# .9999... equals 1?

• 123 Replies
• 23350 Views ##### Re: .9999... equals 1?
« Reply #90 on: March 11, 2007, 01:25:41 PM »
speaking of decimal points, isn't this like saying that 9r is equal to 10r?
Maybe in the sense that they are both infinity.

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#### Kasroa Is Gone

• 6869 ##### Re: .9999... equals 1?
« Reply #91 on: March 11, 2007, 01:28:40 PM »
I think the same methods for proving 0.9r = 1.0r can be applied to 9r = 10r. Read the FAQ and lurk more.

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#### Kasroa Is Gone

• 6869 ##### Re: .9999... equals 1?
« Reply #92 on: March 11, 2007, 01:32:11 PM »
So can something be infintely small as long as it's not a decimal number? Read the FAQ and lurk more.

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#### RESOCR

• 416
• I argue for stress relief! ##### Re: .9999... equals 1?
« Reply #93 on: March 11, 2007, 01:37:19 PM »
well by your logic 9r=10r, even though 9r-10r=-1 (no matter how many nines or zeros at the end, as long as it is an equal amount which it should be (infinity=infinity, duh) they would subtract to be 1, or -1) and NOT 0.
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Quote from: Tom Bishop
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#### EnragedPenguin

• The Elder Ones
• 1004 ##### Re: .9999... equals 1?
« Reply #94 on: March 11, 2007, 02:10:42 PM »
no, i didnt mean a number smaller than zero, I meant after as in next on the number line, greater than, etc.

Right.There is no smallest number after zero. You don't even need to resort to irrationals to see that, since between any two numbers there are not only infinitely many irrationals, there are infinitely many rationals as well. This is pretty obvious in the case of rationals; because if you have any number bigger than zero, you can divide that number by two and get a smaller number bigger than zero.

By this, of course, Skeptical meant no smallest number after zero on the number line.
A different world cannot be built by indifferent people.

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#### RESOCR

• 416
• I argue for stress relief! ##### Re: .9999... equals 1?
« Reply #95 on: March 11, 2007, 02:15:25 PM »
no, i didnt mean a number smaller than zero, I meant after as in next on the number line, greater than, etc.

Right.There is no smallest number after zero. You don't even need to resort to irrationals to see that, since between any two numbers there are not only infinitely many irrationals, there are infinitely many rationals as well. This is pretty obvious in the case of rationals; because if you have any number bigger than zero, you can divide that number by two and get a smaller number bigger than zero.

By this, of course, Skeptical meant no smallest number after zero on the number line.

Well whatever he meant, he is right, and it is not what I meant. Maybe it would be better explainable as the smallest non-zero number?

Anyway, ho would you display it? I've been (trying) to display it as 0.0r1, because thats how it makes sense to me.
« Last Edit: March 11, 2007, 02:40:27 PM by RESOCR »
Quote from: ice wall gard 469320
Quote from: Tom Bishop
Atmosphere gets thinner with altitude

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#### Kasroa Is Gone

• 6869 ##### Re: .9999... equals 1?
« Reply #96 on: March 11, 2007, 02:16:39 PM »
You can't have the smallest number after zero because you can always divide it by 2 again to get a smaller number. Read the FAQ and lurk more. ##### Re: .9999... equals 1?
« Reply #97 on: March 11, 2007, 02:23:42 PM »
I was confused with what you were originally saying, Kasroa.  You mean 9.9r = 10.0r and not an infinite amount of nines on the left of the decimal place, right?

well by your logic 9r=10r, even though 9r-10r=-1 (no matter how many nines or zeros at the end, as long as it is an equal amount which it should be (infinity=infinity, duh) they would subtract to be 1, or -1) and NOT 0.
I don't think that's right, Resocr.
9 - 10 = -1, but
9.9 - 10.0 = -0.1,
9.99 - 10.00 = -0.01, etc.
So 9.9r 10.0r =/= -1.

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#### Kasroa Is Gone

• 6869 ##### Re: .9999... equals 1?
« Reply #98 on: March 11, 2007, 02:30:07 PM »
I don't know what I'm saying I don't have any further education in Maths  Read the FAQ and lurk more.

