for the limiting process, zeno never reaches one, he is always an infinitesimal away. Correct?

I guess Einstein never reaches c either!


The strange thing to me is that certain ratios dont have the '...' and others do. For the ones that do, odd things like this topic arise.

How come I have the nagging suspicion that trisecting an angle has something to do with this?:uhh:

[vanesch]

OK Integral,

Explain why

1.00000...=0.99999...

Please!



The decimal representation of real numbers is indeed redundant.
A decimal representation is a mapping from Z into the finite set of symbols {"0","1",..."9"} (if we work in basis 10), which you can call n(z), with the property that there exists a number z0 such that n(z>z0) = "0".
You can define the "partial sum":
S(m) = Sum_{k = 0}^m 10^{z0 - k} n(z0-k)
The real number that is represented by n(z) is then given by the limit:
R[n] = lim_{m \rightarrow \infty} S(m)

We have now two different "decimal representations":

n1 has z0 = 1 and n1(1) = "0", n1(0) = "1", n1(-1) = "0", n1(-2) = "0"....

n2 has z0 = 0 and n2(0) = "0", n2(-1) = "9", n2(-2) = "9",....

It turns out that, according to the above procedure, both n1 and n2 have the real number associated to them equal to 1 = R[n1] = R[n2].

cheers,
Patrick.

[polyb]

'Cause it's defined that way for the reals. There are no infinitesimally small numbers in the reals, so the two representations point to the same real number, the multiplicative identity.

If you won't take anyone's word here, in addition to the dozens of demonstrations, please read the following completely. It's wriiten by the 1998 Field's Medal winner, Tim Gowers.

http://www.dpmms.cam.ac.uk/~wtg10/decimals.html

Thanks Gokul, that will give me something to mull over. On brief glance I like his addressing the subtlety involved on this topic.

As for infinitesimals, I realize that there are no infinitesimally small real numbers but these objects are often useful for dealing with things. Please excuse my lack of proper etiquette with this subject but that is why I'm here, to gain some understanding in a chatroom through some random strangers that use pseudonyms!


As for taking others words for it: all I am asking is for is some enlightenment on the subject and hence the posting, so far I have had some pretty curt responses while others have been more helpful. My gratitude is with the later.

The decimal representation of real numbers is indeed redundant.
A decimal representation is a mapping from Z into the finite set of symbols {"0","1",..."9"} (if we work in basis 10), which you can call n(z), with the property that there exists a number z0 such that n(z>z0) = "0".
You can define the "partial sum":
LaTeX graphic is being generated. Reload this page in a moment.
The real number that is represented by n(z) is then given by the limit:
LaTeX graphic is being generated. Reload this page in a moment.

We have now two different "decimal representations":

n1 has z0 = 1 and n1(1) = "0", n1(0) = "1", n1(-1) = "0", n1(-2) = "0"....

n2 has z0 = 0 and n2(0) = "0", n2(-1) = "9", n2(-2) = "9",....

It turns out that, according to the above procedure, both n1 and n2 have the real number associated to them equal to 1 = R[n1] = R[n2].

cheers,
Patrick.

Thanks Patrick, so now I have to ask:

So can we say R[n1]-R[n2]=0? Perhaps only in the case where the partial sum's upper bound goes to infinity?

[Integral]

I find it interesting that , "because it is obvious" is your only proof, while from us you demand formal proof. Perhaps you should try proving your claims.

Here is my version of the proof (http://home.comcast.net/~integral50/Math/proof2a.pdf ) if you consider my construction carefully you may get a glimpse of why this equality holds.

[Gokul43201]

As for infinitesimals, I realize that there are no infinitesimally small real numbers but these objects are often useful for dealing with things.Look into the hyperreals (http://mathworld.wolfram.com/HyperrealNumber.html) then. Within the reals though, the only infinitesimally small number is 0.

[vanesch]

Here is my version of the proof (http://home.comcast.net/~integral50/Math/proof2a.pdf ) if you consider my construction carefully you may get a glimpse of why this equality holds.

