Math: .99repeating = 1?

[Flat U. Lance]

Is the repeating decimal .9999999 equal to 1?
.33333 repeating is equal to one third.
.6666 repeating is equal to two thirds.
.333333 + .66666 = 1
Some math nerds like bigneck say that .9999 repeating is not equal to 1.

------------------
Don't listen to bigneck.

[bigneck]

Flat,
.87632 = .87632
17 = 17
1 = 1
.99999999 = .99999999

you = dork

bigneck

[AWB]

Yes, it's the same. E.g.,
<BLOCKQUOTE><font size="1" face="Verdana, Arial">code:</font><HR><pre>
Fractions:
2/7 + 5/7 = 7/7 = 1
Decimals:
______ ______ ______
0.285714 + 0.714285 = 0.999999

Fractions:
2/9 + 7/9 = 9/9 = 1
Decimals:
_ _ _
0.2 + 0.7 = 0.9
[/code]


------------------
Wrong thinking is punished, right thinking is just as swiftly rewarded. You'll find it an effective combination.

[AWB]

[b]AAUUGGHH! I fed the trolls! Now they'll proliferate!



------------------
Wrong thinking is punished, right thinking is just as swiftly rewarded. You'll find it an effective combination.

[bigneck]

Flat,
.87632 = .87632
17 = 17
1 = 1
.99999999 = .99999999

you = dork

bigneck

[bigneck]

Sorry for posting that last one twice - if I did. Actually, .333333 + .66666 = .999993

[handy]

We already answered this one down below.

What is it with the logic people anyway?

1 does not & is not a 9 period.

[Flat U. Lance]

Hey Handy, where was this answered before?

[RussellM]

zero point...repeating

Russell

[RussellM]

Last discussed Is 0.999...=1 ? (http://boards.straightdope.com/ubb/Forum3/HTML/007229.html)

The last entry in this thread was only 5 days ago.

Russell

[C K Dexter Haven]

Another way to approach this, is that .99999... = SUM (9*10^-n) as n --&gt; infinity.

It can easily be shown that this infinite sum is 1.

[Arnold Winkelried]

CKDExtHavn, you sound like Fermat in your last post.

[C K Dexter Haven]

Hey, Jinx! Think!!

1 = 1.
4 = 4.
.25 = .25

Therefore, 1/4 can't equal .25, can it?

Sheeeeeeesh. [/sarcasm]

.333 is not equal to 1/3
.333333333333 is not equal to 1/3
.33333333[1 billion 3's]33 is not equal to 1/3

but .3333.... where the decimal 3 repeats infinitely many times IS in fact, equal to 1/3.

[CheapBastid]

Again with the Infinities (http://www.greylabyrinth.com/Puzzles/puzzle006.htm)...

[tony1234]

Again, I suggest we be a little more charitable. Many people have the understandable intuition that .9 repeating does not equal 1. This intution can be modelled in nonstandard number systems without contradition (or so I have heard). The existence of infintesimals is no more problematic than the existence of imaginary numbers.

Tony

------------------
Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[pipefitter]

Now ask a machinist if .9999 is equal to one.
Most likely he will say no, because they are used to working in tollerences where that .0001 might make a difference in the way a part fits

------------------
Hand me that wrench. No, the one that looks like a hammer.
Sig Courtesy of Walley

[Monty]

But, you see, pipefitter, not only are we not talking about machinists and fittings, we're also not talking about .9999. We're discussing 0.99999... ("Zero point Nine (with the 9 repeating ad infinitum)"). Thus, I think I'll go with the answer posted by the professional mathematician above who's already stated for us that not only does 0.99999... ("Zero point Nine (with the 9 repeating ad infinitum)") equal 1, but that it's easily proved.

[Red 2]

test

------------------
"I'm hungry, let's get a taco." - Mr. White

[Kupek]

.3333 is a decimal approximation of 1/3. It does not equal 1/3.

In the same vein, .9999 does not equal 1.

If you want to say that the limit as the number of decimal places approaches infinity equals 1, fine, knock yourself out, but a functions limit is not an algebraic value.

Let me stress: The values obtained from limits are not algebraic numbers. They are just that: limits. They tell you what a graph tends to do as you approach a certain number.

.9999999 will never equal 1. What about the limit? Like I said, that value is not an algebraic number, it simply tells us what the tendency of .999999 will be. Since we can never reach infinity, .999999 will never equal 1. (That's why taking the limit of something as we "approach" infinity is so useful; it tells us what happens to the function when x increases without bounds.)

------------------
Kupek's Den (http://www.geocities.com/~tortolia/kupek)

[Cabbage]

Let me stress: The values obtained from limits are not algebraic numbers. They are just that: limits. They tell you what a graph tends to do as you approach a certain number.

However (oversimplifying it a lot), the real numbers are defined/constructed as the limits of rational numbers.

Most people have an intuitive feel for what real numbers are, but not a rigorous understanding. That, in part, is what leads to misconceptions like this.

------------------
...ebius sig. This is a moebius sig. This is a mo...
(sig line courtesy of WallyM7)

[Jinx]

THINK! For goodness sake!

Ask yourself if 0.333 = 0.3333?
Is 0.3333 repeating = 0.333?
What is 1/3? Without saying "repeating", there would be NO exact decimal equivalent!

You see that, from this example,
0.9999 repeating ad infinitum is still &lt; 1!
These are two distinct numbers!



------------------
"They're coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time... :)" - Napoleon IV

[The Ryan]

Originally posted by Kupek:
.3333 is a decimal approximation of 1/3. It does not equal 1/3.

In the same vein, .9999 does not equal 1.

If you want to say that the limit as the number of decimal places approaches infinity equals 1, fine, knock yourself out, but a functions limit is not an algebraic value.


.9999999 will never equal 1. What about the limit? Like I said, that value is not an algebraic number, it simply tells us what the tendency of .999999 will be. Since we can never reach infinity, .999999 will never equal 1. (That's why taking the limit of something as we "approach" infinity is so useful; it tells us what happens to the function when x increases without bounds.)


The phrase ".9 reapeating" is shorthand for "the limit as n goes to infinity of .[n nines]". Since the former can be easily comprehended by laypeople and is easier to say than the latter, it is usually how it is expressed. However, the unfortunate result of this is that most people incorrectly belief that ".9 repeating" represents a decimal point followed by an infinite number of nines, that it means that you keeping on adding nines until it equals one. This is ridiculous. You can't put an infinite number of nines after a decimal place, let alone get one by doing so. It's a metaphor, darn it! A METAPHOR. Sheesh. You don't think that when someone says "sunrise was at six thirty" they means the sun actually moved above the Earth, do you?


Let me stress: The values obtained from limits are not algebraic numbers. They are just that: limits. They tell you what a graph tends to do as you approach a certain number.
What?!!! Limits are "special" numbers? A number is a number. It doesn't matter where you got it. Where did you learn this? Whoever taught you this should be fired.