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#### RESOCR

• 416
• I argue for stress relief! ##### Re: .9999... equals 1?
« Reply #99 on: March 11, 2007, 02:34:39 PM »
I was confused with what you were originally saying, Kasroa.  You mean 9.9r = 10.0r and not an infinite amount of nines on the left of the decimal place, right?

well by your logic 9r=10r, even though 9r-10r=-1 (no matter how many nines or zeros at the end, as long as it is an equal amount which it should be (infinity=infinity, duh) they would subtract to be 1, or -1) and NOT 0.
I don't think that's right, Resocr.
9 - 10 = -1, but
9.9 - 10.0 = -0.1,
9.99 - 10.00 = -0.01, etc.
So 9.9r 10.0r =/= -1.

uh, I was more referring to 9-10=-1
99-100=-1
999-1000=-1
etc.

I did not add any decimals. 9r (99999999999999999999....)-10r(100000000000000000000....)=-1.
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Quote from: Tom Bishop
Atmosphere gets thinner with altitude

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#### Kasroa Is Gone

• 6869 ##### Re: .9999... equals 1?
« Reply #100 on: March 11, 2007, 02:36:22 PM »
I think once you add the "..." it represents something slightly different. Like 0.999... = 1. But 0.999 = 0.999 Read the FAQ and lurk more. ##### Re: .9999... equals 1?
« Reply #101 on: March 11, 2007, 02:44:38 PM »
well by your logic 9r=10r, even though 9r-10r=-1 (no matter how many nines or zeros at the end, as long as it is an equal amount which it should be (infinity=infinity, duh) they would subtract to be 1, or -1) and NOT 0.
You can't do this with real numbers, because real numbers have decimal expansions with a leftmost digit, but not necessarily a rightmost so neither 9r nor 10r represent real numbers. Incidentally, p-adic numbers have base p representations with a rightmost, but no leftmost digit. So in the 10-adics, ...999 is an actual number, (although it's important to realize that there is a last 9 but no first, unlike 9r which one would think of as having a first but no last) and it is in fact true, in the 10-adics, that ...999=-1. (If you find this confusing, don't worry about it. The p-adic numbers are a very abstract algebraic concept which only professional mathematicians care about anyways.)

On the subject of the "first number after zero on the number line", as I've said before, there is no such object. The real numbers are dense, which means that between any two real numbers there is another real number. In fact, this is obvious, because if a and b are real numbers, so is their average, (a+b)/2, which lies half way in between them on the number line. As a corollary to this, there is no first number after zero on the number line, because if there were, half of it would lie between it and zero, so it wouldn't actually be the first.
« Last Edit: March 11, 2007, 02:47:00 PM by skeptical scientist »
-David
E pur si muove!

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#### RESOCR

• 416
• I argue for stress relief! ##### Re: .9999... equals 1?
« Reply #102 on: March 11, 2007, 02:52:46 PM »
well by your logic 9r=10r, even though 9r-10r=-1 (no matter how many nines or zeros at the end, as long as it is an equal amount which it should be (infinity=infinity, duh) they would subtract to be 1, or -1) and NOT 0.
You can't do this with real numbers, because real numbers have decimal expansions with a leftmost digit, but not necessarily a rightmost so neither 9r nor 10r represent real numbers. Incidentally, p-adic numbers have base p representations with a rightmost, but no leftmost digit. So in the 10-adics, ...999 is an actual number, (although it's important to realize that there is a last 9 but no first, unlike 9r which one would think of as having a first but no last) and it is in fact true, in the 10-adics, that ...999=-1. (If you find this confusing, don't worry about it. The p-adic numbers are a very abstract algebraic concept which only professional mathematicians care about anyways.)

On the subject of the "first number after zero on the number line", as I've said before, there is no such object. The real numbers are dense, which means that between any two real numbers there is another real number. In fact, this is obvious, because if a and b are real numbers, so is their average, (a+b)/2, which lies half way in between them on the number line. As a corollary to this, there is no first number after zero on the number line, because if there were, half of it would lie between it and zero, so it wouldn't actually be the first.

9r has a leftmost digit, being 9. 10r has a leftmost digit, being 1.

And if you say there is no first number after 0 on a number line, that is saqying no numbers even exist, that 0 is as far forward as you can go.

and if 0.9r is a real number, and 1 is definately a real number, there would be numbers between them.
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#### Kasroa Is Gone

• 6869 ##### Re: .9999... equals 1?
« Reply #103 on: March 11, 2007, 02:56:12 PM »
Sometimes you just have to accept that things can make sense mathematically but not make sense logically.