That's neat :approve:

[Integral]

That's neat :approve:
Thank you, I think so too. A significant point of interest is that I never do arithmetic on digits "at infinity" unlike the usual demonstrations (3 * .333... etc. ) . The worst part is having to resort to the nested interval theorem, This is where the lack of mathematical sophistication that I mentioned in my earlier post comes in. I have really only proved this for those who understand that theorem, ie preaching to the choir.

The key to understanding this is that the symbol .999... means an infinite number of 9s. Not some finite number of 9's approaching infinity but an infinite number. Perhaps the lack of understanding comes from the concept of "you can never reach infinity" While this is true in a physical sense, mathematically we can represent the concept of having reached infinity, one way to express this is the ellipsis at the end of a pattern of digits.

[cdm1a23]

In skimming over these posts I have seen people argue with the evidence that: .999... + 9 = 9.999...

Is this correct math?

By the way, I understand the answer to the subject question of .999... = 1, I just was wondering the above question as a separate topic.

[Human Being]

Hi, forgive me if I'm being redundant...

Whatever happened to the uber-simple:

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

?? If that hasn't been discussed yet, is it "correct"?

[cdm1a23]

Hi, forgive me if I'm being redundant...

Whatever happened to the uber-simple:

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

?? If that hasn't been discussed yet, is it "correct"?

This is basically what I am asking about... is this a legitimate "proof"? It seems like there is something wrong about it that I can't quite see... like it is right because it is not actually proving anything, but only manipulating its own given information? Does anyone know?

[Integral]

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1


Sure this is correct, but it is not a proof it is a demonstration. The Theorems of Real Analysis which show that to be meaningful are the same ones which are needed in the proof of 1 = .999... Mainly you need to show that .999... is indeed a fixed number. if you assume that is a legal operation you have essentially assumed the result, therefor you are using circular logic.

Again, the proof I have posted above is nearly fundamental. First I show that .999... is locked inside of a set of nested intervals. I have specified the theorem used to prove that there can only exist 1 real number in the intersection of a infinitely nested set of intervals. The fact that both 1 and .999... exist inside of the SAME set of nested intervals means that the occupy the same location on the Real Number line, this means that they are equal.

[Gokul43201]

Hi, forgive me if I'm being redundant...

Whatever happened to the uber-simple:

x = 0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1

?? If that hasn't been discussed yet, is it "correct"?It's correct because it works. It's not obvious though, how one multiplies numbers that are represented by infinitely long sequences of digits, even if you're just multiplying by 10.

[Hurkyl]

Most importantly, this example is a good demonstration of why mathematicians define the decimals the way they do.

[polyb]

Thank you to everyone that responded to correct my thinking on this topic. I apologize for the delayed reply. I now understand why you say 1=0.999...! This short discourse has given me a opportunity for an overdue, albeit brief, review of limit and number theory. THANK YOU!:biggrin:
I would like to review my thinking in order to make sure my thought is correct.
Using vanesch's suggestion, it appears that I was constructing both numbers as follows:
For 1.0..., I was thinking
S(m)_{(1.0...)} = \sum^m_{k=0}10^{z_{0}-k}n(z_{0}-k)
where
n_{1} has z_{0} = 1 and n_{1}(1) = "0", n_{1}(0) = "1", n_{1}(-1) = "0", n_{1}(-2) = "0"....
and for 0.9..., I was thinking
S(m)_{(0.9...)} = \sum^m_{k=0}10^{z_{0}-k}n(z_{0}-k)
where
n_{2} has z_{0} = 0 and n_{2}(0) = "0", n_{2}(-1) = "9", n_{2}(-2) = "9",....
Of course the flaw in my thinking had more to do with the fact that it was finite and I was only doing a patial sum. Hence why I said 'it is obvious', at least so I thought, that:
S(m)_{(1.0...)} \neq S(m)_{(0.9...)}
because
S(m)_{(1.0...)} - S(m)_{(0.9...)}\neq 0
which is true if you are only considering a finite or partial sum. This led me to conclude:
S(m)_{(1.0...)} - S(m)_{(0.9...)} = p^m
Where we would could say for this case that p=0.1, which would be true as long as the sum was partial. But this is not the case, for as pointed out by vanesch. For in order to qualify these as real numbers, the limit of m had to go to infinity, or
R(n)=lim_{m \rightarrow \infty} S(m)
in which the difference that I was seeing would go to zero, or
p^m \rightarrow 0
I suspect that this number p^m is correlated to the nested intravel proof (http://home.comcast.net/~integral50/Math/proof2a.pdf) that Integral so thoughtfully provided(BTW, thanks Integral, this really helped me wrap my mind around this nugget plus it was quite neat! :biggrin:).
So there you have it, hopefully I have corrected my thinking. If you think that further correction is needed please post your thoughts!
There is one thing I did come across that would finalize it for me, which is the following:
Does
lim_{m \rightarrow \infty} [1^m = (0.9...)^m]?
Other thoughts:
Regarding the trisection of angle, after some thought I came to realize what had been nagging me. Though one can bisect an angle it seems that the only way to trisect an angle is to use a limiting proceedure of bisections. If anyone has any thoghts please let me know. I am also curious as to wether any of the greeks had worked out limiting proceedures to any extent.
As for the physical case of the speed of light and Einstein's wish to go that fast, I am curious how fast he would have to go in order to be in that intravel such that nature agrees to say c=0.99...*c. Are these the same energies needed to create a black hole? Is this a feasable thought? I realize that this is another topic that belongs elsewhere but I had to finish the thought.
Once again THANK YOU everyone that helped in this exercise!
Some Peanuts for your thoughts! (http://simpler-solutions.net/pmachinefree/thinkagain/comments.php?id=1589_0_3_0_C)