[emarkp]

Sigh. Is 0.999...=1 ? (http://boards.straightdope.com/ubb/Forum3/HTML/007229.html)

Yes, they're equal. Go see the other thread.

[Monty]

One minor problem with your theory, there, Ryan. It's wrong, as in "incorrect."

The very definition of "repeating" in this sense is that, in fact, an infinite number of 9s follows the decimal point. As mathematics is rife with infinites and definitions, perhaps you'd care to explain exactly what the little bar over the 9 means in mathematical notation? Last I checked, it meant the number under the bar repeated ad infitum as in "an infinite number of 9s" in this case.

[C K Dexter Haven]

There is a theologic or philosophic point underlying this, of course, that Kupek raises.

There are numbers -- let's say, measurements -- in the real world, like 7, 2, and .333, that people can measure and use. These numbers have tolerances and measurement-errors. Two people measuring the same quantity with the same equipment may get slightly different answers within the tolerance of the measuring tools.

Mathematics, however, extracts a theoretic underlying construct from the real world situation, where numbers "exist" in their own right, not tied to any measurable object. There is advantage in developing this theoretic construct.

We've discussed elsewhere on this board, how many digits of pi do we need? At the point that you are talking about measuring the circumference of something down to less than the width of an electron (in terms of error/ tolerance), you probably got way more digits than is reasonable. In the theoretic world, the fact that pi is an infinite, non-repeating decimal is quite interesting. In the real world, if I want to approximate the ratio of circumference to diameter and I use 3 as the approximation, there are plenty of situations where that's just dandy (within 5%).

So, the question we ask is to what extent the question ("Is .999... = 1?) refers to the "real" world, in which case I agree with Kupek, you can never measure far enough for it to equal 1.... Or to the theoretic world of abstract mathematics, where you can take the limit of the partial sums to equal 1.

Sticking to the real world has problems, of course, such as the famous (1/3)+(1/3)+(1/3) may not equal 1 if the computer doesn't round properly.

Sticking to the abstract world of math has problems, too, of course, since we're apt to forget that there is no such thing as a "line" without width, in the real world.

Innerestin' turn, here.

[bj0rn]

two facts:
1. its called "rounding"

2. go about it the other way. take "1" and divide it in half, then divide the outcome in half as well etc...
the number of times you divide the outcome is equal to the number of 3s, 6s or 9s in 0.xxrepeating.

bj0rn - chickens for sale...!

[handy]

Anyone remember when the Pentium chip would process 1 to .989999 or so? remember what happened? Yep.

[Kupek]

Here's one of my problems with saying that .99999 repeating equals 1. If that is true, then so is this:

.99(rept.) = 1.00(rept.)

That just don't sit right.

Cabbage: I don't exactly know the process by which one would define the real numbers using ration numbers, but I think I can venture a guess.

The difference between that and are example is that those (I'm assuming) are the cases when the limit as x approaches b equals f(b). That doesn't hold true in our case, because you simply can't evaluate f(infinity).

In our case, the function will forever get closer and closer to 1, but never actually reach it. That's why we use limits. The limit is 1, but we never get there. When I make a distinction between algebraic numbers and a resultant of a limit, I am keeping in mind what these values represent: a plain ol' 2 represents 2 of something, but the 2 that is a limit of something represents the tendency of the function. There's a difference, and in this case, I think it's important.

Case in point: f(x) = (x^2 - 6x + 9)/(x - 3)

lim x -&gt; 3 of f(x) = 0
f(3) = undefined

This limit has a removable discontinuity, namely (x - 3). The graph of the function looks like a straight line, except there is a hole at x = 3, because the function divides by zero.

The way I see it, saying that .99(rept.) equals 1 is the same as saying that f(3) = 0. The closer and closer x gets to 3 from either the left or the right, the closer the functions gets to 0. It will never be zero, however, since the function is not defined there.

Just because a function approaches a value does not mean it has to reach that value. The limit as x approaches infinity of our .99(rept.) function is 1, but it will never reach one, just get closer the larger x gets. If it never reaches it, then I don't see how it can equal 1. It's damned close, but that's not the same thing.

Oh, and Ryan, the sun does move around the Earth if you use Earth as your reference frame. All we have to do is zero the Earth's velocity, and calculate the sun's velocity in reference to the Earth, and we have the sun moving around the Earth.

------------------
Kupek's Den (http://www.geocities.com/~tortolia/kupek)

[Cabbage]

Well basically, you can kind of think of it like this. Using just rational numbers, we can define pi, for example, to be the limit of the sequence 3, 3.1, 3.14, 3.141,... You get the idea. The limit doesn't exist in the rationals, but this is a way of making the rationals complete--by throwing in all the possible values that rationals "can" converge to, such as pi.

If we have an infinite decimal expansion (as with pi), how else can you say what it's equal to, without the idea of a limit? The sequence .9, .99, .999,... converges to 1, which certainly seems to me equivalent to saying .999... (infinitely) is 1.

------------------
...ebius sig. This is a moebius sig. This is a mo...
(sig line courtesy of WallyM7)

[Kupek]

I see your point. That fact that

.99(rept.) = 1.00(rept.)

would still hold true prevents me from agreeing with you. With that, we've said that two different numbers equal each other.

The problem with what you presented is that pi is irrational, whereas 1 is rational. One can be expressed exactly on our normal ten based number system, whereas pi can not. (I suppose you could make a number system based on pi, which would make every number that is not a multiple of pi irrational, but I think we've been assuming we're working exclusively on a ten based system.)

I mean, we don't say what pi is equal to; that's why it's represented as pi and not some number. It's irrational and can not be expressed exactly on our number system. We just approximate it, knowing that we aren't using an exact value.

------------------
Kupek's Den (http://www.geocities.com/~tortolia/kupek)

[Cabbage]

The problem with what you presented is that pi is irrational, whereas 1 is rational. One can be expressed exactly on our normal ten based number system, whereas pi can not.

I see what you mean, but I could claim that pi actually can be expressed as a decimal, it's just that the decimal expression would have to be infinitely long. (And when I say "expressed", I'm not meaning actually written down, but, rather, just the idea that the infinitely long stretch of digits is the exact value of pi. We can only find it to a finite number of places, but, theoretically, we can find it as far out as we want to). Similarly, I would also say that .999... is just another expression for 1. It's limiting value is certainly 1. It's an expression that can't actually be written down, but it's still equal to 1.

------------------
...ebius sig. This is a moebius sig. This is a mo...
(sig line courtesy of WallyM7)

[C K Dexter Haven]

Try it this way, Kupek.

If .999... and 1.000... (infinite repeating decimals) are different numbers, then you can find a number (in fact, you can find infinitely many numbers) BETWEEN THEM.

Go ahead. Find a number BETWEEN .99999... and and 1.00....

Granted, you can find lots of numbers between .999999999999999 and 1.000000000. That's no problem, because these are two different numbers. But if I let the decimal repeat infinitely, there is no number between the two.

