What is .99 repeating as a fraction?
Quote: weaselmanIf you believe that Pi is a "real" number,
Easy boss, when did I say that Pi was a real number???
Quote: YoDiceRoll11Easy boss, when did I say that Pi was a real number???
Ah, my apologies. So then, you do not believe in Pi after all?
Quote: YoDiceRoll11Easy boss, when did I say that Pi was a real number???
That pi is in R follows from the definition of R. On what basis do you reject the definition of R?
Quote: MoscaIn base pi, pi is 1. And all the other numbers go on forever.
"In Communist Russia..." ah never mind.
Quote: AcesAndEights"In Communist Russia..." ah never mind.
And in 1890s Indiana, pi is 3.2...
Quote: JyBrd0403I completely disagree with you here. .... The proof fails for ex. 2/2. Which proves that .999... does not equal 1 for all fractions. It still blows my mind that it .999... does equal 1 for certain fractions.
".999... does not equal 1 for all fractions". I can't even begin to fathom what that means, and as a result I can't possibly refute it. Nice work, sir!
a != b for all values of x.
Quote: YoDiceRoll11I might have misspoken. I meant that it is irrational rather than non real. Need more sleep.
So, you do know that most real numbers take an infinite number of digits to be represented in base-10 notation. What is it that's so special about 0.999... to you, or 0.333 ... for that matter, that you deny them the right to be multiplied by an integer?
Quote: weaselmanSo, you do know that most real numbers take an infinite number of digits to be represented in base-10 notation. What is it that's so special about 0.999... to you, or 0.333 ... for that matter, that you deny them the right to be multiplied by an integer?
In fact, it's "almost all" real numbers that have an infinite (non-terminating) representation. It's also "almost all" rational numbers, and almost all reals are irrational, so it's very few numbers which have a terminating decimal representation. In fact, the only numbers with a terminating decimal representation are the set of numbers
x / 2^a*5^b for all integers x, a, b. The rational number 1/3 doesn't fit that formula, so it has a non-terminating but repeating decimal representation 0.333... But 0.333... still equals 1/3, and multiplying both sides by 3 still works.
Quote: MathExtremistIn fact, it's "almost all" real numbers
Yes. More over, almost all real number cannot actually be notated or otherwise described in any way whatsoever at all :)
Quote: MathExtremistIt's also "almost all" rational numbers,
This I don't agree with. Rationals are countable (aleph-null), so, for terminating decimals to be a null-subset, there would have to be a finite number of them, which is obviously not so. I would say, there are "equal amounts" of terminating and non-terminating decimals among rational numbers (both subsets are aleph-null)
Rational numbers x / 2^a * 5^b are obviously a subset of
rational numbers x / 2^a * 3^b * 5^c * 7^d * 11^e ... ad infinitum for all prime factors
but only the former have terminating decimal representations. Is it nevertheless proper to say those sets are equal in cardinality because both are countable?
Quote: MathExtremistI thought "almost all" meant "all but a countably-infinite subset".
If that was true, you could argue that "almost all integers are less than 10" :)
Quote:Does it require that the larger set be uncountable?
"Almost all" means all but, perhaps, a zero-measured subset.
Now, a zero-measured subset can be defined more or less formally, depending on the context, but, in simple terms, for infinite sets, you can think of it as a subset of a smaller cardinality.
So, when talking about a continuum, "almost all" means "all except, maybe, finite or countable number of elements", as you said, but in application to countable sets, "almost all" means "all, but, maybe, a finite number".
Quote:
Rational numbers x / 2^a * 5^b are obviously a subset of
rational numbers x / 2^a * 3^b * 5^c * 7^d * 11^e ... ad infinitum for all prime factors
but only the former have terminating decimal representations.
Is it nevertheless proper to say those sets are equal in cardinality because both are countable?
Yes. Like all countable sets, they have the cardinality Aleph-null.
BTW, having a (proper) subset of equal cardinality is the definition of an infinite set.
Quote: MathExtremist".999... does not equal 1 for all fractions". I can't even begin to fathom what that means, and as a result I can't possibly refute it. Nice work, sir!
a != b for all values of x.
I'll take what I can get.
Quote: MathExtremist"
a != b for all values of x."
You got it! Now try it with x = unknown = 2. Please don't respond to this post. I'll go with my previous post, I'll take what I can get.
Drinking again tonight.
The natural numbers are a countable infinite set, which are a subset of the real numbers, an uncountable infinite set. The difference of which is directly related to this question.
With natural numbers I can pick some number k1, and another number k2, in which there is exactly one natural number between them, or similarly no natural number.
However, I can pick some real number r1 and another number r2. Because real numbers are uncountably infinite, I can find another number r3 between r1 and r2. Then I can find another, and another, and through induction I can ALWAYS find another number between r1 and the last r I picked. Thus I can always find a real number between 1 and .999... and thus 1 does not equal .999 repeating.