Like I was reading the other day a solution to that old paradox where a question such as "Is 'no' the answer to this question?". They managed to find a mathematical "inbetween" answer that can be both yes and no at the same time. A load of rubbish it seems to be, but it makes sense mathematically. Read the FAQ and lurk more. ##### Re: .9999... equals 1?
« Reply #104 on: March 11, 2007, 03:02:20 PM »
9r has a leftmost digit, being 9. 10r has a leftmost digit, being 1.

And if you say there is no first number after 0 on a number line, that is saqying no numbers even exist, that 0 is as far forward as you can go.

and if 0.9r is a real number, and 1 is definately a real number, there would be numbers between them.
Except that 0.9r IS 1. ##### Re: .9999... equals 1?
« Reply #105 on: March 11, 2007, 03:09:10 PM »
9r has a leftmost digit, being 9. 10r has a leftmost digit, being 1.
Sorry, I should have said that decimal expansions have a leftmost digit, and finitely many digits before the decimal place, but potentially infinitely many afterwards. (p-adics have finitely many digits after the decimal, and perhaps infinitely many before, and may have no first digit) A decimal expansion is shorthand for the limit of the things you get when you take finite decimal expansions, so 0.999... is the limit of the sequence 0.9, 0.99, 0.999, 0.9999, ..., and the limit of this sequence is 1. p-adic representations are the same way, only the limit is taken in the p-adic numbers instead of the real numbers, so in the p-adics, ...999 is the limit of the sequence 9, 99, 999, 9999, ..., which is -1 (in the p-adic numbers).

Quote
And if you say there is no first number after 0 on a number line, that is saqying no numbers even exist, that 0 is as far forward as you can go.
No, it's not saying this at all. There are lots of real numbers after zero, it's just that none of them is the first. This is no different from the fact that there is no last natural number. Surely you would agree that natural numbers exist, and that there are in fact lots of them, but none of them is the largest? It is equally true that positive real numbers exist, and there are lots of them, but none of them is the smallest.
-David
E pur si muove! ##### Re: .9999... equals 1?
« Reply #106 on: March 11, 2007, 03:14:19 PM »
Sometimes you just have to accept that things can make sense mathematically but not make sense logically.
No. If something doesn't make sense logically, it can't make sense mathematically either. What you mean to say is that something can be counterintuitive, but still make sense mathematically; this is certainly true.

Quote
Like I was reading the other day a solution to that old paradox where a question such as "Is 'no' the answer to this question?". They managed to find a mathematical "inbetween" answer that can be both yes and no at the same time. A load of rubbish it seems to be, but it makes sense mathematically.
It sounds like a load of rubbish to me too, but probably the rubbish was not the mathematics that was being done, but rather the reporter trying to rephrase the mathematics in terms of ordinary language, which can be fantastically wrong even when the mathematics is correct. Ordinary language is sometimes unsuitable for expressing mathematics, and this is very often the case when paradoxes arise. The question, "Is 'no' the answer to this question?" simply has no answer, which is fine when you are talking about language. In order to do mathematics, you must somehow translate the question into a mathematical framework, and so the mathematicians were probably just saying that they had found a neat mathematical framework in which "questions" analogous to that question could be "asked", and would have a well-defined "answer" in their mathematical framework.
-David
E pur si muove!

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#### Kasroa Is Gone

• 6869 ##### Re: .9999... equals 1?
« Reply #107 on: March 11, 2007, 03:17:10 PM »
It was quite scary given that if we are ever invaded by robot aliens we won't be able to confuse them and blow up their heads by shouting paradoxical questions at them. Read the FAQ and lurk more.

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#### RESOCR

• 416
• I argue for stress relief! ##### Re: .9999... equals 1?
« Reply #108 on: March 11, 2007, 05:58:16 PM »
It was quite scary given that if we are ever invaded by robot aliens we won't be able to confuse them and blow up their heads by shouting paradoxical questions at them.

Reminds me of the Dark Tower series.

You just have to ask 'Why did the chicken cross the road?'

back to topic, I think I get what your trying to say, that you progress it enough to the point they difference is indistinguishable, and being no solid difference, they are the same. Maybe? So it would hold true that 0.89r=0.9, 1.09r=1.1, etc?

But I still don't believe there is no real number that would be first, we'd just have to apply a range of 'tolerance', how many decimal places are accounted for. As long as you increase but maintain that there will be a smaller number.

Although someone pointed out a number line plotted on a graph, X axis being the number and -Y axis being the number of decimal places it has, or something to that effect. I liked that.
Quote from: ice wall gard 469320
Quote from: Tom Bishop
Atmosphere gets thinner with altitude ##### Re: .9999... equals 1?
« Reply #109 on: March 11, 2007, 07:45:13 PM »
It was quite scary given that if we are ever invaded by robot aliens we won't be able to confuse them and blow up their heads by shouting paradoxical questions at them.
Like Santa Claus (Futurama)?

~D-Draw

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#### Kasroa Is Gone

• 6869 ##### Re: .9999... equals 1?
« Reply #110 on: March 12, 2007, 03:02:59 PM »
I don't understand some of the terms used in the explanations when they start talking about "limits" etc. They say 0.999... reperesents the limit of "insert lots of fractions being added together here". What does that mean exactly? Can anyone explain the definition of decimals in layman's terms. I understand how 0.999... = 1 but I don't understand why exactly. Read the FAQ and lurk more. ##### Re: .9999... equals 1?
« Reply #111 on: March 12, 2007, 03:16:57 PM »
Limits are just what happens to a function as it gets closer to a set point.

For example if we have f(x)=x+1  and we want to know limit x->3 then we can easily calculate that as limit f(x)=4.  Limits obviously have much more significance in more advanced problems.  For example the other week I was calculating how long it would take my team at work to update the credit card details of 1000 members.  I knew that we make about 20 calls per hour and that we resolve 29% of calls.  However if you graph that equation, you'll quickly see that we would end up calling a small amount of people a large number of times.  For this reason, we imposed the limit of 5 calls maximum per person, and found that we would resolve something like 80% of people within 5 calls.

Often in maths, you'll set the limit at infinity (despite the hilarious claim that maths cannot deal with infinity) and see what happens to a function as it approaches an infinite value for x.

That's about as basic and understandable as I can put it, and hopefully not too incorrect  ##### Re: .9999... equals 1?
« Reply #112 on: March 12, 2007, 03:17:44 PM »
My explanation from page 1:
Masterchief, are you serious, or are you joking? The identity .99999...=1 is a well known fact. Every so often people doubt it, but such people don't even know the definitions.

First question: what is meant by the representation .98604, or .33333..., or .9999....?

Answer: each of these is shorthand for a sum, where the nth digit is the multiple of 10-n that we are adding. So .98604 is simply shorthand for 9*10-1+8*10-2+6*10-3+0*10-4+4*10-5, and .3333... is shorthand for 3*10-1+3*10-2+3*10-3+3*10-4+....

This immediately raises a second question: what is meant by an infinite sum? You can't add infinitely many things, can you?

Answer: yes, you can, in some situations. You can add 1+0+0+0+0+0+..., because after the first term you aren't changing things at all. You can add 1/2+1/4+1/8+1/16+..., because such a sum can be represented "physically" on the number line as follows: first you go half way from 0 to 1, then you go half way from where you are to 1, then you go half way from where you are now to 1, etc. The sum is 1, because that's exactly where you get when you move half way to one infinitely many times (this is not, of course, a proof, but acts as motivation for the idea of limits.) You cannot, however, take the sum 1-1+1-1+1-..., because that sum alternates between 1 and 0, and never settles on any one number (anyone who tells you that sum exists is either completely nuts or has been staring at the zeta function too long). The definition that makes this work is the definition of limits. A sequence of numbers x_n is just that: a sequence x_1, x_2, x_3, x_4, .... Such a sequence is said to converge to a limit x if for any e>0, there is an N>0 so that the differences between x and x_n are less than e for n>N. An infinite sum is just the limit of the partial sums of the initial terms, provided that limit converges. So the decimal number .999999... really is the limit of the sums 9/10, 9/10+9/100, 9/10+9/100+9/1000, etc.

They don't teach you all this when they teach you decimals in school, because it's complicated, so they usually just gloss over it, but it is what decimal representations actually are, and so any demonstration that .999...=1 has to use this fact (of course, there are some tricks people sometimes use to "cheat", but those involve taking things on faith, which are true, but which are exactly the things that Erebos is denying above). The lack of such actual definitions is why so many people are confused, or try to deny that .9999...=1. But what's true is that decimal numbers are really just representations of real numbers, and are not always unique. The fact that 1 can also be written as .9999.... is just one example.

Proof that 1=.99999...:
We know that .999... is, by definition, the sum 9/10+9/100+9/1000+... which is the limit of the sums 9/10, 9/10+9/100, 9/10+9/100+9/1000, etc. Simply finding common denominators tells us that 9/10+9/100+9/1000+...+9/10n=99...9/10n=(10n-1)/10n=1-1/10n, and since for any e>0, 1/10n is eventually smaller than e, this limit is 1.

p.s. I decided to change my avatar in honor of this thread.
If there are specific things from this explanation which are unclear, let me know which and I'll explain further.

Beast's post is dealing with the limit of a function at a point (and something about phone banking which doesn't really seem relevent), and he doesn't define it at all. Decimals are defined as the limit of a sequence, which is different (but related). You can see the definition of the limit of a sequence in the post I quoted above.
« Last Edit: March 12, 2007, 03:21:03 PM by skeptical scientist »
-David
E pur si muove!

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#### Kasroa Is Gone

• 6869 ##### Re: .9999... equals 1?
« Reply #113 on: March 12, 2007, 03:28:40 PM »
My explanation from page 1:
Masterchief, are you serious, or are you joking? The identity .99999...=1 is a well known fact. Every so often people doubt it, but such people don't even know the definitions.

First question: what is meant by the representation .98604, or .33333..., or .9999....?

Answer: each of these is shorthand for a sum, where the nth digit is the multiple of 10-n that we are adding. So .98604 is simply shorthand for 9*10-1+8*10-2+6*10-3+0*10-4+4*10-5, and .3333... is shorthand for 3*10-1+3*10-2+3*10-3+3*10-4+....

This immediately raises a second question: what is meant by an infinite sum? You can't add infinitely many things, can you?

Answer: yes, you can, in some situations. You can add 1+0+0+0+0+0+..., because after the first term you aren't changing things at all. You can add 1/2+1/4+1/8+1/16+..., because such a sum can be represented "physically" on the number line as follows: first you go half way from 0 to 1, then you go half way from where you are to 1, then you go half way from where you are now to 1, etc. The sum is 1, because that's exactly where you get when you move half way to one infinitely many times (this is not, of course, a proof, but acts as motivation for the idea of limits.) You cannot, however, take the sum 1-1+1-1+1-..., because that sum alternates between 1 and 0, and never settles on any one number (anyone who tells you that sum exists is either completely nuts or has been staring at the zeta function too long). The definition that makes this work is the definition of limits. A sequence of numbers x_n is just that: a sequence x_1, x_2, x_3, x_4, .... Such a sequence is said to converge to a limit x if for any e>0, there is an N>0 so that the differences between x and x_n are less than e for n>N. An infinite sum is just the limit of the partial sums of the initial terms, provided that limit converges. So the decimal number .999999... really is the limit of the sums 9/10, 9/10+9/100, 9/10+9/100+9/1000, etc.

They don't teach you all this when they teach you decimals in school, because it's complicated, so they usually just gloss over it, but it is what decimal representations actually are, and so any demonstration that .999...=1 has to use this fact (of course, there are some tricks people sometimes use to "cheat", but those involve taking things on faith, which are true, but which are exactly the things that Erebos is denying above). The lack of such actual definitions is why so many people are confused, or try to deny that .9999...=1. But what's true is that decimal numbers are really just representations of real numbers, and are not always unique. The fact that 1 can also be written as .9999.... is just one example.

Proof that 1=.99999...:
We know that .999... is, by definition, the sum 9/10+9/100+9/1000+... which is the limit of the sums 9/10, 9/10+9/100, 9/10+9/100+9/1000, etc. Simply finding common denominators tells us that 9/10+9/100+9/1000+...+9/10n=99...9/10n=(10n-1)/10n=1-1/10n, and since for any e>0, 1/10n is eventually smaller than e, this limit is 1.

p.s. I decided to change my avatar in honor of this thread.

I bolded the parts that I don't understand. Read the FAQ and lurk more. ##### Re: .9999... equals 1?
« Reply #114 on: March 12, 2007, 03:54:37 PM »
Okay, the parts you bolded were the definition of a sequence, the definition of a limit, and the proof that the limit of the sequence of partial sums of the series represented by .999... is 1 (don't worry if you didn't understand that last sentence right now.)