The Real Numbers can be shown to be dense (I hope I'm remember the right term): that is, between any two real numbers, you can find a bunch of other numbers. The rational numbers are likewise: between any two rational numbers, you can find infinitely many OTHER rational numbers.

So, although the two numbers "look" different, they are in fact the same number.

[tony1234]

The difference between 1 and .99 repeating is an infintesimal quanity. For information about such quantities, hyperreal numbers and nonstandard analysis, see
http://neuronio.mat.uc.pt/crcmath/math/math/n/n161.htm.

Please do not reply that "we're talking about the standard number system here." The original question was not phrased in terms of any particular number system, not even implictly. Rather, the original question, as I interpreted it, rested on the false assumption that there was only a single "true" number system. That idea, of course, is untrue to the nature of mathematics.

By the way, I think the discussion would be much simplified if, instead of discussing whether .99-repeating equals 0, you discussed whether 99-repeating equals infinity, or the the limit of the sequence 1, 2, 3 . . . equals infinity. They are "reciprocal" issues and the conventionality of the answer would become more apparent.

Tony

------------------
Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[ZenBeam]

The difference between 1 and .99 repeating is an infintesimal quanity.

From your site, an Infinitesimal is
A quantity which yields 0 after the application of some Limiting process. The understanding of infinitesimals was a major roadblock to the acceptance of Calculus and its placement on a firm mathematical foundation.

This never says 1 - 0.999repeating is an infinitesimal; merely defining the term doesn't make it so. 0.999repeating is not a limit any more than 1.000repeating is a limit because neither are defined by a limiting process. 0.999repeating is an alternate representation of 1, and is exactly equal to 1.

The difference is seen in this example: (I'll use the notation 0.9[n] to represent the last digit repeating n times, and 0.9 to represent the last digit repeating infinitely many times.)

If I have 9.9[inf], I can't tell whether I got it by adding 9 to 0.999[inf], by multiplying 0.9[inf] by 10, by adding 9.9 to 0.1 * 0.9[inf]. They all give the same result. If I set D = 1-0.9[inf], then 10 - 9.9[inf] = D, 10*D, and 0.1*D depending on how I got 9.9[inf]. So D = 10*D = 0.1*D. This means D=0, plain and smple.

If I have 9.9[n], I get different representations: 9.9[n]9.9[n-1], 9.9[n+1].
If E[n] = 1-0.9[n], 10-9.9[n] = E[n], 10-9.9[n-1] = 10*E[n] = E[n-1], 10 - 9.9[n+1] = 0.1+*E[n] = E[n+1]. I'm able to keep track of how I got the term E[n].


------------------
[i]It is too clear, and so it is hard to see.

[tony1234]

Zenbeam,

Hyperreal numbers, see the link in my previous posting, are numbers less than 1/n as n approaches infinity. It seems to me that 1-.9[inf] falls under this definition. The proofs you present are interesting, but they only suggest to me that there must be distinct rules for the applying the operations of multiplication and addition to hyperreals to avoid a contradition or the erroneous conclusion they equal zero.

I admit I could be wrong: 1-.9[inf] may not be a hyperreal number (even though it looks like the limit of .1-sum[1/9*10n] as n approaches infinity). I don't know enough about nonstandard analysis, however, to know for sure.

Tony



------------------
Two things fill my mind with ever-increasing wonder and awe: the starry skies above me and the moral law within me. -- Kant

[The Ryan]

Monty posted 04-13-2000 07:35 AM
The very definition of "repeating" in this sense is that, in fact, an infinite number of 9s follows the decimal point.
Have you ever seen an infinite number of nines following a decimal point? Does such a thing exist? Does it have any meaning? No,no, and no.
As mathematics is rife with infinites
Actually, mathematicians are very careful about infinities; they don't just go throwing them around.
explain exactly what the little bar over the 9 means in mathematical notation?
I thought I already did.
.(N barred)=
lim
n -&gt;infinity of

N*10^(-m)+N*10^(-2m)+...N*10^(-nm)
where m=number of digits in N



Last I checked, it meant the number under the bar repeated ad infitum as in "an infinite number of 9s" in this case.
Yeah, that sounds like a rigorous mathematical definition :rolleyes:

CKDextHavn posted 04-13-2000 09:37 AMSo, the question we ask is to what extent the question ("Is .999... = 1?) refers to the "real" world, in which case I agree with Kupek, you can never measure far enough for it to equal 1.... Or to the theoretic world of abstract mathematics, where you can take the limit of the partial sums to equal 1.
You have it backwards. We can never measure far enough for 1 not to equal .9r. Think about it. If one bar is one inch, and the other is .9r inches, what degree of accuracy do you need to meaure a difference? An infinitly small degree of accuracy (or infinitly high, dpending on how you look at it) because they are the same number. If we are measuring to the closest one thousandth of an inch, what would the .9r inch bar be measured as? Well, the closest multiple of inch/1000 to .9r is... 1.000. So we get that the bar is the same length as the one inch bar.

Kupek posted 04-13-2000 07:30 PM
Just because a function approaches a value does not mean it has to reach that value. The limit as x approaches infinity of our .99(rept.) function is 1, but it will never reach one, just get closer the larger x gets. If it never reaches it, then I don't see how it can equal 1. It's damned close, but that's not the same thing.
"Function"? "x"? Where did these come in? .9r isn't approaching one; it is the limit of a sequence that approaches one. .9r has one value and one value only. It can't approach one because it isn't "moving". It already is one; it doesn't have to approach one.

Kupek posted 04-13-2000 11:02 PM

I see your point. That fact that .99(rept.) = 1.00(rept.) would still hold true prevents me from agreeing with you. With that, we've said that two different numbers equal each other.
No, we've said that one number is equal to itself. It's the same number, just expressed differently. Do you see a contradiction in the equation 2/4=1/2?

[bigneck]

CKDextHavn
Administrator
Posts: 2018
Registered: Feb 99
posted 04-14-2000 08:07 AM
--------------------------------------------------------------------------------
Try it this way, Kupek.
If .999... and 1.000... (infinite repeating decimals) are different numbers, then you can find a number (in fact, you can find infinitely many numbers) BETWEEN THEM.

Go ahead. Find a number BETWEEN .99999... and and 1.00....

Granted, you can find lots of numbers between .999999999999999 and 1.000000000. That's no problem, because these are two different numbers. But if I let the decimal repeat infinitely, there is no number between the two.

--------------------------------------------
What is the number between
.98999999.... and .9999999......?
Are they equal as well?
Does that make .98999999... = to 1?

[23skidoo]

What number is in between 0.98999... and 0.999...? Quite a few, actually.

For example, 0.99.
And 0.997.
And 0.9903.
And 0.99008.
And you get the idea.

0.98999... and 0.999... have numbers inbetween them. 0.999... and 1 do not.

[Konrad]

CK sez:
Another way to approach this, is that .99999... = SUM (9*10^-n) as n --&gt; infinity.

