.999999999999 repeating = 1

[Robert J. Kolker]

The problem is there is no remainder. 1/3 has no remainder.

In the ring of integers (where quotients are generally not defined)

1 = 0*3 + 1, so the remainder is 1.

For two integers m and n, where m > n, we can write m = q*n + r where q and r are integers

and r lies between 0 and m-1. If m < n, m = 0*n + m so m is the "remainder".

Bob Kolker

[Robert J. Kolker]

Ah, that's actually very interesting Jake. I've been arguing this whole time against a proof that was already false for completely different reasons, it appears?

From what I observed, everyone was taking this:

.33333... * 3

to be valid math. You're the first person to ever tell me that it's not. I assumed that the infinite sequence of 3's was automatically multiplied, every single one of them.

The concept I was trying to create and define was meant to do basically what you just said. To be exact and just perform the math on the fraction instead of being inexact and doing it on the infinite sequence of numbers.

The notation 0.(3) repeating 3s is an abomination. It causes confusion. If I had a dollar for every time I have seen the 0.999... thingy I would probably be on my way to wealth beyond my dreams of avarice.

The notation, as I pointed out in another posting on this thread is a (bad) abbreviation of an infinite series which is an infinite sum which the the limit of finite partial sums, all well defined mathematically. The 19th century was THE century in which limits, series and sums were finally defined with rigor so that honest to goodness theorems could be proved about them. The big names are Cauchy, Weirstrass, Kroenecker if you wish to do the historical research.

This entire thread is a testimony to how badly mathematics is taught in the U.S. (I assume most of the posters to this thread are United Stateseans). In Europe this would not happen. Whatever else is wrong with Europe, they do know how to teach mathematics in the lower grades, the gymnasia and the universities.

I used to have a full head of hair, but I tear my hair out whenever I see endless conversations on 1 = 0.999... . Now I have only ragged patches of hair on my scalp.

Bob Kolker

[John Link]

Given the apparent difficulties of understanding that 0.999...=1.0, much of which seems to be related to difficulty with the concept of infinity, I wonder what sort of discussion might ensue if I started a thread about the various orders of magnitude of the infinities? How about this: There are infinitely many integers, infinitely many rational numbers, and infinitely many real numbers. The infinity of the integers is equal to the infinity of the rational numbers, while the infinity of the real numbers is greater than the infinity of the integers and rational numbers.

Just trying to cause trouble!

John Link

[John Link]

You're confusing "representation" with division. You cannot "represent" the result in base 10 with a finite number of digits 1/3 = 0.333... WITH NO REMAINDER, because "..." means INFINITE.

No, "..." does not mean infinite, it means infinitely repeating.

John Link

[Jake_Ellison]

After reading Robert J. Kolker's post, it's becoming more and more obvious that I simply forgot how the 0.(9) , 0.(3) etc. were defined in mathematical analysis (and that they indeed come from where he's saying they come from). I seem to remember using them before learning math. analysis, and that's why I got confused about their definition, and came up with a definition that doesn't involve the limit of that SUM.

Having read it, his explanation makes perfect sense, and mine no longer does. But hopefully, now, everyone will stop speculating and realize that according to his definition, 0.(9) does indeed equal 1, and there's no longer the need to accept that it "just does, by definition", since he calculated the limit of that SUM, and the result was 1.

Edited December 10, 2009 by Jake_Ellison

[brian0918]

Having read it, his explanation makes perfect sense, and mine no longer does. But hopefully, now, everyone will stop speculating and realize that according to his definition, 0.(9) does indeed equal 1, and there's no longer the need to accept that it "just does, by definition", since he calculated the limit of that SUM, and the result was 1.

Of course, those denying this equality will simply say that the "limit" is not the same as the "actual".

[Jake_Ellison]

Of course, those denying this equality will simply say that the "limit" is not the same as the "actual".

The notation refers to the limit (where n -> infinity), not an actual, given n. If there is an actual n, there's no need for any knowledge of limits, it's simple algebra. So let's hope no one will do that

[Amaroq]

I'm back. I grow weary of debating this thing, but I'll try one last time to explain myself. In the end, I'll probably just give up, but I doubt I'll ever accept something that sounds so nonsensical to me. Unless I learned the math required to accept this equation, I'd just be accepting it on faith anyway.

So far my argument has been based against the relatively simple proof that I can understand. That 1/3 = .333... and both multiplied by 3 equals 1 and .999 simultaneously.

My basic argument is that there's two ways to arrive at .333... One, by simply stating that you have a series of repeating 3's. Two, by division and/or fraction.

