Re: [Perm Unresolved] 0.999... = 1?
Have you heard of the Peano Axioms? The 5th is PMI = Principle of Mathematical Induction. These Axioms define the natural number set, which does not contain infinity (as it isn't actually a number). Therefore, you can't use PMI as it was above to prove anything apart from things with natural numbers.
There are other versions of Induction, like backwards (n => n-1 is true) and negatives (n => -n is true) which allows proofs for all integers, as well as one for rationals (n => n+1 and 1/n are true), but nothing that allows you to prove for infinites.
Re: [Perm Unresolved] 0.999... = 1?
Re: [Perm Unresolved] 0.999... = 1?
Quote:
Originally Posted by zaza
How am I wrong?
zaza
I don't know
Re: [Perm Unresolved] 0.999... = 1?
sorry if i misinterpret you, but the actual value (x-1)/(x^2-1) ISN'T 0.5 for x = 1...
that's like saying (a-b)(a+b) = (a-b) for a=b=1, therefore (a-b)(a+b)/(a-b) = (a-b)/(a-b) and after cancelling: a+b = 1, so 1+1 = 1.
It just doesn't work like that, you can't cancel 0.
Alternatively, are you saying that because 5 * 0 = 7 * 0, then 5 = 7?
Re: [Perm Unresolved] 0.999... = 1?
Quote:
Originally Posted by sql_lall
sorry if i misinterpret you, but the actual value (x-1)/(x^2-1) ISN'T 0.5 for x = 1...
Well, it just turns out IT IS 0.5 and you can calculate it by application of the "L'Hopital rule". I just can't remember how it is exactly formulated in calculus but it sure applies to this sample function I've chosen.
If you define f(x) = x -1 and g(x) = x2 -1 then the (finite) limit of f/g as x approaches 1 is the same as that of f'/g', the ratio of the derivatives. Then:
lim (x->1) (x - 1) / (x2 -1) = lim (x->1) 1 / 2x = 0.5
Re: [Perm Unresolved] 0.999... = 1?
Hi,
Can we please stop this stuff about limits? Limits apply to Functions and Series. They do not apply to numbers. The number we are talking about here is 0.9 recurring. This is also known as 1. It does not "tend" to anything, because there is no variable x to do the tending.
If, on the other hand, we consider the series 0.9 + 0.09 + 0.009 + ..., then we can consider the limit. When you sum this GP to infinity, you get 1. I showed this earlier.
As far as the "being wrong" bit goes, that was with regard to yrwyddfa's comment. Regarding the "infinity <> undefined comment", I also agree - as a physicist I tend to get lazy and use "infinity" as a generic word to mean poles, singularities and all sorts of other stuff, particularly when not obviously conversing with mathematicians. Sorry for the confusion.
But, I would point out that the Limit of y = (x-1)/(x^2-1) as x->1 is not the same as evaluating it at x=1, hence the confusion caused by saying "though the actual value is 0.5" is unnecessary. L'Hopital's rule is used when you can't obviously determine the limit in the first instance, and hence is applicable here. However, it still applies to the limit, not to the evaluation.
I think there has been a lot of confusion over this topic arising from improper use of language. I am guilty myself (see above). But I hope there is mostly understanding now.
Cheers
zaza
Re: [Perm Unresolved] 0.999... = 1?
Quote:
Originally Posted by zaza
Hi,
Can we please stop this stuff about limits?
Well, after you settled the question in one of the earliest posts by substracting 0.999... from 9.999... the rest of the thread has just been beating around the bush. I mean, no point in further discussing, 0.999... = 1 and 2+2 is 4, and that's all there is to that.
Re: [Perm Unresolved] 0.999... = 1?
If 0.9r=1 then why do we have two symbols for it. I could understand it if one said 0.9r approximated 1, or 0.9r is so close to 1 it might as well be 1.
But why have multiple homogenous mathematical symbols for it? Why would anyone every use 0.9r when they can use 1 - if they are equal they are interchangeable. Why does 0.9r exist?
What fraction produces 0.9r and can you use simply algebra to show that that fraction is equal to 1?