Quote: ewjones080
The natural numbers are a countable infinite set, which are a subset of the real numbers, an uncountable infinite set. The difference of which is directly related to this question.
Not really.
Quote:
With natural numbers I can pick some number k1, and another number k2, in which there is exactly one natural number between them, or similarly no natural number.
This is true for natural numbers, but not for rational numbers, which is a countable set too. Thus, the property you mentioned is simply a curiosity specific to natural numbers, not a fundamental property of countable sets.
Quote:
However, I can pick some real number r1 and another number r2. Because real numbers are uncountably infinite, I can find another number r3 between r1 and r2. Then I can find another, and another,
No, not because they are uncountably infinite. A set with such property is properly called a "dense set". Both real and rational numbers are dense sets, while latter is countable, the former is not.
Quote:and through induction I can ALWAYS find another number between r1 and the last r I picked.
While entirely irrelevant here, the question of using induction on uncountable sets is actually a very non-trivial one. Within the boundaries of classical calculus you cannot do that at all.
Quote:Thus I can always find a real number between 1 and .999... and thus 1 does not equal .999 repeating.
This is not true, and does not follow from the above chain of reasoning. In the above, your r1 and r2 were different numbers to begin with, but 0.999... and 1 are the same number, and therefore, there is nothing in between.
A fundamental property of real numbers is that you can subtract one from another and get another finite real number, which is the difference between the two. Subtracting 0.999... from 1 yields zero, which means that the two numbers are actually one.
Quote: ewjones080However, I can pick some real number r1 and another number r2. Because real numbers are uncountably infinite, I can find another number r3 between r1 and r2. Then I can find another, and another, and through induction I can ALWAYS find another number between r1 and the last r I picked. Thus I can always find a real number between 1 and .999... and thus 1 does not equal .999 repeating.
That's circular. You're starting with the premises that r1 = 1 and r2 = 0.999... and r1 != r2, and concluding that r1 != r2.
___________
1
.99999..........
No, you can't "round off" .999... because it is equal to 1. There is nothing to round off.Quote: JyBrd0403Can't you just simply say that .999... rounded off is 1? You can't round off 1, at least I don't think you can, but you can round off .999..., so they are not the same number. If you round .666... off you get .667 correct? So .667 = .666...?
Quote: s2dbakerNo, you can't "round off" .999... because it is equal to 1. There is nothing to round off.
Somehow, I knew that was going to get argued with here.
Are you sure you can't round off .999...? Can you round off 1? Come on, if you can round off .666... you can round off .999... right? I dare say you have to round off .999... the number never ends.
Quote: JyBrd0403Somehow, I knew that was going to get argued with here.
Are you sure you can't round off .999...? Can you round off 1? Come on, if you can round off .666... you can round off .999... right? I dare say you have to round off .999... the number never ends.
Do you know what "rounding" is? Rounding is a process of approximation and truncation. 1 is not an approximation of 0.999..., so you can't round 0.999... to 1. It is simply a shorter representation. You can round 0.666... to 0.667 because 0.667 is an approximation and truncation of 0.666...
Suppose you did want to "round off" .999... to 1. How much do you think you would round off to make them exactly equal?
Quote: DocJyBrd:
Suppose you did want to "round off" .999... to 1. How much do you think you would round off to make them exactly equal?
A number > 0, doc. Wouldn't waste my time figuring out what that number is, at least not for this discussion. It's a number greater then zero, which means .999... is not equal to 1, for the purposes of this discussion that would be sufficient.
Quote: JyBrd0403Somehow, I knew that was going to get argued with here.
Are you sure you can't round off .999...? Can you round off 1? Come on, if you can round off .666... you can round off .999... right? I dare say you have to round off .999... the number never ends.
You can round '1'.
1.00... rounded down is 1.
Throw 1.00 into the round function of excel if you don't believe me.
But that is hardly a proof of why .999... == 1.
How about this one (no one has tried it)...
11/12 = 0.9166...
(1/12) = 0.0833...
1100/12 = 91.666...
100/12 = 8.333...
what is (1100+100)/(12)?
I'll give you a clue it's equivalent to (91+8) + (.666... + .333...)
And that's where everyone else disagrees with you. If you don't think this number is exactly equal to zero as the others do, then do you have some means of representing its non-zero value?Quote: JyBrd0403A number > 0, doc.
Have any of you read this?
Seems like this thread is an imcredible waste of time if you haven't
Quote: MathExtremistDo you know what "rounding" is? Rounding is a process of approximation and truncation. 1 is not an approximation of 0.999..., so you can't round 0.999... to 1. It is simply a shorter representation. You can round 0.666... to 0.667 because 0.667 is an approximation and truncation of 0.666...
So, does .667 = .666... ? You can argue this all day long, but you rounded the dame number off. And, just by the way, you're really bad at this, you're just making a jackass out of yourself. You didn't round off .999...? Some guy is now saying he rounded off 1.000 to 1. Would you mind straightening him out for me, he's going off the deep end.