Consider that if you divided 1 by 3, and only did short division, your answer will be .3 with a remainder of 1. Or of .1, or something like that. I'm only used to dealing with whole remainders. If you then move into longer division, you get an increasingly long series of 3's with an increasingly smaller remainder.

At face value, .333... defined by this sequence:

.3 + .03 + .003...

and .333 arrived at by division (Or fraction, I equate the results of both), appear the same. But this second one has that infinitesimally small remainder. When the numbers were finite, you could observe the difference. But as they stretched on and became infinite, that remainder was lost in the infinity.

I think it is fallacious to use this proof:

1/3 = .333...

*3

3/3 = .999...

1 = .999...

Because while it appears to work, it fails to take that remainder I mentioned into account. The remainder was lost in the infinity, but technically, that remainder is there somewhere...

I would like this argument of mine to be addressed directly. Please don't run me around in circles asking me questions about math that I don't understand how to retaliate to. The only thing I've heard against my argument that I've been willing to accept so far is one user who said it isn't valid to multiply .333... by any number without redefining multiplication.

[Jake_Ellison]

I would like this argument of mine to be addressed directly. Please don't run me around in circles asking me questions about math that I don't understand how to retaliate to. The only thing I've heard against my argument that I've been willing to accept so far is one user who said it isn't valid to multiply .333... by any number without redefining multiplication.

The answer is what Robert J. Kolker wrote, but I can give you a very basic and obvious proof:

0.(9) = 9/10+9/10^2+ 9/10^3 +.... (this is obvious, you don't need any math knowledge beyond fractions to follow it).

That equals one, here's the proof:

0.(9) = 9/10+9/10^2+9/10^3+... (we substract both sides from 1)

1-0.(9) = 1-9/10-9/10^2-9/10^3-... (we solve the first substraction on the right side of the equation, 1-9/10 :

1-0.(9) = 1/10-9/10^2-9/10^3-... (we manipulate the right side of the equation:

1-0.(9) = 1/10-1/10 * [9/10+9/10^2+....] (we note that the expression in the [] bracket is the same as the one we started with, and it equals 0.(9), so we make the switch:

1-0.(9) = 1/10 - 1/10*0.(9) (we multiply both sides by 10:

10-10*0.(9)=1-0.(9) (we solve the equation the usual way, with 0.(9) as the unknown)

9=9*0.(9)

0.(9)= 9/9

0.(9) = 1

Q.E.D.

[Jake_Ellison]

I just noticed that I needlessly complicated the above proof, by doing the whole "we substract both sides from one" thing....I was looking at 1-0.(9), because you mentioned that remainder. The above proof is valid, but a simpler version is:

0.(9) = 9/10+ 9/10^2+.... (as I said above)

We extract 1/10 from the right side of the equation, and get:

0.(9)= 1/10*[ 9+9/10+9/10^2+....] (the stuff in the [] brackets is 9+0.(9), so we make the switch)

0.(9)=1/10*[9+0.(9)]

If we solve that simple equation with the unknown 0.(9), we get 0.(9)=1, Q.E.D.

Edited December 21, 2009 by Jake_Ellison

[John Link]

I think it is fallacious to use this proof:

1/3 = .333...

*3

3/3 = .999...

1 = .999...

Because while it appears to work, it fails to take that remainder I mentioned into account. The remainder was lost in the infinity, but technically, that remainder is there somewhere...

Do you accept that 1/3 = .333... ? (If not, then we've really got to start at the beginning.)

Do you accept that .333... + .333... + .333... = .999... ? (If not, what do you say the value of .333... + .333... + .333... is?)

I'm sure you accept that 1/3 + 1/3 + 1/3 = 1

Assuming you answered YES to my two questions we have the following:

1 = 1/3 + 1/3 + 1/3 = 333... + .333... + .333... = .999...

So 1 = .999...

[Jake_Ellison]

Do you accept that 1/3 = .333... ?

...

Do you accept that .333... + .333... + .333... = .999... ?

He shouldn't accept either, unless you first define .333...., and prove that the two statements are true.

If we define .333... as the SUM (n=1,inf) [ 3/10^n], then the first statement can be proven the same way I poved that 0.(9)=1, above, while the second statement can be proven by using the appropriate theorems that describe the correct way of adding two or more SUMs together.

But that's way more work than directly proving that the SUM represented by 0.(9) equals 1.

[John Link]

He shouldn't accept either, unless you first define .333...., and prove that the two statements are true.

