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  #1  
Old 04-12-2000, 08:59 AM
Flat U. Lance Flat U. Lance is offline
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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.

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  #2  
Old 04-12-2000, 09:26 AM
bigneck bigneck is offline
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Flat,
.87632 = .87632
17 = 17
1 = 1
.99999999 = .99999999

you = dork

bigneck
  #3  
Old 04-12-2000, 09:27 AM
AWB AWB is offline
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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]


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  #4  
Old 04-12-2000, 09:28 AM
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  #5  
Old 04-12-2000, 09:29 AM
bigneck bigneck is offline
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Flat,
.87632 = .87632
17 = 17
1 = 1
.99999999 = .99999999

you = dork

bigneck
  #6  
Old 04-12-2000, 09:31 AM
bigneck bigneck is offline
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Sorry for posting that last one twice - if I did. Actually, .333333 + .66666 = .999993
  #7  
Old 04-12-2000, 09:38 AM
handy handy is offline
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We already answered this one down below.

What is it with the logic people anyway?

1 does not & is not a 9 period.
  #8  
Old 04-12-2000, 09:41 AM
Flat U. Lance Flat U. Lance is offline
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Hey Handy, where was this answered before?
  #9  
Old 04-12-2000, 09:47 AM
RussellM RussellM is offline
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zero point...repeating

Russell
  #10  
Old 04-12-2000, 09:55 AM
RussellM RussellM is offline
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Last discussed Is 0.999...=1 ?

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

Russell
  #11  
Old 04-12-2000, 10:00 AM
C K Dexter Haven C K Dexter Haven is offline
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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.
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Old 04-12-2000, 11:04 AM
Arnold Winkelried Arnold Winkelried is offline
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CKDExtHavn, you sound like Fermat in your last post.
  #13  
Old 04-12-2000, 01:53 PM
C K Dexter Haven C K Dexter Haven is offline
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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.
  #14  
Old 04-12-2000, 04:14 PM
CheapBastid CheapBastid is offline
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Again with the Infinities...
  #15  
Old 04-12-2000, 05:20 PM
tony1234 tony1234 is offline
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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

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  #16  
Old 04-12-2000, 07:24 PM
pipefitter pipefitter is offline
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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

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  #17  
Old 04-12-2000, 10:03 PM
Monty Monty is online now
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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.
  #18  
Old 04-12-2000, 10:17 PM
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test

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Old 04-12-2000, 11:37 PM
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.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.)

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  #20  
Old 04-13-2000, 12:12 AM
Cabbage Cabbage is offline
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Quote:
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.

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  #21  
Old 04-13-2000, 12:15 AM
Jinx Jinx is offline
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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!



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  #22  
Old 04-13-2000, 12:27 AM
The Ryan The Ryan is offline
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Quote:
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?


Quote:
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.
  #23  
Old 04-13-2000, 12:49 AM
emarkp emarkp is offline
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Sigh. Is 0.999...=1 ?

Yes, they're equal. Go see the other thread.
  #24  
Old 04-13-2000, 07:35 AM
Monty Monty is online now
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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.
  #25  
Old 04-13-2000, 09:37 AM
C K Dexter Haven C K Dexter Haven is offline
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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.
  #26  
Old 04-13-2000, 09:52 AM
bj0rn bj0rn is offline
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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...!
  #27  
Old 04-13-2000, 09:55 AM
handy handy is offline
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Anyone remember when the Pentium chip would process 1 to .989999 or so? remember what happened? Yep.
  #28  
Old 04-13-2000, 07:30 PM
Kupek Kupek is offline
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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.

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  #29  
Old 04-13-2000, 10:31 PM
Cabbage Cabbage is offline
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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.

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  #30  
Old 04-13-2000, 11:02 PM
Kupek Kupek is offline
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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.

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  #31  
Old 04-14-2000, 12:58 AM
Cabbage Cabbage is offline
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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.

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  #32  
Old 04-14-2000, 08:07 AM
C K Dexter Haven C K Dexter Haven is offline
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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.
  #33  
Old 04-14-2000, 08:37 AM
tony1234 tony1234 is offline
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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

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  #34  
Old 04-14-2000, 11:04 AM
ZenBeam ZenBeam is offline
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Quote:
The difference between 1 and .99 repeating is an infintesimal quanity.
From your site, an Infinitesimal is
Quote:
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[inf] 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].


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  #35  
Old 04-14-2000, 07:19 PM
tony1234 tony1234 is offline
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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



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  #36  
Old 04-14-2000, 08:04 PM
The Ryan The Ryan is offline
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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.
Quote:
As mathematics is rife with infinites
Actually, mathematicians are very careful about infinities; they don't just go throwing them around.
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



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

CKDextHavn posted 04-13-2000 09:37 AM
Quote:
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.
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
Quote:
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

Quote:
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?
  #37  
Old 04-15-2000, 01:36 PM
bigneck bigneck is offline
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CKDextHavn
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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?
  #38  
Old 04-15-2000, 03:52 PM
23skidoo 23skidoo is offline
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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.
  #39  
Old 04-15-2000, 05:51 PM
Konrad Konrad is offline
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CK sez:
Quote:
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.
  #40  
Old 04-15-2000, 06:25 PM
Kamino Neko Kamino Neko is offline
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Quote:
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.



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Eschew Obfuscation
  #41  
Old 04-16-2000, 01:06 PM
Monty Monty is online now
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The Ryan posted 04-14-2000 08:04PM the stuff between the quotes; my responses are in between the quotations.
Quote:
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:
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:
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:
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:

Quote:
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
&

[quote]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.
  #42  
Old 04-17-2000, 04:48 AM
RM Mentock RM Mentock is offline
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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
  #43  
Old 04-17-2000, 06:36 AM
Jinx Jinx is offline
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Quote:
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...




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  #44  
Old 04-17-2000, 06:38 AM
Jinx Jinx is offline
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I can see why Cecil feels like there is no hope for the teeming millions!

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  #45  
Old 04-17-2000, 07:32 AM
ZenBeam ZenBeam is offline
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Quote:
I'd really like to see a long division where 1 divided into itself can yield 0.9999...
Oh, this is relevant.

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It is too clear, and so it is hard to see.
  #46  
Old 04-17-2000, 08:06 AM
C K Dexter Haven C K Dexter Haven is offline
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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.
  #47  
Old 04-17-2000, 08:39 AM
Jinx Jinx is offline
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Quote:
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?




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  #48  
Old 04-17-2000, 08:48 AM
Jinx Jinx is offline
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Quote:
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!




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"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
  #49  
Old 04-17-2000, 09:27 AM
C K Dexter Haven C K Dexter Haven is offline
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&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.
  #50  
Old 04-17-2000, 09:38 AM
kbutcher kbutcher is offline
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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...?
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