What is .99 repeating as a fraction?
(and are aware that some still argue against that.)
There are proofs everywhere.
I like this one.
1/3 = .333...
Multiply each side by 3.
since
fraction 1/9 = .111...
fraction 2/9 = .222...
we can see a pattern
can .999... be expressed as a fraction?
I said 9/9.
Sally
Otherwise, it's "One".
He is right. There are many fractions that can express .999... as a fraction.Quote: mustangsally
can .999... be expressed as a fraction?
I said 9/9.
A real smart math person at my work says yes, but he has no social skills. A real geek!
He said that is one answer and went back to his work.
My favorite:
1/1
or is it 4/4?
Quote: IbeatyouracesI thought .999 = 999/1000.
You are correct.
The above people are smoking crack.
You're right.Quote: IbeatyouracesI thought .999 = 999/1000.
You'll note the three dots in the original post:
Quote: mustangsallyI am sure most know that .999... = 1.
What is implied is
0.9999999999999999999999999999999999999999999 . . . .
I.E. A repeating decimal
By the way, Wikipedia has an entire article on .999
http://en.wikipedia.org/wiki/0.999…
And cute little proofs whereby you get to a whole integer by the process of transition or removal is just as ridiculous.
Quote: YoDiceRoll11It doesn't matter. You can't just take an infinite repeater and imply that it equals a whole integer. Put the pipe down. ;)
And cute little proofs whereby you get to a whole integer by the process of transition or removal is just as ridiculous.
We were taught (by assuredly non-pipe smoking Algebra profs) that indeed .9999 (with a bar over it) was = 1. I remember half the class did have a big problem with it. It didn't bother me either way, though I was intrigued by the argument "There is no number between the two, so they are the same number"
Quote: cclub79We were taught (by assuredly non-pipe smoking Algebra profs) that indeed .9999 (with a bar over it) was = 1. I remember half the class did have a big problem with it. It didn't bother me either way, though I was intrigued by the argument "There is no number between the two, so they are the same number"
The problem with this, is that there is a number between the two, an infinitely smaller and smaller exponentially decreasing number. There is ALWAYS a number between .999999... and 1. And there always will be.
I'm curious -- if you don't think the infinite repeating decimal .999... is equal to 1, how much do you think they differ by? Is that difference something that does not equal zero? On the chance that you think they differ by something equivalent to zero, how is that different from being equal?Quote: YoDiceRoll11It doesn't matter. You can't just take an infinite repeater and imply that it equals a whole integer. Put the pipe down. ;)
And cute little proofs whereby you get to a whole integer by the process of transition or removal is just as ridiculous.
Quote: YoDiceRoll11The problem with this, is that there is a number between the two, an infinitely smaller and smaller exponentially decreasing number. There is ALWAYS a number between .999999... and 1. And there always will be.
An exponentially decreasing number? What does that mean?
Let X = 1
and Y = 0.999...
then X-Y is not an "exponentially decreasing number". It is a constant, because both X and Y are constants. Pop quiz: what is X-Y?
Quote: DocI'm curious -- if you don't think the infinite repeating decimal .999... is equal to 1, how much do you think they differ by? Is that difference something that does not equal zero? On the chance that you think they differ by something equivalent to zero, how is that different from being equal?
Please re-read my last post. I can't quantify how much .999..., is different to 1. Because there is always an infinitely smaller number between it and 1. Plain and simple.
.999 does not equal 1.
.99999999999999999999999999999999999 does not equal 1
.999999999999999999999999999999999999999999999999999999999999999999999999999999 does not equal 1.
Extrapolate to your heart's content.
Quote: YoDiceRoll11Please re-read my last post. I can't quantify how much .999..., is different to 1. Because there is always an infinitely smaller number between it and 1. Plain and simple.
.999 does not equal 1.
.99999999999999999999999999999999999 does not equal 1
.999999999999999999999999999999999999999999999999999999999999999999999999999999 does not equal 1.
Extrapolate to your heart's content.
None of the numbers you just wrote are equal to 0.999...
They are equal to some finite number of 9s following the decimal. Your idea is correct for any finite number of 9s following the decimal, but not for 0.999... which has an infinite number of 9s following the decimal.
Quote: YoDiceRoll11
.999 does not equal 1.
