You are not logged in.

- Topics: Active | Unanswered

**Traesk****Member**- Registered: 2005-04-04
- Posts: 1

The "proof" that says 0.999... = 1 goes like this:

(1) x = 0.999...

(2) 10x = 9.999...

(3) 9x = 9

(4) x = 1

(5) 0.999... = 1

In (1) it's stated that x is equal to 0.999... with an infinite number of 9s. When you multiply it by 10 in (2) you get 9.999... with an infinite number of 9s. But this can't be true. In (2) you must get 9.999... with 1 less than infinite number of 9s. If "infinite" was 3 for example, you would have x = 0.999 in (1), and 9.99 in (2). 9.99 - 0.999 isn't 9. It's 8.991. Divide that by 9 in (4), and you get 0.999, not 1.

By saying x = 0.999... in (1), you have already defined a number of decimals, which is arbitrarily large. (2) can't have the same number of decimals; it must be (infinite - 1) number of decimals. For this proof to be true, you have to add another decimal in (2). 9.999... - 0.999... will not become 9 if the number of decimals aren't the same.

Just a thought.

Joakim

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,611

And that is a good point. I guess it depends on how you look at it.

Taking your argument, they will always be one decimal place out, but that difference is at infinity.

So the difference is "infinitesimal", which means infinitely small, or "approaching zero as a limit".

So, **0.999... = (1 - infinitesimal)**, perhaps? Is that the same as saying **0.999... = 1** ?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**Roraborealis****Member**- Registered: 2005-03-17
- Posts: 1,594

I agree in a way, but infinite means 'no end'. No matter if you take the decimal away or not, the nines will still go on forever and ever.

And if there will always be a decimal place out, and difference is at infinity (which means: the difference is at the end) then there can't be since there is NO end.

School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?

Offline

**jam-pot****Member**- Registered: 2005-04-20
- Posts: 142

Yes, I agree infinity is a number or theorum that does not stop.

It is, like only one factor of things in our universe, can carry on forever past time and space with out restrictions or minipulating things like that.

Just a thought......

the allmighty spatula * want a tip* dont eat yellow snow: the meaning of life is a number and that number is 1

Offline

**baaba****Member**- Registered: 2006-03-03
- Posts: 1

The proof fails because of circular logic. For the proof to be valid, it must be known that 0.999... is a rational number, however the only way to know that is to prove that 0.999... equals 1 in another way, thus making the "easy" proof obsolete.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I never did buy that proof, but I like this one better.

Consider the distance between .999... and 1. We will call this distance d.

We know d can't be less than 0, as this would result in a number that is less than .999...

Let's assume d > 0. By the Archimedian Principle, there exists an n∈N such that 1/n < d. But d is the distance from 0.999... to 1. That is, 0.999... + d = 1. However, I believe we can all agree (although I'm unsure how to strictly prove) that 0.999... + 1/n ≥ 1. But 1/n < d. So 0.999... + d > 1. Thus, d is not the distance, contradiction.

We shown that d is not greater than or less than 0, so the only thing we are left with is d=0:

0.999... + d = 1

0.999... = 1.

*Last edited by Ricky (2006-03-03 12:13:42)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Going over that proof, I would have to say it is valid.

For the proof to be valid, it must be known that 0.999... is a rational number

I believe the proof is also valid for irrational numbers. For example, you can subtract pi from integers with no problem.

As long as 0.999... = 0.999..., then I see no problem.

Traesk, your objections stem from a misconception of infinite. Infinite shares almost no common properties as finite numbers. That you compare the properties of finte and assume they apply to infinity is sorely mistaken.

Infinity can be just plain weird. Roraborealis was right when he (she?) said that there would always just be another 9 to step up and take it's place.

infinite - 1 = infinite. Like I said, weird, but true.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Ricky, I think just because you prove something is not greater or less than zero, doesn't prove it is zero. Because infintesimal might be between zero and greater than zero.

**igloo** **myrtilles** **fourmis**

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

If a number is not 0, it is either greater than or less than zero. In other words, a number must be greater than 0, equal to it, or less than. There is no other possibility.

Edit: In other words, show me a number that is not greater than 0, less than 0, or 0. It simply does not exist.

*Last edited by Ricky (2006-03-03 15:38:22)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Yes, but infintesimal is possibly BETWEEN zero and greater than zero.

I'm just pulling your leg

**igloo** **myrtilles** **fourmis**

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,611

To paraphrase your proof, you are asking "what is the distance **d** between 0.999... and 1"

And then saying it cannot be *less than 0* and it cannot be *more than 0*, so it *is* 0, right?

Now, the *less than 0* case makes sense.

But the *more than 0* case relies on:

Ricky wrote:

... I believe we can all agree (although I'm unsure how to strictly prove) that 0.999... + 1/n ≥ 1.