It can easily be shown that this infinite sum is 1.

I'm not disagreeing with you about whether .999... = 1 but I don't think this is the best way to approach the problem. Aren't all integrations/infinite sums off by an infintesmal?

This entire question depends on whether or not you accept an infintesmal as a number.

.999... can be different from 1 when you're dealing with badly-behaved functions where the function doesn't exist at 1. ( (x-1)/(x-1)^2 ) Of course, this still means dealing with infintesmals.

[Tengu]

Originally posted by 23skidoo:
What number is in between 0.98999... and 0.999...? Quite a few, actually.

For example, 0.99.


Hmm....it seems to me that 0.99 would be equal to 0.98999..., for the same reason 1 is equal to 0.999....

Just a nitpick, the rest I completely agree with.



------------------
Eschew Obfuscation

[Monty]

The Ryan posted 04-14-2000 08:04PM the stuff between the quotes; my responses are in between the quotations.
Monty posted 04-13-2000 07:35 AM
quote:
-----------------------------------
The very definition of "repeating" in this sense is that, in fact, an infinite number of 9s follows the decimal point.
-----------------------------------

Have you ever seen an infinite number of nines following a decimal point? Does such a thing exist? Does it have any meaning? No,no, and no.

Nobody can actually see an infinite number of things, nor can anyone see the square root of a negative number, let alone see a negative number; however, in mathematics there are the concepts of infinites (transfinites), imaginary numbers (square roots of negative numbers), and negative numbers (numbers less than zero). Since we're discussing a purely mathematical issue and not a physical issue, your snide comment is irrelevant, not to mention that it shows you've no idea of what you speak. But, to answer your questions correctly: No (but I and others have postulated them, that's part of mathematics), Yes (they exist in mathematics, the topic under consideration in this thread), and Yes (they have meaning in the realm of mathematics, the topic under consideration in this thread).

quote:
-----------------------------------
As mathematics is rife with infinites
-----------------------------------

Actually, mathematicians are very careful about infinities; they don't just go throwing them around.

Well, apparently according to you, there's no such thing with wich to be careful about throwing around. BTW, do you always contradict yourself in the very same posting? Anyway, discuss why there's a whole notational system for infinities when "they don't just go throwing them around," please. I missed reading the funnies this morning and need a dose of humour which I'm sure your description of Aleph, Aleph-null, etc., will bring to the discussion. [My snide comment here, feel free to point it out.]


quote:
-----------------------------------
explain exactly what the little bar over the 9 means in mathematical notation?
-----------------------------------

I thought I already did.
.(N barred)=
lim
n -&gt;infinity of
N*10^(-m)+N*10^(-2m)+...N*10^(-nm)
where m=number of digits in N

Nope, see ** below for an authorative answer on this.


quote:
-----------------------------------
Last I checked, it meant the number under the bar repeated ad infitum as in "an infinite number of 9s" in this case.
-----------------------------------

Yeah, that sounds like a rigorous mathematical definition.

**From Merriam-Webster's online dictionary:

repeating decimal (noun)

First appeared 1773

: a decimal in which after a certain point a particular digit or sequence of digits repeats itself indefinitely -- compare TERMINATING DECIMAL

&

terminating decimal (noun)

First appeared circa 1909

: a decimal which can be expressed in a finite number of figures or for which all figures to the right of some place are zero -- compare REPEATING DECIMAL

From "Number," Microsoft (R) Encarta. {bolding and description of pictures in the article brought to you by Monty}

[quote]It can be shown that every rational number can be represented as a repeating or periodic decimal; that is, as a number in the decimal notation, which after a certain point consists of the infinite repetition of a finite block of digits. Conversely, every repeating decimal represents a rational number. Thus, 617/50 = 12.34000..., and 2317/990 = 2.34040... . The first expression is usually written as 12.34, omitting the infinite repetition of the block consisting of the single digit 0. The second expression is frequently written as {picture of 2.340 with a bar over the 40, picture of 2.340 with a bar under the 40, picture of 2.340 with a short dash over the 4 and a short dash over the 0} to indicate that the block of two digits, 4 and 0, is repeated infinitely.

That rigorous enough for you?

Here's a hint or two for you:

1) The professional mathmetician (CKDext) has already shown that your take on this issue is incorrect
2) If you really don't know what you're talking about, keep quiet and try to learn.

[RM Mentock]

Actually, I don't see much wrong with anything that The Ryan, Monty, or Dex said. Except when you say that the other is wrong.

I thought you guys were on the same side.

------------------
rocks

[Jinx]

Originally posted by CKDextHavn:
Hey, Jinx! Think!!

1 = 1.
4 = 4.
.25 = .25

Therefore, 1/4 can't equal .25, can it?

Sheeeeeeesh. [/sarcasm]

.333 is not equal to 1/3
.333333333333 is not equal to 1/3
.33333333[1 billion 3's]33 is not equal to 1/3

but .3333.... where the decimal 3 repeats infinitely many times IS in fact, equal to 1/3.

What kind of reply is this? You're just reiterating my point! Do you understand the concept of a fraction?

Of course 1/4 = 0.25 why should it not?
And yes, 1/3 = 0.3333 repeating - get it?
As for 1 = 0.999 repeating...

Please go back to basic math class if:
a) You cannot comprehend the number 1
b) You fail to understand the meaning of the fraction 1/1 and how to set that up as a long division.

I'd really like to see a long division where 1 divided into itself can yield 0.9999...




------------------
"They're coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time... :)" - Napoleon IV

[Jinx]

I can see why Cecil feels like there is no hope for the teeming millions!

------------------
"They're coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time... :)" - Napoleon IV

[ZenBeam]

I'd really like to see a long division where 1 divided into itself can yield 0.9999...

Oh, this is relevant. :rolleyes:

------------------
It is too clear, and so it is hard to see.

[C K Dexter Haven]

Proper treatment for trolls, IMHO: answer ONCE, to be sure that you are not dealing with honest ignorance, then drop it.

One more time, and then I'm done with this.

Several people have offered proof that .999...(repeating infinitely) = 1. The simple multiplication proof is usually sufficient:

(1) x = .99999....
(2) 10x = 9.999999...
Subtracting (2)-(1):
(3) 9x = 9, hence x = 1.

OK, that's a proof. The testimony of every professor of mathematics in the country would be another sort of proof, I suspect.

You don't buy those? OK, then, if you thinkf .999... is different from 1, then you should be able to pick a number BETWEEN them. Go ahead, tell us one.

For any two distinct real numbers, there are infinitely many real numbers between them. For instance, the midpoint. Go ahead, name one. We're not talking about theoretical numbers, now, like "infinitesimals" or whatever, we're talking real, honest-to-god, write-downable numbers. Name one, between .999... and 1.

If you can't, then you have to concede that .99999... is just another way of writing 1, in the same way that .333333... is just another way of writing 1/3 or that .11111.... is another way of writing 1/9.