.99999999999999999999999999999999999 does not equal 1
.999999999999999999999999999999999999999999999999999999999999999999999999999999 does not equal 1.
Extrapolate to your heart's content.
You are right as long as there is only a finite number of nines. If there are infinitely many of them, then obviously there is nothing bigger than it yet smaller than 1, because you can't add more digits to it, and you can't increase any of them.
In fact, any (non-zero) terminating rational number has two alternative decimal representations, not just 1 for example:
1.25 = 1.24999999... etc.
An important thing to understand is that this is not two numbers that happen to be equal, but rather two different ways to write down the same number.
Much like 1.25 = 5/4 = 1+1/4 etc.
Note, that this is not some kind of mind bugling fundamental property of mathematics, just a curios artefact of the positional numerical notation, that we use to express decimals. Roman numerals do not have such an oddity and neither do simple fractions.
Consider 5/11. In decimal notation, that equals 0.4545... (45 repeating). Since that is the definition of how to denote 5/11 as a decimal number, it would be silly to turn around and suggest that 0.4545... is somehow not equal to 5/11 based on the logic that 0.45 != 5/11, 0.4545 != 5/11, 0.45454545454545 != 5/11, etc.
Quote: YoDiceRoll11Please re-read my last post. I can't quantify how much .999..., is different to 1. Because there is always an infinitely smaller number between it and 1. Plain and simple.
.999 does not equal 1.
.99999999999999999999999999999999999 does not equal 1
.999999999999999999999999999999999999999999999999999999999999999999999999999999 does not equal 1.
Extrapolate to your heart's content.
I did not ask about the difference between 1 and any of those decimals with a finite number of digits. I asked how much 1 differs from the infinite repeating decimal .999... and whether you think that difference is something other than zero. I think you deliberately evaded answering the questions that were asked.
Oh, by the way, I did re-read your last post where you made the erroneous statement (repeated in your next post):
Quote: YoDiceRoll11There is ALWAYS a number between .999999... and 1. And there always will be.
At least that statement is false to anyone who understands what an infinite repeating decimal is.
x * 10 = 9.9999999..... = 9 + x
x*10 - x = 9x
=> 9x = 9 + x - x
=> 9x = 9
=> x = 1
QED.
.666... = 2/3
.999... = 1
Quote: MathExtremistNone of the numbers you just wrote are equal to 0.999...
They are equal to some finite number of 9s following the decimal. Your idea is correct for any finite number of 9s following the decimal, but not for 0.999... which has an infinite number of 9s following the decimal.
You are correct. My point was that the extrapolation though, there will always be another 9. I'm just on the other side of the fence on this one.
Edit: I already know all the arguments for .999...= 1, don't bother.
Quote: YoDiceRoll11You are correct. My point was that the extrapolation though, there will always be another 9. I'm just on the other side of the fence on this one.
An infinite number of 9s precludes the option for there to be another 9.
Quote: YoDiceRoll11Only if you stop the process of infinity, which presents several problems......
Like what for example?
So far, the only problems we see with infinity are those caused exactly by what you are trying to do - attempting to extrapolate some properties of finite quantities (such as the existence of a larger quantity) and erroneously assuming that an infinity must possess such property too.
Infinity is not a number, and, even though, sometimes it is possible to operate it as if it was, you need to remember that not everything you know about numerical quantities also holds for infinity. In particular, there is nothing greater than it.
You cannot add another nine if you've already got an infinite number of them. This, in fact, is a definition of the infinity. If you can increase the amount of anything (like nines after the decimal place) by adding another instance (or in fact any countable number of instances), that means that you did not have an infinite amount to begin with.
Quote: weaselmanLike what for example?
So far, the only problems we see with infinity are those caused exactly by what you are trying to do - attempting to extrapolate some properties of finite quantities (such as the existence of a larger quantity) and erroneously assuming that an infinity must possess such property too.
Infinity is not a number, and, even though, sometimes it is possible to operate it as if it was, you need to remember that not everything you know about numerical quantities also holds for infinity. In particular, there is nothing greater than it.
You cannot add another nine if you've already got an infinite number of them. This, in fact, is a definition of the infinity. If you can increase the amount of anything (like nines after the decimal place) by adding another instance (or in fact any countable number of instances), that means that you did not have an infinite amount to begin with.
Yep, I'm already aware of this. I'm not assuming infinity possesses any real number quantity. That's ridiculous.
Of course you can't add another nine. I'm not saying that. I'm saying there is already another 9, forever. And since there is already a 9 there....forever...it will forever get closer to 1 and never be 1.
Just like Dividing a number in half. You can do that infinitely but you will never get to zero, correct? Otherwise you might as well divide .99999... by 0.
Edit: My argument is the same as the .999... is not a rational number crowd. 1/3 doesn't equal = .333333, it is .333..., so adding 1/3+1/3+1/3 to equal 1 is not the same as (.333...)(3). You can't express 1/3 as a rational decimal, .999... isn't a rational decimal, it can't equal 1.
Quote: YoDiceRoll11Yep, I'm already aware of this. I'm not assuming infinity possesses any real number quantity. That's ridiculous. Of course you can't add another nine. I'm not saying that. I'm saying there is already another 9, forever. And since there is already a 9 there....forever...it will forever get closer to 1 and never be 1.
It seems that you are confusing a "number" (as a mathematical quantity) with the process of writing it down in a particular representation. Yes, if you try to write down all those 9s one by one, you'll never get a number written down, that is equal to 1. But just because you can't write a number down in a particular representation does not mean it does not exist. PI has an infinite number of digits in decimal representation too, and so does 1/3.
Moreover there are some exotic base systems in which all rational numbers (including all integers) have an infinite number of digits. It does not of course mean that integers do not exist in those representations.
Quote:Just like Dividing a number in half. You can do that infinitely but you will never get to zero, correct?
You can say that repeatedly dividing a number in half and getting the result closer to zero is akin to the process of writing down nines one-by-one, and getting it closer and closer to 1, yes. But that is not what we are talking about.
Not the process of writing down a number, but the number itself.
Quote:My argument is the same as the .999... is not a rational number crowd. 1/3 doesn't equal = .333333, it is .333..., so adding 1/3+1/3+1/3 to equal 1 is not the same as (.333...)(3). You can't express 1/3 as a rational decimal, .999... isn't a rational decimal, it can't equal 1.
Yes, you can: 0.(3) or 0.3333.... or "0.3 with a bar" are all rational decimals, representing 1/3. They are not approximations (as you seem to think now), they are the exact and accurate representations of that exact number.
And yes, 1/3+1/3+1/3 is exactly the same as 0.333...*3.
Quote: YoDiceRoll11Edit: My argument is the same as the .999... is not a rational number crowd. 1/3 doesn't equal = .333333, it is .333..., so adding 1/3+1/3+1/3 to equal 1 is not the same as (.333...)(3). You can't express 1/3 as a rational decimal, .999... isn't a rational decimal, it can't equal 1.
You're contradicting yourself: you have
1: 1/3 = 0.333...
2: 1 != (0.333...)*3
Either both are true, or neither are true. All you're doing is multiplying both sides of the first equation by 3 to arrive at the second. If 1/3 = 0.333... then 1/3 * 3 = 0.333... * 3. Since we know 1/3 * 3 = 1, so does 0.333... * 3.
Quote: YoDiceRoll11And since there is already a 9 there....forever...it will forever get closer to 1 and never be 1..
I'm really enjoying this thread. Not to be a smart ass here, but how did 1 become 1? For mathematical purposes, .99999 is not getting any closer to 1 and is not getting any further away from 1. It's staying exactly at .999999 forever. For smart ass purposes 1 became 1 because everybody agreed that it was 1 and stopped counting to infinity.
Quote: mustangsally
I like this one.
1/3 = .333...
Multiply each side by 3.
since
fraction 1/9 = .111...
fraction 2/9 = .222...
we can see a pattern
can .999... be expressed as a fraction?
I said 9/9.
I love that.
Quote: MathExtremistSimilarly, 1.0 and 0.999... have the same value but different notations.
1.0 and 0.99999 don't have the same value. Unless previously agreed to for sanity purposes I assume LOL
Quote: WizardI would liken 0.999... as the probability of not hitting pi if you threw a dart between 0 and 10.
LOL I dig this thread. That's great.
Quote: weaselman
Moreover there are some exotic base systems in whi They are not approximations (as you seem to think now), they are the exact and accurate representations of that exact number
Incorrect assumption. I don't think they are approximations. I was merely summarizing the initial portion of the argument I agree with Please don't mistake that for the entire reasoning.