First, what is a sequence of real numbers?
A sequence is a list of numbers indexed by the natural numbers (1, 2, 3, 4, 5, etc.), or equivalently it can be thought of as a function from natural numbers to real numbers. Often a sequence is written x1, x2, x3, x4, ... meaning that the first number in the sequence is x1, the second is x2, etc. The numbers in the sequence are called the terms of the sequence; for instance, the nth term of the sequence x1, x2, x3, ... is the number xn. (You can also talk about sequences of things other than real numbers, such as sequences of complex numbers, or sequences of points in the plane, etc., but for our purposes all we need are sequences of real numbers.)

Some example sequences:
The constant sequence zero: 0, 0, 0, 0, ...
The sequence 1/n: 1, 1/2, 1/3, 1/4, 1/5, ...
The sequence of prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, ...
The sequence of powers of 1/10: 1/10, 1/100, 1/1000, ...
etc.

Second, what is a convergent sequence, and what is a limit?
A sequence of real numbers x1, x2, x3, ... is said to converge to a limit l if for every real number ε greater than zero, there is a natural number N such that for every n>N, the distance |xn-l| between the nth term of the sequence and the limit is smaller than ε. Roughly speaking, a sequence converges to l if the points on the number line defined by the terms of the sequence eventually get as close to l as you want.

Examples:
The constant sequence zero: 0, 0, 0, 0, ... converges to 0, because it is always as close as you could possibly want to zero.
The sequence 1/n: 1, 1/2, 1/3, 1/4, 1/5, ... converges to 0, because given ε>0, we can consider the real number 1/ε. There is some natural number N larger than 1/ε (this is sometimes called the Archimedian property of the real numbers). Then for n>N, 1/n<1/N<ε, so |1/n-0|<ε. This is exactly what it means for the sequence to converge to 0.

Third, what is an infinite sum?
An infinite sum, usually called an infinite series, is simply the sum of an infinite number of terms arranged in a sequence, and is defined by means of limits. If the terms of the series are y1, y2, y3, ..., then we define the nth partial sum sn of the series as the finite sum y1+y2+y3+...+yn. The series is said to converge to a limit s if the sequence of partial sums s1, s2, s3, ... converges to s (as defined above for a sequence), in which case we write y1+y2+y3+...=s.

Now does the proof make some sense? (By the way, can you see this symbol: ε? I want to make sure it displays properly on your browser. It's the lower case Greek letter epsilon, and it is almost always used as I have used it here in the context of limits in mathematics.)

p.s. The contents of this post constitute material which is not generally taught except in college math courses, and some high school calculus classes.

Edit: I decided to define infinite series in more detail as well.

You can also check out the Wikipedia pages on limits (of sequences) and series.
« Last Edit: March 12, 2007, 04:07:56 PM by skeptical scientist »
-David
E pur si muove! #### cmdshft

• The Elder Ones
• 13129
• swiggity swooty ##### Re: .9999... equals 1?
« Reply #115 on: March 13, 2007, 05:11:40 PM »
tl;dr

Use limits. It's pre-calc High School math.

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#### Kasroa Is Gone

• 6869 ##### Re: .9999... equals 1?
« Reply #116 on: March 13, 2007, 05:16:28 PM »
I lose it when all the maths lingo pops up. Read the FAQ and lurk more. #### Wendy

• 18492
• I laugh cus you fake ##### Re: .9999... equals 1?
« Reply #117 on: March 19, 2007, 01:36:17 AM »
A simple way of explaining it is that there is no number in between 0.9 repeat and 1, and because of that they are the same number. And to whoever said that 0.3 repeat isn't exactly 1/3, you're wrong. There is no exact number for a third, if you don't take the concept of infinity into account.
Here's an explanation for ya. Lurk moar. Every single point you brought up has been posted, reposted, debated and debunked. There is a search function on this forum, and it is very easy to use.

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#### SPrinkZ

• 762 ##### Re: .9999... equals 1?
« Reply #118 on: March 20, 2007, 12:00:16 AM »
Shit, I can't believe Masterchief thought that .9 (repeating) didn't equal 1. I knew that when I was like...crap, I don't even recall how young I was.
ah. ##### Re: .9999... equals 1?
« Reply #119 on: March 20, 2007, 12:19:09 AM »
Shit, I can't believe Masterchief thought that .9 (repeating) didn't equal 1. I knew that when I was like...crap, I don't even recall how young I was.
I bet your mother thought you were cool.

~D-Draw