Now, I don't need a full proof (I am a trusting kinda guy) but some reason to believe it would be nice.

(BTW, when in doubt I think 1 = 3 × (1/3) = 3 × 0.333... = 0.999...)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Anothed proof based on the demical system:

0.999...=9/10+9/100+...+9/10^n=lim(n->oo; Sum 9/10^i)=lim(n->oo;9*Sum1/10^i)=9*lim(n->oo;Sum1/10^i)=

=9*lim(n->oo;(1/9)*(10^n-1)/10^n)=9*(1/9)lim(n->oo;((10^n/10^n)-(1/10^n))=lim(n->oo;1-(1/10^n))=1-lim(n->oo;1/10^n)=1-0=1;

Simple limits.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Ancient Greeks did not have 0.111... nor 0.999... . What they had are rational numbers. 3/7 is a rational member. 0.1 can be translated into a rational number 1/10 from our decimal system.

but 0.999... is actually defined by 0.9+0.09+0.999+...=translated to=9/10+9/100+9/1000+...

and is it a rational number? At least NO by its BASIC FORM.

in the rational number world, 0.999... is not a number, and by only finite translation, it's only the sum of a INFINITE series.

0.999...?=1 is a question that whether the infinitesmall 1/10^n is equal to zero. And it's a question that infinitesmall?=0

And it had been a tough question for Newton and his following calculus mathematists. It was famous in history as Berkeley's Attack.

Cantor is the Neo who "dealed" this problem. (my point of view) He defined a real number to a infinite set including all the rational numbers ahead of it. And by his definition REAL number 1 is 0.999... to some extent, since 0.999... can represent the latest rational number before 1.

Maybe you've noticed the latest rational number before 1 or any real number cannot be explicitly pointed out. And that's the true problem--he just moved infinity from calculus and infinite series outside number system to inside it! He actually hided infinitesmall into the definition of a real number!

So 0.999... ,can be translated into a sum of rational number series by finite translation, then can be translated into real number 1, by infinite translation.

Real number 1 isn't the integer 1 or the rational number 1. Can a set of rational numbers define a rational number?This is Russel's Attack.

So your problem'sabug of real number system.

**X'(y-Xβ)=0**

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Ancient Greeks did not have 0.111... nor 0.999... .

0.111... = 1/9, which is rational.

0.999... as argued above, is one, and so maybe the greeks didn't write it that way, but they certainly had it.

Real number 1 isn't the integer 1 or the rational number 1.

Are you saying that 1 is different than 1/1 is different than 1.0?

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

You reminded me of another possible proof.

Between any two real numbers exists a real number.

So let's assume 0.999... does not equal 1. Then there must exist a number between 0.999... and 1. It is clear that no such number exists.

Therefore, contradiction, 0.999... = 1.

Edit:

Hmm, on further review, this seems to be almost exactly the same as my previous proof, just reworded.

*Last edited by Ricky (2006-03-13 17:31:06)*

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

0.111...?=1/9 --this was taught during primary school, but itsn't neccesarily true.

Ancient Greeks only had two types of number systems: the integer system and the rational numbers one. By "rational" as it literally means, an integer devided by an integer. all 0.1 0.01 0.001 can directly refer to rational numbers 1/10 1/100 1/1000, so the difference is small and superficial, or nominal. However, 0.111...cannot be written into this form.

how do you get 0.111...? rethink the procedure

1/9= (1*10/9) /10 = (1+1/9)/10= 1/10+1/9/10= continue to compute= 1/10+(1*10/9)/10/10= 1/10+(1+1/9)/100=1/10+1/100+1/900=11/100+10/900/10=...

Here Implicit induction is used despite in primary school. But don't take it for granted-

The induction only proves when you continue to compute to next number place, 0.11..1 gets a new 1 and the leftover shrinks by tenth!

Where does the leftover finally go? Infinitesmall again. So 1/9=0.111... is actually proved by induction and limit theory! 1/9=0.111...+infinitesmall. Berkeley's ghost is still on the way.

Of course primary school textbook won't go this deep. Better rethink it.

the gap between integer system and rational numbers system is a "/", but the gap between real number system and real number system is far more fundamental. real number system can't get away with infinite, whether it be infinitesmall or "...".

To conclude, the difference between integer system and rational number system is infinity, decimal system without ... is a special rational number system, but the difference between rational number system and real number one is INFINITE. Redefining 1 to 1/1 include a "/",but redefining 1 to a infinite set makes a huge difference.

Your proofs (the latest proof Dedekind's cut) are all based on real number system, and they are quite true within this system, these rules. But this rules are not commonly believed by mathematists.