If we define .333... as the SUM (n=1,inf) [ 3/10^n], then the first statement can be proven the same way I poved that 0.(9)=1, above, while the second statement can be proven by using the appropriate theorems that describe the correct way of adding two or more SUMs together.

But that's way more work than directly proving that the SUM represented by 0.(9) equals 1.

Agreed.

John Link

[Jake]

My basic argument is that there's two ways to arrive at .333... One, by simply stating that you have a series of repeating 3's. Two, by division and/or fraction.

Those aren't really two different ways. I think I can show you why.

Consider that if you divided 1 by 3, and only did short division, your answer will be .3 with a remainder...

The remainder was lost in the infinity, but technically, that remainder is there somewhere...

The long division process that you are talking about can be expressed as a series (see expression below).

I've shown the result of this finite series for n = 4. You can see the (finitely) repeating 3s and the remainder. Note though that 1/3 is equal to the repeating 3s plus the remainder divided by 3. I think this might be something you missed.

Below is the same expression with limits added. While for any actual (finite) n, the remainder is non-zero, the limit of the remainder as n approaches infinity is zero.

To any mathematicians, my apologies for using the radical symbol. I wanted to show that I was representing the math behind long division, but the MS Word equation editor lacks the proper symbol.

Any complex mathematical concept is going to be vastly more difficult to tackle if you don't have a good grasp of the formal definitions first. The expression below shows the equivalence of the three different notations, their definition (as the limit of an infinite series), and the result of calculating that limit.

These three notations are all shorthand and are defined as the limit as n approaches infinity of the series. The well-defined, calculable value of this limit is 1, therefore 0.(9) = 1 (because it is defined as the limit). 0.(9) is NOT equivalent to actually summing the series to infinity; this is not possible in reality, and is therefore a poorly constructed expression^* (see below). It's really important to understand that the notation represents a limit, not an actual summation.

I've never thought about this equality, and I was unconvinced by many of the "proofs" offered in previous posts, but once I took a few minutes to write out the problem formally, it was clear. The wikipedia page.. that another poster mentioned does indeed have a great bit of info on this.

^*I'm not a professional mathematician, so I'm not sure that this is considered an invalid expression (it should be though).

Edit: Fixed an error in the first two equations

Edited December 23, 2009 by Jake

[John Link]

To any mathematicians, my apologies for using the radical symbol. I wanted to show that I was representing the math behind long division, but the MS Word equation editor lacks the proper symbol.

If you're going to be dealing with limits of infinite series and expect to be taken seriously I suggest you do not use the radical sign to indicate long division. The radical sign has a well-established meaning, and that meaning is not long division! All you needed to write was "1/3 =" and say that you were expressing that ratio by considering the long division of 3 into 1.

Also, any sum with an infinite number of terms is defined as the limit of the partial sums, so the expression to which you object is perfectly fine.

John Link

Edited December 23, 2009 by John Link

[RationalBiker]

My brother posed me this proof that says that .9999999 repeating = 1. Doesn't this violate the law of identity?

I looked through a majority of this thread, and if someone mentioned this, I missed it. I am, by far, not steeped in mathematical knowledge.

However, Ayn Rand stated the law of identity as A is A, not A = A. There is a difference. The values on either side of an "=" sign do not share the same identity, even if they are equivalent mathematically.

[John Link]

The values on either side of an "=" sign do not share the same identity, even if they are equivalent mathematically.

Not so. The equal sign "=" is used in mathematics to indicate identity. See http://en.wikipedia.org/wiki/Equal_sign

To anticipate objections to what I have written, consider the following true equation:

3+2 = 13-8

The expressions on the two sides of the equal sign (i.e., "3+2" and "13-8") are not the same, but the numbers on the two side of the equal sign (i.e., 3+2 and 13-8) are the same.

John Link

[John Link]

Wikipedia has an article about .999...

http://en.wikipedia.org/wiki/0.999...

John Link

Edited December 24, 2009 by John Link

[Schmarksvillian]

Ayn Rand stated the law of identity as A is A, not A = A. There is a difference. The values on either side of an "=" sign do not share the same identity, even if they are equivalent mathematically.

That has been answered in this thread, but I reiterate that answer since this concerns a quite fundamental point in mathematics. It is a terrible misconception that the equality sign (identity sign) '=' does not ordinarily in mathematics express, indeed, identity. The ordinary mathematical understanding of '=' is indeed that the two terms on each sign of the sign denote the very same object. So, e.g. in this immeditate context, the expression "1 = .999..." is understood to mean that 1 and .999... are identical, that is, the same object, that is, 1 IS .999..., not merely that they are equivalent in some sense less than that of being identical.