Oh, and BTW, Monty, I'm no longer a professional mathematician. I did that back in the 70s, but then I became an actuary (pay was better) and now I'm a more generalized consultant. Contrary to the comment made in another thread, I did not predate Riemann, although I did have a class under Zygmund.

[Jinx]

Originally posted by ZenBeam:
Oh, this is relevant. Referring to: I'd really like to see a long division where 1 divided into itself can yield 0.9999...


Your sarcasm shows you've missed the point.
Excuse me, but it is relevant! I can demonstrate that 0.25 equates to 1/4. I can demonstrate that 0.3333 repeating equates to 1/3. Can you demonstrate that 1.0 equates to 0.9999 repeating (without rounding)?

Technically, these are two distinct points on the numberline. If you receive a paycheck, would you accept anything less than 1.0 times the amount you are owed?




------------------
"They're coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time... :)" - Napoleon IV

[Jinx]

Originally posted by CKDextHavn:
One more time, and then I'm done with this.

(1) x = .99999....
(2) 10x = 9.999999...
Subtracting (2)-(1):
(3) 9x = 9, hence x = 1.

OK, that's a proof...[/B]

Faulty logic, prof! Excuse me, but if you're so sharp on mathematics, you should recognize the error in your logic! Here, you are attemtping to employ methods of solving simultaneous equations! The ground rule for aplying this method is that the two equations MUST BE independent! Since formula (b) is 10 times forumla (a), these are NOT independent equations! Thus, you have proved nothing!

Thus, it is improper to subtract (a) from (b)! Algebra cannot be applied at whim! ;)




------------------
"They're coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time... :)" - Napoleon IV

[C K Dexter Haven]

&lt;&lt; The ground rule for aplying this method is that the two equations MUST BE independent! &gt;&gt;

Baloney. The ground rule is that the two equations must be consistent, or you will get paradoxical results. GENERALLY, given two dependent equations in two variables, you cannot reach a solution. However, these are two equations in ONE variable, and I have obtained one from simple algebraic manipulation of the other.

A long division? Sure. (.99999...)/1 = .99999...

Happy?

The way that you toss out incorrect statements and ignore the comments of any others, leads me to suggest that this conversation is finished. Fighting ignorance is one thing; fighting stubborness is something else.

[kbutcher]

What's interesting about this thread to me, is the logic. Many have issued mathmatical proofs that .999... = 1.

The detractors, on the other hand, have offered no mathmatical proof whatsoever. In fact, it would seem that the mathmaticians have won the point.

The only avenue left to those who do not beleive, is using "logic" to state why the proofs are wrong, not showing any mathmatical flaws in the proofs themselves.

Am I wrong? Did I miss the post that had a valid mathmatical proof that 1 is not equal to .999...?

[ZenBeam]

I can demonstrate that 0.25 equates to 1/4. I can demonstrate that 0.3333 repeating equates to 1/3. Can you demonstrate that 1.0 equates to 0.9999 repeating (without rounding)?

CKDextHavn demonstrated 0.999 repeating = 1, so I won't bother repeating him. I'd like to see you demonstrate that they are unequal.

Kbutcher writes
Am I wrong? Did I miss the post that had a valid mathmatical proof that 1 is not equal to .999...?

Nope. So far, all we've seen is handwaving.

------------------
It is too clear, and so it is hard to see.

[Boris B]

Well, I tried this prove on the other thread on this subject, but I'll try it again. I don't know if it's mathematically valid or not.

1 - 0.999... = D
D x 10 trillion = D
Therefore, D = 0

The idea is, multiply the difference between one and point nine repeating by the highest finite number you can think of, and the product will be no different. When multiplying yields no change, one of your factors is zero. QED

[sloth]

Jinx -

If 1/3= .3repeating (as you agree)

Is not .3repeating + .3 repeating + .3 repeating = .9 repeating true?

If so by substitution aren't 1/3+1/3+1/3 = .9 repeating and since 1/3 * 3 = 1, 1=.9 repeating?


An

[Jinx]

I know this is beating a dead horse, but like cold fusion fiasco, this is math chicanery!

Originally posted by CKDextHavn:
&lt;&lt; The ground rule for aplying this method is that the two equations MUST BE independent! &gt;&gt;

Baloney. The ground rule is that the two equations must be consistent, or you will get paradoxical results. GENERALLY, given two dependent equations in two variables, you cannot reach a solution. However, these are two equations in ONE variable, and I have obtained one from simple algebraic manipulation of the other.

A long division? Sure. (.99999...)/1 = .99999...

Happy?

a1) So, like what if I used 8x instead of 10x? I could fudge "x" to equal whatever I wanted! Ever studied matrices and determinants? What's the value of a determinant when the equations are not unique?

a2) If your initial premise is let x=0.999...
then how can x &lt;&gt; x?

b) Your "long division" fails to prove that 1/1 = 0.999...

c) Equations must be consistent? No, the term is independent...otherwise, "x" can be shown to equal whatever value you wish by yielding x&lt;&gt;x.

------------------
"They're coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time... :)" - Napoleon IV

[Jinx]

Originally posted by CKDextHavn:
&lt;&lt; The ground rule for aplying this method is that the two equations MUST BE independent! &gt;&gt;

Baloney. The ground rule is that the two equations must be consistent, or you will get paradoxical results. GENERALLY, given two dependent equations in two variables, you cannot reach a solution. However, these are two equations in ONE variable, and I have obtained one from simple algebraic manipulation of the other.

A long division? Sure. (.99999...)/1 = .99999...Happy?

Sorry if this might double-post because my first posting didn't show for some reason. Anyway, I know this is beating a dead horse, but what we have here is math chicanery!

a1) The equations must be independent! What if a selected 8x instead of 10x? What would the solution be, then? I can "fudge" x to equal whatever I wish! You have simply proven x&lt;&gt;x...an old math mindbender.

a2) Even used matrices? What is the value (solution) would the determinant of your pair of equations have? And, then ask yourself, what is the significance of this solution? What is trying to tell me? It's trying to tell you that the second equation is just a scalar multiple of the first (i.e.: not independent).

b) As for your example of long division, you failed to show that 1/1 = 0.999... All you have shown is that any number divided by unity is equal to itself! Hmmf!



------------------
"They're coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time... :)" - Napoleon IV

[Jinx]

Originally posted by kbutcher:
Am I wrong? Did I miss the post that had a valid mathmatical proof that 1 is not equal to .999...?

It all depends on who taught you math. If you have ever worked with matrices and determinants, you will instantly recognize that two equations MUST BE unique (independent) in order to subtract one equation from the other.

If you understand the concept of why two equations are not unique if one is a mulitple of the other, then you would agree that x&lt;&gt;x is no proof.

Again, I'll ask you, what if I used 8x instead of 10x in the example "proof" given by the Administrator on page 1 of this thread?