Actually lots of mathematists questioned this defination, the first was Cantor's teacher, and later Russel. Russel's attack was fundamental-he questioned an element in the set( a rational member) can be defined by another set...As for here, i would use his argument to say a real 1 is not a rational 1.

Umm, 0.999... between 1? 0.999...5 might be the answer, i add 5 on the right of all your 9, nomatter how many or increasing.

*Last edited by George,Y (2006-03-13 20:12:01)*

**X'(y-Xβ)=0**

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

0.999...5

Because ... goes on for infinity, 5 is *never* included in that number. Thus:

0.999...5 = 0.999...

Where does the leftover finally go? Infinitesmall again. So 1/9=0.111... is actually proved by induction and limit theory! 1/9=0.111...+infinitesmall.

Can you show me an actual proof?

When you try to divide 1 by 9, you are always left with a remainder of 1, and this never changes. Thus, it's just .111111111111111111111111111 for however many times you want to go out. This is represented by .111... I don't see where this infinitesmall comes into play.

And even if it does, limits show that infinitely small is 0.

But this rules are not commonly believed by mathematists.

Which rules? I think the only one I used is that between every two real numbers, there exist another real number. I can prove this, it isn't a rule.

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

FORMER PROOF of

What the Quotient and the Remainder of the dividend 1/9 are

as soon as the Nth digit after the decimal point of the Quotient Comes Out

Hypothesis:(H(n)) the quotient is 0.11...1 (n 1s) and the remainder is 0.0...0(n 0s so far)1=1/10^n

Proof:

1) n=1

0.1

____

9) 1.0

9

_____

1------show the remainder is 0.1

H(1) is true

2) If H(k) is true, when k>=1, deduct H(k+1)

the quotient is 0.11...1(k 1s) and the remainder 0.00...(altogether k 0s)1, continue the division

kk+1 th digit after decimal point

| |

0.11...11

_______

9)1.0

...

...

__

10

9

__

*10

9

__

1---the remainder is 0.00..0(altogether k+1 0s)1

* the remainder in H(k) is 0.00...(altogether k 0s)1 with the form 1 corresponding the nth digit after the decimal point in the quotient. To further compute, add 0 on its right.

When H(k) is true, H(k+1) is also true

Conculusion) for n>=1, H(n) is true.

Proof of the error of the quotient to 1/9 is 1/900...0(n 0s)when the former has n digits after decimal point---

1/9-0.11...1(n 1s) =

10^n/(9*10^n) - 99...9/(9*10^n)

= 1/(9*10^n)=1/900...0(n 0s)

0.11...1 still not a rational number, the only rational way up to now is to express it as a series

{S(n)}={0.1, 0.11, 0.111, ...},

Limit[S(n)]=1/9 when n->infinitity

since the error has been derived, proof is omitted to the following

Error |S(n)-1/9|=1/(9*10^n) forever shrinks when n->infinity

Limit|S(n)-1/9|=0 when n->infinity

another expression for above is

S(n)->1/9, when n->infinitity

|S(n)-1/9|->0, when n->infinitity

Clearly |S(n)-1/9|=1/(9*10^n) is the infinitesmall in this case. it's a function of integers. it's not ZERO. it approaches to it. Any infinitesmall is a VARIABLE induced by n, x or some other varible.

If you say infinitesmall is zero here inorder to say 0.111... =1/9 Berkeley's Attack is far more than enough to tear the logic of calculus and limit theory.

if infinitesmall is zero ds/dt has a denomirator 0, no sense

if infinitesmall is not zero 0.111... never equal to 1/9, and realtime velocity can never be approximate:

V(t)=ds/dt=(k(t+dt)²-kt²)/dt=k+kdt kdt?

"And what are these fluxions? The velocities of evanescent increments. And what are these same evanescent increments? They are neither finite quantities, nor quantities infinitely small, nor yet nothing. May we not call them ghosts of departed quantities? "

---The analyst: or a discourse addressed to an infidel mathematician, written by Berkeley, published in 1734.

http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Berkeley.html

the former proof implies it's not controversial to rewrite 0.111... as {S(n)}, since it's a series covergent to 1/9, there always exists another series convergent more rapidly to 1/9.

Almost all theroems in real numbers come from one origin, Cantor's definition of real numbers.

"There always exist a number between two numbers" is saying real numbers are thick. rational numbers are thick, too.

But can you even say 0.99... or 0.11 a real number? (the first time i doubted this)

"This division will never be exact. We will keep getting 090909... To round it off to three places, we have calculated the digit in the fourth place. It is 9. Therefore, we will add 1 to the previous digit:

0.09090.091

One sometime sees, 1 ÷ 11 = 0.090909... The three dots, called ellipsis, mean "and so on for as far as you please according to the indicated pattern, or according to the rule."