Quote: YoDiceRoll11You guys are free to believe what you want to. I Just disagree with the majority. Sue me.
There is two things you are allowed to disagree on without looking like an idiot:
1. Someone's opinion
2. Something that is incorrect...
The fact of that matter is that .999... (.9 repeating) and 1 are the same value. This has been proven with various methods.
Your argument is that at some point, there will always exist another 9. This is only true in a case where the number is not .999..., but instead .999......9, which is a finite number.
The difference is Σ (k=1 to n) 9/(10^k) and Σ(k=1 to ∞) 9/(10^k)
n is a real, finite number. Infinity is not...
All that aside, I guaruntee you'll never obtain a job where it is crucial that you don't round .9999...9 to 1, so what's the difference?
Quote: YoDiceRoll11You guys are free to believe what you want to.
I Just disagree with the majority.
Sue me.
Me Sally, you Sue
Are you a Boy named Sue?
Can you tell me where you disagree in the steps below? This is basically the same argument presented previously by ME. (I assume we now agree that the notation .xxxx... means that the x's continue for an infinite number of digits.)
(A) 1/3 = .3333... (Yes or no?)
(B) 3*(1/3) = 3*(.3333...) (Yes or no?)
(C) 3*(.3333...) = .9999... (Yes or no?)
(D) 3*(1/3) = 1 (Yes or no?)
(E) 1 = .9999... (Yes or no?)
I thought that you had agreed to (A), though maybe not, and you continue to disagree with (E). In my view, the lines in between just lead from what you agree with to what you disagree with, so I am curious as to where the disagreement arises.
Quote: TriplellThere is two things you are allowed to disagree on without looking like an idiot:
1. Someone's opinion
2. Something that is incorrect...
The fact of that matter is that .999... (.9 repeating) and 1 are the same value. This has been proven with various methods.
Your argument is that at some point, there will always exist another 9. This is only true in a case where the number is not .999..., but instead .999......9, which is a finite number.
The difference is Σ (k=1 to n) 9/(10^k) and Σ(k=1 to ∞) 9/(10^k)
n is a real, finite number. Infinity is not...
All that aside, I guaruntee you'll never obtain a job where it is crucial that you don't round .9999...9 to 1, so what's the difference?
While I'm up drinking tonight, there's tons and tons of things you can disagree with without looking like an idiot, ex. that was a good movie, it's too early to go to sleep, that light is too bright, etc. etc.
(.9 repeating) and 1 are not the same value. This has been proven with a tiny bit of logic.
n is the definition of an infinite number (until it is defined).
Infinity is a finite number (when it has been defined).
All that aside, I GUARANTEE that you WILL obtain a job where it is crucial that you round .9999...9 to 1, so that's the difference.
Quote: Doc... often when I find myself completely disagreeing with someone on a "simple" math problem, I develop the opinion that that person is either an idiot or is putting me on. ....
I have just been reminded that there are multiple examples in this thread, as well as in a thread about a series of wins/losses.
Quote: JyBrd0403While I'm up drinking tonight, there's tons and tons of things you can disagree with without looking like an idiot, ex. that was a good movie, it's too early to go to sleep, that light is too bright, etc. etc.
Those all fall under 1. Someones opinion.
Quote: JyBrd0403
(.9 repeating) and 1 are not the same value. This has been proven with a tiny bit of logic.
It has been proven that they are the same
Quote: JyBrd0403
n is the definition of an infinite number (until it is defined).
Infinity is a finite number (when it has been defined).
n is only an infinite number when it is defined as being infinity
infinity is not a finite number...ever
Quote: JyBrd0403
All that aside, I GUARANTEE that you WILL obtain a job where it is crucial that you round .9999...9 to 1, so that's the difference.
Care to elaborate?
Quote: TriplellThose all fall under 1. Someones opinion.
It has been proven that they are the same
n is only an infinite number when it is defined as being infinity
infinity is a transient number, never a finite number
Care to elaborate?
Someones opinions? Are those opinions Correct or Incorrect? That 1 they all fall under is a big damn 1. LOL
Someone has proven that .99999 and 1 are exactly the same? Who was this? And is he correct? (check my previous post, if you want to define the definition of infinity)
n is only an infinite number when it is defined as being infinity. Exactly, it has to be defined. n has an infinite number of possibilities, until it has been defined. There's an infinite number of ways of being wrong, but probably only one way of being right :) LOL
infinity is a transient number , never a finite number. Now, I'm drunk tonight, and I'm too tired to look up the definition of transient, but I'll guarantee you that if infinity is defined as a finite number, it would be a finite number. To clarify the point, I will now define infinity as being 10,000 trials. Infinity now equals 10,000 trials. Get the point. If not, forget math, try logic.