------------------
"They're coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time... :)" - Napoleon IV

[Jinx]

Sloth, the limit of 0.9999... is 1.0; I'd agree to that. In short, it will continually approach, but never reach, the value of 1.0

------------------
"They're coming to take me away ha-ha, ho-ho, hee-hee, to the funny farm where life is beautiful all the time... :)" - Napoleon IV

[zut]

Originally posted by Jinx:
If you have ever worked with matrices and determinants, you will instantly recognize that two equations MUST BE unique (independent) in order to subtract one equation from the other.

Untrue; I'm afraid you're extrapolating the requirement that matrices be non-singular in order to solve the system of equations beyond its applicable range. This requirement doesn't mean you can't subtract two dependent equations, it just means you can't solve a system of n equations in n variables if any of the equations are dependant. Not applicable to the current problem.

Originally posted by Jinx:
Again, I'll ask you, what if I used 8x instead of 10x in the example "proof" given by the Administrator on page 1 of this thread?

OK, easy enough,
(1) x = .99999....
(2) 8x = 7.999999...
Subtracting (2)-(1):
(3) 7x = 7, hence x = 1.

[JonF]

If you have ever worked with matrices and determinants, you will instantly recognize that two equations MUST BE unique (independent) in order to subtract one equation from the other.

No, N linear equations in N unknowns must all be independent for a unique solution to exist. They need not be independent in order to validly combine the equations arithmetically.

If you have N independent linear equations in N unknowns then you may multiply any equation by a constant and/or add any equation to itself or any other equation without changing the solution (except you can't multiply an equation by -1 and then add it to itself). This applies equally to the case N = 1, which is the case in the previous messages.

Demonstration for N = 1:

Any linear equation in one unknown "X" may be written as:

aX = b

The matrix representation of this system is:

{a}{x} = {b}

The determinant of the coefficient matrix is:

a

and is non-zero if "a" is non-zero.

The determinant of the coefficient matrix with the coefficients of X replaced by the constant vector from the right side of the equation is:

b

Therefore, by Cramer's rule, if a &lt;&gt; 0 then:

X = b/a

(gee, we really needed matrix algebra for that one, didn't we?) {grin}

Now multiply the original equation by a constant "C" and add the result to the original equation:

aX = b
+ CaX = Cb
---------------
aX + CaX = b + Cb

Simplifying:

(1+C)aX = (1+C)b

(can you see where this is going?)

By similar reasoning, the determinant of the coefficients is (1+C)a and the determinant with the coefficients of X replaced by the constant vector from the right side of the equals sign is (1+C)b, so the solution to the "system" of "equations" is, by Cramer's rule:

(1+C)b
--------
(1+C)a

or, provided C &lt;&gt; -1,

b/a

which is the same as the solution of the original problem.

Therefore, a linear equation in one unknown may be multiplied by any constant (except -1) and added to itself without changing the solution.

The proof for larger values of N is left as an excercise for the student {grin}.

------------------
jrf

[ZenBeam]

two equations MUST BE unique (independent) in order to subtract one equation from the other.

If A = B and C = D, then A-C = B-D. Always.

I'm still waiting for your proof that 1 and 0.999repeating are not equal.

------------------
It is too clear, and so it is hard to see.

[C K Dexter Haven]

Agreeing with those who say that you can always subtract one equation from another. Equals subtracted from equals are equal, so to speak.

If you are subtracting equations that are not independent, then you may not be able to find a unique solution (assuming n equations in n unknowns.

But we're starting with simple math, here.

x = .9999999....(infinte repetition)
10x = 9.99999....

You ask, what if you use 8x instead of 10?
Then you get
8x = 8*(.99999999...) which doesn't help much. But it's still a valid equation.

If x = y
Then ax = ay and
Therefore (ax - x) = (ay - y)
Or (a-1)*x = (a-1)*y.

In most instances, this is ho-hum, it won't help much; but in some cases, it can help solve the system.

This has hit a point of being silly. On the one side, there are mathematicians and the like, citing proof after proof. On the other side, there are people who "don't believe it" or "don't like it." Enough.

[23skidoo]

Sorry if my math is simple and I don't know big 50 cent words. But I was under the impression that you could multiply one equation by a variable to get a new equation, then find the difference to help you solve a problem. Like this one: Find the sum of all the terms in the series (or is it sequence?) that has 1 as its first term, and every other term 1/2 of the preceding term.

r= (1 + 1/2 + 1/4 + 1/8 + 1/16 + ...)

2r= (2 + 1 + 1/2 + 1/4 + 1/8 + ...)

2r - r = 2

r = 2

If you can't do this, how else do you solve this problem?

[DrMatrix]

I don't want to give another proof that
1 = 0.9
We have seen enough proofs of that here, But I want to say something about non-standard analysis. I like non-standard analysis and I think it is very helpful in understanding calculus. However, the standard decimal notation [i]cannot represent infintesimals. The difference between 1 and 0.9 is [i]not infintesimal, it is zero!

I agree with CKDextHavn. This has gotten silly.

------------------
Virtually yours,

"Feynman was wrong.
I understand Quantum Physics completely.
Anybody seen my drugs?" - A WallyM7™ .sig

[C K Dexter Haven]

23skidoo: Yes, you can multiply and add equations as you've suggested, subject to the following TECHNICAL POINT: ONLY if you know that the series converges (to a finite sum.)

For instance, following your example,
x = 1 + 1 + 1 + 1 + 1 ...
x = 1 + (1 + 1 + 1 + 1 + ...)
Subtracting the two equations:
0 = 1
In this case, you can't subtract the two equations because the series don't converge.

[FrankGrimes]

As has been proven, .99... is equal to one. The reason why some people don't agree is because it "doesn't look pretty" as someone wrote earlier (sorry forgot who). This is because, as has also been mentioned, we are using the decimal system which does not handle the infinite very well.

However the naysayers do raise an interesting point that is more philosophy than math.

0.9 is a whole with a tenth missing. add 0.09 and you almost fill the missing tenth but there's still a hundredth missing. etc, etc. So by adding 9s at the end you are filling the hole in the whole in progressively smaller increments until the whole becomes infinitely small.

So the question is, is an infinitely small hole a hole? I guess the answer is no but I understand why it's hard to get past.

[CrueltyToFoliage]

Face it, folks, with our flawed number system, .999... = 1

But it doesn't look like it should, I know. Then again, you'd think you could divide by zero. Or take zero to the zeroeth power. To me, that last is the biggest offender against our system of numbers, because on one hand, we have a rule saying that anything to the zeroeth power is one, and on the other we have a rule saying that zero to any power is zero. We live in a flawed world.

------------------
"Give a man a fire and he's warm for a day, but set fire to him and he's warm for the rest of his life."

--Terry Pratchett

[C K Dexter Haven]

Why do you say the number system is flawed?

The number system, independent of our representation of it, works just fine. The number 1 (the multiplicative identity) is uniquely determined. The real number system is a complete ordered field (each of those terms is defined in mathematics) and there are mathematical proofs that it is unique -- any other number system is really the same, but might use different representation (for instance, a base 12 or base 8 representation is NOT a different number system.)