!!!!This means that 1 ÷ 11 cannot be expressed exactly as a decimal. !!!! However, we may approximate it to as many decimal places as we please by following the pattern 090909."

http://www.themathpage.com/ARITH/divide-whole-numbers-2.htm

between two !!!! 's(added by me) intrigued my doubt.

Maybe i should reread some books before exactly illustrating my point of view on Cantor's definition

**X'(y-Xβ)=0**

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Your proof breaks down here:

Proof of the error of the quotient to 1/9 is 1/900...0(n 0s)when the former has n digits after decimal point---

1/9-0.11...1(n 1s) =10^n/(9*10^n) - 99...9/(9*10^n)

= 1/(9*10^n)=1/900...0(n 0s)

0.11...1 still not a rational number, the only rational way up to now is to express it as a series

You just went from finite and tried to apply it to infinite. Sure it will work for every single positive n that exists. * But infinity is not a number*.

Clearly |S(n)-1/9|=1/(9*10^n) is the infinitesmall in this case. it's a function of integers. it's not ZERO. it approaches to it. Any infinitesmall is a VARIABLE induced by n, x or some other varible.

But we must go infinitly far out. Since this is the case, what's the difference between approaching 0 and being 0?

In other words, whats the distance of 1/n as n goes to infinity. Pick any number less than 0, and I'll call you an idiot (that was a joke ). Pick any number greater than 0, and I can find a number less than it. Thus, since it isn't less than or greater than 0, it must be 0. Since the distance is 0, it is 0.

if infinitesmall is zero ds/dt has a denomirator 0, no sense

ds will be approaching 0 as dt approaches 0, and thus, we are left with 0/0, an *indeterminate form*, and so it does make sense.

*Last edited by Ricky (2006-03-14 16:39:40)*

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Ricky wrote:

.But infinity is not a numberwhat's the difference between approaching 0 and being 0?

Pick any number greater than 0, and I can find a number less than it. Thus, since it isn't less than or greater than 0, it must be 0. Since the distance is 0, it is 0.

if infinitesmall is zero ds/dt has a denomirator 0, no sense

ds will be approaching 0 as dt approaches 0, and thus, we are left with 0/0, an

indeterminate form, and so it does make sense.

I think it's time to make up some agreement and brief the points.

1-0.999...=infinitesmall by k/10^n type

1/9-0.111...=infinitesmall by k/10^n type

1/infinitesmall is Exactly Stable 0,so that .999...=Exactly1 , 0.111...=Exactly 1/9, or else you admit infinitesmall isn't as SAME as 0, THOUGH "CLOSE"

between 2 real numbers exists a real number.

By the way, no easy excuses, "you cannot simply apply finite to infinite" is one of them, it will be considered an excuse, not one explaination. Because the question "Why?" "How?" will challenge it.

INFINITE is not a CARD GAME, with whatever set-up rules in mathematists' imaginary world.

It should be LOGICALLY CONSISTENT .

Next time i will explain it in more detail.

**X'(y-Xβ)=0**

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

1-0.999...=infinitesmall by k/10^n type

1/9-0.111...=infinitesmall by k/10^n type

1/infinitesmall is Exactly Stable 0,so that .999...=Exactly1 , 0.111...=Exactly 1/9, or else you admit infinitesmall isn't as SAME as 0, THOUGH "CLOSE"

between 2 real numbers exists a real number.

Unless I misunderstand your wording, I believe I agree with all of those.

By the way, no easy excuses, "you cannot simply apply finite to infinite" is one of them, it will be considered an excuse, not one explaination. Because the question "Why?" "How?" will challenge it.

Sure, and I gave a reason behind it. Infinity is not a number. You were using variables, which must be some real number.

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Do you agree real numbers are defined as variables, and they are mobiles?

I could give a example that approaching is not being even in real number system.

What's the difference between [0,1) and [0,1]?

**X'(y-Xβ)=0**

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

George, I agree that approaching is different than being. But that's not what we are talking about. 0.111... isn't approaching anything. It's a notation, and nothing more. It means "0.1 with an infinite number of ones after it." Now, I think you are trying to prove that 1/9... is different than "0.1 with an infinite number of ones after it." Now if that is so, then please tell me which decimal place does not have a '1'.

What's the difference between [0,1) and [0,1]?

One's closed, one's neither closed nor open.

*Last edited by Ricky (2006-03-16 16:58:26)*

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Do you agree real numbers are defined as variables, and they are mobiles?

Can you define mobiles?

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,611

I have a theory. Infinity is sometimes seen as something that "grows" which is misleading. I have seen Infinity defined as being "1 bigger than any number you can think of", which gives the idea that infinity grows and grows. Also saying "it goes on and on" can also give that idea.

But infinity is already full grown.

For example, a circle with infinite radius is a straight line.

So there is no end to 0.999...

Right? (He asks tentatively)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline