Math: .99repeating = 1?

[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.)

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Kupek’s Den

[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.

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[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.

[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 -> 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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Kupek’s Den

[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.

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[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.

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Kupek’s Den

[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.

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[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

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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[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, 10D, and 0.1D depending on how I got 9.9[inf]. So D = 10D = 0.1D. 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].

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

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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 ->infinity of

N10^(-m)+N10^(-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 AM

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

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.

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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 --> 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.

[Kamino_Neko]

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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