The problem is that we have decimal representations of these real numbers. Some of those decimal representations are hard to imagine, like 1/3 being represented by .333333... with the 3 repeated.

One way to deal with those "hard" representations is by thinking of the "true" value as the limit of an infinite sequence:

.3 + .03 + .003 + .0003 + ...

That can be a very useful representation, but let's not pretend that is anything other than 1/3... unless, of course, you STOP somewhere along the way, and then it's different, it's no longer the repeating-3.

We can describe this in different ways; we can say the infinite repeating decimal equals 3, we can say the series converges (adds) to 3, we can say that the partial sums converge to 3, we can say that the series IS 3. There is no difference between these various representations of the real number 3, except the way we look at them.

Similarly, we can write
e = 1 + 1/1 + 1/2! + 1/3! + 1/4! +...

That infinite series converges to the unique real number e. That is a representation of a uniquely defined real number e.

And so with .99999... Whether you want to say it "converges to 1" or "partial sums converge to 1" or "is 1", is saying the same thing. So, have we got this thing moved to a matter of language and taste? I'm really tired of posting on this one.

[jayron 32]

You people miss an important concept here. Doing something an infinite number of times does not mean the same thing as doing it as often as you'd like. 0.9 repeating is not the same as writing a zero with as many nines as you can write as your hand gets tired. It is possible to have a number which is a REAL, RATIONAL, number that cannot be expressed as a decimal since it repeats infinitely. It is not possible to write 1/3 as a decimal without noting that it is 0.3 repeating (which is NOT the same thing as a zero, a decimal point, and writing threes until your hand gets tired). 0.9 repeating is equal to one, zero with a decimal point and as many nines as you feel like writing is never equal to one.

Look at it this way. If x-y=z, and z=0, then x=y. It has to be. If x=1 and y=.9 repeating, what does z equal? It can't equal anything but zero, since there cannot be a digit following an infinitely repeating digit. .9 repeating equals 1

[manhattan]

::test post::

[Mousseduck]

I think it depends on what you are doing with your numbers.
For most intents and purposes, 0.99... is close enough to 1 to say it equals 1. However, some people do maths (and I won't go into the uses of it) where they use limits, and a decimal will approach a number but will never equal that number. You can kind of get an idea of that with this teaser...

Set an imaginary finish line somewhere (eg, a point on the wall 5 metres away from you). Find a method of advancing so that even though you always move ahead, you never reach the finish line.

The answer is - always advance by half the remaining distance.

By the time the remaining distance is a millionth of a micrometre, most people would say you're there. So, 0.9999... can be close enough to 1 that it makes no practical difference, or you can be using a type of maths that says it does make a difference. Use your own judgement.

[DrMatrix]

::yawn:: Zeno's paradox. You can never get from point A to point B, because you must travel one half of the distance in a finite amount of time, the half the remaining distance in a finite amount of time, ... The sum of an infinite number of positive numbers is obviously not finite. Therefore movement is not possible. Replace one half in the above with nine tenths, and you have the arguement that 0.9... cannot equal 1.0.

If you believe 0.9... is not equal 1.0, do you also believe that movement is not possible?

------------------
Virtually yours,

DrMatrix
"Feynman was wrong.
I understand Quantum Physics completely.
Anybody seen my drugs?" - A WallyM7™ .sig

[DrMatrix]

::sigh:: In math two quantities are either equal or not. Close enough isn't good enough. If you start at the point A and place the point B half way to your goal, you will never get half-way there, so you certainly will never get to where you are going. I'd say that pretty much says you can't go anywhere.

As far as a function that approches 1 but never reaches 1, how about the 1 - 1/10^n. This equals 0.9[n]. But the limit (as n goes to infinity) does reach one. 0.9... is not close enough to 1. It is 1.

------------------
Virtually yours,

DrMatrix
"Feynman was wrong.
I understand Quantum Physics completely.
Anybody seen my drugs?" - A WallyM7™ .sig

[DrMatrix]

I swear by Georg Cantor and all that's Holy that I shall not post to this topic again. If I feel like posting I will run into a brick wall until the feeling passes. But don't worry, I will not get hurt because I will only go 0.9... of the way to the wall.

::THUD::

------------------
Virtually yours,

DrMatrix
"Feynman was wrong.
I understand Quantum Physics completely.
Anybody seen my drugs?" - A WallyM7™ .sig

[wolfman]

Here's my attempt to formalize what others have been saying as an iteratize proof(many years since i did one, go easy on me).

conjecture. 1 is not equal .999(rep)

principle1. if two numbers are different then there is a measureable difference between them.
reason. if x-y =0 then x=y by definition

principle2. .999(rep) can also be written as 9/10 +9/100 +9/1000 ..... or 9/10^1 +9/10^2 +9/10^3 ....+9/10^k


extrapolation. Therefore there is a measurable difference between 1 and .999(rep)


For any arbitrary number n hypothesized to be the difference between 1 and .999(rep)
we can take the first 1/n terms of the sequence leaving a difference that is smaller then the hypothesized difference.

result: for any n hypothesized to be the difference. we can prove that it is infact not the diference, and the difference is smaller.

Therefore there is no number that can be shown to be the difference, other than 0.
1 and .999(rep) are equal. The conjecture was false.

------------------
Do you ever get the feeling that everybody thinks you're paranoid?

[Jinx]

Originally posted by CKDextHavn:
If you are talking practical measurements, that you can make with a ruler or some very fine measuring device, then .999... gets as close to 1 as you can measure.

I thought CKDextHavn was going to pose the opposite argument. I thought he was going to say for practical purposes, like measuring, you can say 0.999... equals 1.0 I'd agree, for practical real-world applications. Just like we'll carry out pi only to a few decimal places.

But, technically speaking, we realize such numbers continue ad infinitum. Well, at least I think we'd agree that 0.999... does!
I'd have to take the stand that, in the abstract world of math where lines and planes have no thickness, etc., I'd hope all could accept the abstract concept that 0.999... goes on and on never quite reaching 1.0 Out of respect for those opinions to the contrary, I will even bend and say "IMHO" if that assuages you dis-believers.

Anyway, it all boils down to this old story. A math teacher lined up all the girls of the class along one wall, and all the boys were lined up along the opposing wall facing each other. The math teacher instructed the boys to walk half way across the room. Then half again, and so on....until they kissed the girl awaiting at the opposite wall. The practical boys got to kiss the girl. The theorists are still trying to complete the task today. ...And that is the rub, ladies and gentlemen, between practical math and theory.

I will also agree with CKDextHavn that, just like the decimal places over which we argue, we could debate this topic ad infinitum! As for me, I'm going to make a quantum leap of faith and grab for 1.0 cool-one ;)

In closing, the bright side is, those of us who believe in 0.9999... &lt; 1.0, we shall never age! I've never really reached my first birthday! I keep getting closer and closer and closer...

-----------------------
It's like deja vu all over again; it's like deja vu all over again!