Take .1999...
.1999... * 10 = 1.999...
1.999... = 1 + (1/3) + (1/3) +(1/3)
Divide both sides by 10.
.1999... = (1/10) + (1/30) + (1/30) + (1/30)
.1999... = (1/10) + (1/10)
.1999... = .2
ps. Try to represent .1999... as a fraction
Quote: JyBrd0403And, just by the way, you're really bad at this, you're just making a jackass out of yourself.
I'm glad you think so. It will save me some time in the future.
Quote: MathExtremistI'm glad you think so. It will save me some time in the future.
Yeah, but do you want to straighten that guy out for me? He's saying 1.00 can be rounded down to 1. The guy really needs your help.
Quote: JyBrd0403Yeah, but do you want to straighten that guy out for me? He's saying 1.00 can be rounded down to 1. The guy really needs your help.
Sorry, I meant 1.000... is rounded down to 1. You do realize that 1.000... = 1, don't you? If you didn't realize, I'm making a mockery of you. Technically, all numbers could be represented as repeating decimals. 1 = 1.000..., 2 = 2.000..., (3/2) = 1.5000...
Get my point.
You don't "round" .999... to 1, because it is 1. Same way, you don't round 1.000... to 1, because it is 1.
PS: Maybe you should straighten me out yourself.
Quote: TriplellSorry, I meant 1.000... is rounded down to 1. You do realize that 1.000... = 1, don't you? If you didn't realize, I'm making a mockery of you. Technically, all numbers could be represented as repeating decimals. 1 = 1.000..., 2 = 2.000..., (3/2) = 1.5000...
Get my point.
You don't "round" .999... to 1, because it is 1. Same way, you don't round 1.000... to 1, because it is 1.
PS: Maybe you should straighten me out yourself.
No offense intended. I wasn't talking to you, I was talking to MathE. You want to straighten him out for me, MathE?
Quote: JyBrd0403No offense intended. I wasn't talking to you, I was talking to MathE. You want to straighten him out for me, MathE?
You weren't talking to me, but you were referencing me. So I clarified my intentions, which you missed.
Furthermore, you are the one who brought rounding into this entire discussion.
When you round .666... to .667, you are changing the value of the number (losing precision).
.666... = (2/3)
.667 = 667/1000.
Completely different numbers(albeit, they are close).
.999... = 1
1 = 1. The same number. Which brings me back to my point, you don't round .999... to 1, because it is 1.
Quote: MichaelBluejayWhat gets me about this discussion is how some people, with minimal training, are confident enough to insist that they're right and all the world's experts are wrong. That just seems a little...arrogant.
I agree. LOL I'll bet you a billion dollars 2 + 2 = 4. Take me up on the bet?
Quote: Triplell
.666... = (2/3)
.667 = 667/1000.
Completely different numbers(albeit, they are close).
Maybe close was not the right number...
All apologies. Drunk again tonight.
It's extremely close.
Quote: TriplellMaybe close was not the right number...
Probably not the right number :)
Quote: TriplellSorry, I meant 1.000... is rounded down to 1.
You just proved my point. The number 1.000... can be rounded down, while 1 itself cannot be.
See, you just can't speak off the top of your head if you're going to be a good B.S.er. See a good B.S.er takes his time and comes up with some B.S. that makes you stop and think for a while. If he just spoke from the top of his head, he'd run into these kinds of problems all the time. You've got to really plan your B.S. out all the way or it will crumble around you. Of course, I can get a 5 year old to tell you eventually the B.S. will get exposed.
See, if you had thought that out, you would have said 1.000... can't be rounded down to 1 because it is the same number. As a matter of fact, that's the B.S. that MathE was running earlier. Whatever you do, don't speak freely and don't make sense. Two big problems for B.S.ers.
Quote: JyBrd0403You just proved my point. The number 1.000... can be rounded down, while 1 itself cannot be.
So, now you are arguing that 1.000... isn't 1 either? :)
How about just 1.0?
Quote:See, you just can't speak off the top of your head if you're going to be a good B.S.er. See a good B.S.er takes his time and comes up with some B.S.
Something tells me that you are a good authority in this area ...
Quote: weaselmanSo, now you are arguing that 1.000... isn't 1 either? :)
How about just 1.0?
You'll have to take that up with triplell. That was his point. But, yes I do agree with it.
Something tells me that you are a good authority in this area ...
I can recognize the species better than most, I assume.
Quote: JyBrd0403You'll have to take that up with triplell. That was his point.
No, it wasn't. He just misspoke about rounding, that's all.
Quote:But, yes I do agree with it.
You "agree" with what? That 1.0 is not equal to 1?
I guess, that's the end of the discussion then. I must admit that you have a very original and unorthodox view of elementary math.
Quote:I can recognize the species better than most, I assume.
That's not what I meant. I was not talking about your species recognition skills at all.