[Gokul43201]

It takes an infinite amount of energy to accelerate something to c, hence it's not possible.

[polyb]

I digressed!:uhh:

[Selak3]

ok, the math works by definitions
but can it exist that A not= B, but A infinitely close to B?
I conjecture something like that does exist in reality.o:)

Selak

[hypermorphism]

ok, the math works by definitions
but can it exist that A not= B, but A infinitely close to B?
I conjecture something like that does exist in reality.o:)
Selak
Define "infinitely close" in a rigorous manner and we'll see. :smile:

[arildno]

I conjecture something like that does exist in reality.o:)
Selak
What has "exist in reality" to do with this? :confused:

[Selak3]

I guess the nature of reality does have something to do with this.

Is there a part of the universe that is indefinitely divisible? Ie: you keeping cutting
a part of reality in half but it never complete disappears? (ie, you get infinitely close
to 0 but never quite reaching it).

Would the definition of 1/x as x approaches to infinity need to be modified?
Would one need to modify mathematics in these cases?
Would one want mathematics to reflect reality?o:)

[arildno]

No, the "nature of reality" has nothing whatsoever to do with mathematical definitions.

[HallsofIvy]

I guess the nature of reality does have something to do with this.
Is there a part of the universe that is indefinitely divisible? Ie: you keeping cutting
a part of reality in half but it never complete disappears? (ie, you get infinitely close
to 0 but never quite reaching it).
Would the definition of 1/x as x approaches to infinity need to be modified?
Would one need to modify mathematics in these cases?
Would one want mathematics to reflect reality?o:)

You have only a rough Idea what mathematics is. Mathematical theories are consistent- that is all we can ask of them. If there is some new fact of "reality" (I'm not sure what you are talking about here- a sort of physics perhaps?) that makes the mathematical model being used not correct, that means you have to change your model. That happens all the time. Mathematics itself stays the same. The way mathematics is applied changes.

(Arildo typed a shorter response and got it in before me! But we are really saying the same thing.)

[hypermorphism]

Would one need to modify mathematics in these cases?
Physics, perhaps. Mathematics is only modified through addition of more concepts, or the finding of a logical flaw in existing concepts. Reference to models of reality is irrelevant.

Would one want mathematics to reflect reality?o:)
No, but that's the exact job description of physics. :biggrin: Mathematics is more explorative in terms of abstract objects and relationships than the stagnation that would result if one had to wait for empirical models of reality.

[arildno]

As a follow-up to my own, and hallsofIvy's comments (and hypermorphism's):
To be sure, "reality" provides a spur to develop new mathematics and "old" mathematics is used in order to create a "reality model".

This, however, does not impinge upon whether or not a given set of axioms defines a consistent or inconsistent mathematics.

[Hurkyl]

Bleh, this archived thread keeps popping up. I think it's time to put an end to this necromancy.