[wolfman]

Here's my attempt to formalize what others have been saying as an iteratize proof(many years since i did one, go easy on me).

conjecture. 1 is not equal .999(rep)

principle1. if two numbers are different then there is a measureable difference between them.
reason. if x-y =0 then x=y by definition

principle2. .999(rep) can also be written as 9/10 +9/100 +9/1000 ..... or 9/10^1 +9/10^2 +9/10^3 ....+9/10^k


extrapolation. Therefore there is a measurable difference between 1 and .999(rep)


For any arbitrary number n hypothesized to be the difference between 1 and .999(rep)
we can take the first 1/n terms of the sequence leaving a difference that is smaller then the hypothesized difference.

result: for any n hypothesized to be the difference. we can prove that it is infact not the diference, and the difference is smaller.

Therefore there is no number that can be shown to be the difference, other than 0.
1 and .999(rep) are equal. The conjecture was false.

------------------
Do you ever get the feeling that everybody thinks you're paranoid?

[ZenBeam]

I'm going to point out that the 1 &gt; 0.999... camp, for all their posting, has still never shown that 1 &gt; 0.999....

------------------
It is too clear, and so it is hard to see.

[Tengu]

Originally posted by Jinx:
But, technically speaking, we realize such numbers continue ad infinitum. Well, at least I think we'd agree that 0.999... does!

And here you've hit on positive proof that 0.999... does equal 1!!!!

0.999... is a decimal point followed by (n) 9s.

The difference between 1 and 0.999... is thus a decimal point followed by (n-1) 0s, and a 1.

However, since (n) equals infinity, (n-1) also equals infinity.

THEREFOR, there is nowhere for the one to go.

0.000(...)1 == 0!

Therefor, the difference between 1 and 0.999... is [b]0[b].



------------------
Eschew Obfuscation

[Mousseduck]

How do you figure from that, that movement is not possible? You can move, you just never reach where you are going.
And no, I said that I don't believe 0.9... is equal to 1, but it is so close it might make no practical difference, and it depends on the applications of the maths before you can definitively answer. In most applications 0.9... can be considered equal to 1 but technically it is not.
I can get back to you with an equation that when graphed will approach a number (like 1) but never reach it if you like.

[C K Dexter Haven]

Dear Lord, here we have a provable mathematical truth (that .999... = 1) and yet people still disbelieving. No wonder that some people doubt less provable stuff, like Evolution or the Existence of God!

OK, one more time. From the top.

If you are talking practical measurements, that you can make with a ruler or some very fine measuring device, then .999... gets as close to 1 as you can measure. If you want to say they are unequal because you stop measuring after (say) a few thousand decimal places, then OK, say that. That's because you can't measure any finer.

Similarly, you'll have to concede that .3333... is never exactly equal to 1/3 (because you stop measuring after a few hundred thousand decimal places).

And you'll have to admit that pi doesn't really exist, since 3.14159... is never equal to pi (because you stop measuring after a few thousand decimal places.)

You want to work in the world of finite measurements, that's fine. Let's say that we stop everything after a hundred thousand decimal places, OK? That's all the further we can measure, that's narrower than the width of an electron. What kind of mathematics do we have, then?

Well, to start, there are only a finite number of numbers between 0 and 1 (just list all those hundred thousand decimal place). And the numbers are discrete. There is no number halfway between (say) .999...[repeat for a hundred thousand places]98 and .999...[repeat for a hundred thousand places]99

((Can I use RFAHTP for "repeat for a hundred thousand places"?))

Just like, if you're dealing with counting numbers, there is no natural number (integer) between 1 and 2.

It is perfectly acceptable to work in this number system. Notice that .333...RFAHTP...3 added to itself three times will be .999...RFAHTP...9, not the same as 1, so you'll need to be careful with fractions like 1/3. That number system will not allow you to divide 1 by 3 and remain in the number system; you can only approximate that the answer is something that should exist between between .333...[RFAHTP]3 and .333...[RFAHTP]4.

So, you can't always remain in that number system when you divide. However, it's a hundred thousand decimals, for God's sake, so who cares what happens in the hundred-thouand-and-one-th decimal place? We just drop it from our system.

With me so far? That's the argument that .999... is not the same as 1.

HOWEVER, if you want to talk about the world of the Real Number system... the mathematical model used by every science, the most common model, the model that gives us the broadest understanding... then you can have infinite decimals, these are well-defined and well-understood.

The Real Number system offers such advantages as:
- 1/3 added to itself three times gives 1
- Division of any real number by another (excluding zero) stays within the real number system
- Pi exists

Note that in the Real Number system, .333... added to itself three times gives .999... but ALSO (in its guise as 1/3) gives 1.

Those two numbers (.999... and 1) are EQUAL. Not just approximate, not just as close as you'd like, but EQUAL, in the same sense that 1 + 1 = 2 or that 1/3 = .33333...

I contend that the Real Number system is a far better system in which to work, even if we cannot actually MEASURE pi exactly.

Are we agreed? Are we done with this? ... I know that I am.

[Boris B]

Originally posted by Mousseduck:
How do you figure from that, that movement is not possible? You can move, you just never reach where you are going....

The point of Xeno's paradox is that it can be applied with equal (zero) validity to any scale. If you can't move thirty feet, you can't move thirty inches, or any distance at all. It doesn't matter where you intend to go; math can't read your mind. Xeno's paradox is a naive way of mixing ostensibly infinite values with finite values.

With a finite speed, you move an infinitely small distance in an infinitely small amount of time. Since measurable amounts of time are just collections of infinitely small amounts of time, and a collection of infinitely small distances is still an infinitely small distance, you can never move. It's bogus as heck, but it persists.

Here's the bogus math:
Finite speed = Distance / Time
Time / Infinity = an instant
Distance / Infinity = zero distance
(Time / Infinity) x a large number = a positive non-zero amount of time
(Distance / Infinity) x a large number = zero distance

You see how that isn't really fair? Xeno seems to be saying that, by taking an instantaneous snapshot of an object with a measurable velocity, you reveal that the velocity is really 0 / 0, which of course is equal to 0! But 0 / 0 is not equal to zero. It is an empty set.

It really makes me want to throw a spear at the guy.

[RM Mentock]

Originally posted by Jinx:
The practical boys got to kiss the girl. The theorists are still trying to complete the task today. ...And that is the rub, ladies and gentlemen, between practical math and theory.

That just goes to show you that almost all mathematicians, deep down, are practical mathematicians.

------------------
rocks

[don willard]

I asked the same question in another post without having done proper research in this one, because I asked the questrion without going first to this here post. Thus ignore my independent post and eliminate it as I now understand the whole thing.

[DrMatrix]

Ummm . . . OK :confused:

I'll just close this since there is an active thread on the subject. Why doesn't .9999~ = 1? (http://boards.straightdope.com/sdmb/showthread.php?s=&threadid=187902&pagenumber=2)

DrMatrix - GQ Moderator