I would say it's true that E(n=1..oo) 9/10^n = 1.
But .999999999 < 1 as long as the number of 9s does not equal infinity.
Or...
Proof: 0.9999... = Sum 9/10^n
(n=1 -> Infinity)
= lim sum 9/10^n
(m -> Infinity) (n=1 -> m)
= lim .9(1-10^-(m+1))/(1-1/10)
(m -> Infinity)
= lim .9(1-10^-(m+1))/(9/10)
(m -> Infinity)
= .9/(9/10)
= 1
Look up Euler's elements of Algebra (1770):
.99999999999.... = 9(1/10) + 9(1/10)^2 + 9(1/10)^3.... = 9 (1/10) / (1 - 1/10) = 1
Or, algebraically, what is 1/3: .33333333333.....
What is 1/3 x 3? Is is .99999999999..... or 1.
Both are the same.
Quote: Doc
Can you tell me where you disagree in the steps below? This is basically the same argument presented previously by ME. (I assume we now agree that the notation .xxxx... means that the x's continue for an infinite number of digits.)
(A) 1/3 = .3333... (Yes or no?)
(B) 3*(1/3) = 3*(.3333...) (Yes or no?)
(C) 3*(.3333...) = .9999... (Yes or no?)
(D) 3*(1/3) = 1 (Yes or no?)
(E) 1 = .9999... (Yes or no?)
I thought that you had agreed to (A), though maybe not, and you continue to disagree with (E). In my view, the lines in between just lead from what you agree with to what you disagree with, so I am curious as to where the disagreement arises.
Thank you for your fair assessment. I will respond in kind, especially since you are uber cool on here.
A: No, it is my assumption, and I join with others that disagree with the entire problem as a whole, that 1/3 does not equal exactly .333..., Just as 3/3 does not equal .999..., because to quantitatively make this exact measurement that 1/3 = exactly .333... (infetesimal math aside), one would have to stop the infinite progression that the majority presents as .333..., and round up to .333333333......................................................4. I can clarify this if you want. Calculators don't show this because they are operating on a different set of
B: No. 3*(1/3)= basically 3*(.33334) rounded up for simplicity, see explanation in A above.
C: Yes of course it does.
D: Yes
E: No, See A.
In summation, .333.... will get EXTREMELY close to 1/3 as it approaches infinity, but will never reach it. Just like a standard exponential halving. And since 1/3 does not truly equal .333... this is where the disagreement lies.
I could re-write your above mini proof as:
Quote:1/9 = 0.111111......
2/9 = 0.222222......
.
.
.
8/9 = 0.888888......
9/9 = 0.999999......= 1
I disagree with this reasoning, and the one you, and unfortunately, the majority, posit.
This is just my opinion. I understand the disagreements. I even understand the sound math that is against it, really I do. Here I disagree on both a mathematical and philosophical stance. Look up the philosophy of math, wow, it is worse than math.
Quote: Triplell
The fact of that matter is that .999... (.9 repeating) and 1 are the same value. This has been proven with various methods.
And there are people that prove it not true, with various methods. But you don't see me calling anyone stupid.
Quote:
Your argument is that at some point, there will always exist another 9. This is only true in a case where the number is not .999..., but instead .999......9, which is a finite number.
Some could boil it down to that. But really it is that you cannot positively reach 1/3 with .333...
Quote:
n is a real, finite number. Infinity is not...
Thanks I had no clue.
Quote: JyBrd0403
(.9 repeating) and 1 are not the same value. This has been proven with a tiny bit of logic.
n is the definition of an infinite number (until it is defined).
Infinity is a finite number (when it has been defined).
This is part of the reasoning of the argument that I agree with.
Quote: Triplell
n is only an infinite number when it is defined as being infinity
infinity is not a finite number...ever
Care to elaborate?
Until you stop it to round up conveniently to make .333... equal exactly 1/3. That's my point, is that you can't make infinity part of the equation unless you quantitatively define it.