There are some technical complications regarding the inability of first order logic to capture identity without a fixed, stipulated semantics for that specific purpose, also regarding certain approaches in alternative mathematics such as certain constructivistic approaches that use the '=' sign to denote an equivalence that might not always be that of identity (and, of course, a mathematician may stipulate in any given context a special use of any typographical symbol). However, in both everyday, ordinary mathematics and ordinary formalizations of such ordinary mathematics, '=' stands for identity.

[Schmarksvillian]

I got confused about their definition, and came up with a definition that doesn't involve the limit of that SUM.

999.... is the limit of the sequence of finite sumS ('sums' plural). In that sense, .999... is referred to as itself an (infinite) sum. That is, the infinite sum is the limit of the sequence of finite sums.

Put more formally:

'.999...' is merely an informal notation to stand for

the limit of the denumerable sequence (with domain starting at, say, 1, for convenience) whose kth term, for every k, is SUM(j = 1 to k) 9/(10^j).

Thus, the problem of showing .999... = 1 reduces to showing

lim(k = 1 to inf)SUM(j = 1 to k) 9/(10^j) = 1,

which was done previously in this thread.

[Drregaleagle]

Even though the mathematical quantities are identical, .9999 repeating and 1 are different conceptually. The fact that we conceive them to be different shows that. '1' conceptually just represents a basic unit of quantity. 0.9999... represents the idea of repeating decimals and infinite series.

[Mindy]

I think an easier way to see that 0.99999999999999 = 1 is the following:

1/3 = 0.33333333333333333 (repeating)

3 * (1/3) = 0.99999999999999999 (repeating)

but 3 * (1/3) can be rearranged to (3*1)/3 and therefore 3/3, which must equal 1 (multiplication is associative and commutative over the real numbers), thus

1 = 0.99999999999999999 (repeating)

It seems counter-intuitive, I know, but so did using a number for zero to the Greeks. By the way, this is not a flawed proof like those hide-the-division-by-zero proofs. If you can find a mistake I'd love to hear it.

The problem is taking an incomplete process of division as complete. One third is not equal to .333333... because to be equal they both have to be specific quantities. An unfinished process of division does not name a specific quantity.

-- Mindy

A=A, not .ZZZZZZ...

[Schmarksvillian]

Even though the mathematical quantities are identical, .9999 repeating and 1 are different conceptually. The fact that we conceive them to be different shows that. '1' conceptually just represents a basic unit of quantity. 0.9999... represents the idea of repeating decimals and infinite series.

1 and .999... are the exact same mathematical object. We may have different notions about the names '1' and '.999...' but they are two names of the same object.

[Schmarksvillian]

An unfinished process of division does not name a specific quantity.

We don't need any unfinished process of division to show that 1 = .999...

[Schmarksvillian]

To whom it may concern: Here again is a rigorous proof. If you do not understand this proof then you don't understand the basics of this subject and you'd do better to STUDY the subject rather than ignorantly SPOUT off about it:

Definition: .999... = lim(k = 1 to inf) SUM(j = 1 to k) 9/(10^j).

Let f(k) = SUM(j = 1 to k) 9/(10^j).

Show that lim(k = 1 to inf) f(k) = 1.

That is, show that, for all e > 0, there exists n such that, for all k > n, |f(k) - 1| < e.

First, by induction on k, we show that, for all k, 1 - f(k) = 1/(10^k).

Base step: If k = 1, then 1 - f(k) = 1/10 = 1(10^k).

Inductive hypothesis: 1 - f(k) = 1/(10^k).

Show that 1 - f(k+1) = 1/(10^(k+1)).

1 - f(k+1) = 1 - (f(k) + 9/(10^(k+1)) = 1 - f(k) - 9/(10^(k+1)).

By the inductive hypothesis, 1 - f(k) - 9/(10^(k+1)) = 1/(10^k) - 9/(10^(k+1)).

Since 1/(10^k) - 9/(10^(k+1)) = 1/(10^(k+1)), we have 1 - f(k+1) = 1/(10^(k+1)).

So by induction, for all k, 1 - f(k) = 1/(10^k).

Let e > 0. Then there exists n such that, 1/(10^n) < e.

For all k > n, 1/(10^k) < 1/(10^n).

So, |1 - f(k)| = 1 - f(k) = 1/(10^k) < 1/(10^n) < e.