I have read the proof that shows that 0.9r=1, and understand how it works, it just doesn't make logical sense even if it makes perfect mathematical sense.
Re: [Perm Unresolved] 0.999... = 1?
Quote:
Originally Posted by yrwyddfa
If 0.9r=1 then why do we have two symbols for it. I could understand it if one said 0.9r approximated 1, or 0.9r is so close to 1 it might as well be 1.
It just turns out the decimal system provides more than one representation for the number 1 -and many other numbers. Try not to think of 0.9r as a number in its own i.e. different from 1.
Quote:
Originally Posted by yrwyddfa
Why would anyone every use 0.9r when they can use 1 - if they are equal they are interchangeable. Why does 0.9r exist?
Of course no one would use 0.9r instead of 1. Why would anyone, as you say.
0.9r is just a representation. Likewise, a sequence like 0.9, 0.99, 0.999, 0.9999,... is a representation of the number one, altogether different because it doesn't use a single number for it. Or, the infinite sum 1/2 + 1/22 + 1/23 + 1/24 + ... is another representation of 1.
0.9r is a representation by a single number yet with infinite digits (like pi has infinite digits).
Quote:
Originally Posted by yrwyddfa
What fraction produces 0.9r and can you use simply algebra to show that that fraction is equal to 1?
No fraction "produces" 0.9r, all you can say is that any fraction of the form A/B with A=B is equal to 0.9r just because 0.9r is equal to 1. If A were not equal to B in that fraction then, using zaza's algebra:
Let 0.9r = A/B, with A<>B
10 * 0.9r = 9.9r = 10A/B
Substracting: 9 = 10A/B - A/B = 9A/B -> 1 = A/B -> A=B in contradiction with the initial assumption.
Quote:
Originally Posted by yrwyddfa
I have read the proof that shows that 0.9r=1, and understand how it works, it just doesn't make logical sense even if it makes perfect mathematical sense.
If it makes mathematical sense then it makes logical sense to me!
Re: [Perm Unresolved] 0.999... = 1?
Still confused: :confused:
1/1, 2/2, 3/3 etc etc are all equal to one.
If the assertion 0.9r=1 that must mean that 1/1 = 0.9r, or 3/3=0.9r ? Is this what you guys are saying?
Re: [Perm Unresolved] 0.999... = 1?
Quote:
Originally Posted by krtxmrtz
Well, it just turns out IT IS 0.5 and you can calculate it by application of the "L'Hopital rule"...lim (x->1) (x - 1) / (x2 -1) = lim (x->1) 1 / 2x = 0.5
As you say, lim (x->1) (x - 1) / (x2 -1) = lim (x->1) 1 / 2x = 0.5
This doesn't mean that the value AT x=1 is 0.5 though.
Check here for an example of the rule. The java applet on that page clearly shows that at the exact value where the fraction = 0/0, the value is indeterminite! (Although, they do later say 9/0 = inf, which is a bit worrying...)
Re: [Perm Unresolved] 0.999... = 1?
Quote:
Originally Posted by yrwyddfa
Still confused: :confused:
1/1, 2/2, 3/3 etc etc are all equal to one.
If the assertion 0.9r=1 that must mean that 1/1 = 0.9r, or 3/3=0.9r ? Is this what you guys are saying?
Of course, since 0.9r = 1 then it's equal to any fraction like A/A for all values of A as well.
Re: [Perm Unresolved] 0.999... = 1?
Quote:
Originally Posted by sql_lall
As you say, lim (x->1) (x - 1) / (x
2 -1) = lim (x->1) 1 / 2x = 0.5
This doesn't mean that the value AT x=1 is 0.5 though.
Check
here for an example of the rule. The java applet on that page clearly shows that at the exact value where the fraction = 0/0, the value is indeterminite! (Although, they do later say 9/0 = inf, which is a bit worrying...)
The value is 0.5 because the function is continuous at that point!
What the applet shows is precisely that the f/g ratio can be calculated (at the limit) as the ratio of the two coloured vertical segments. Since the horizontal segment is the same for both f and g this is equivalent (in the limit) to the ratio of tangents and therefore of derivatives. And these two are well defined.
Btw, why should 9/0 = infinity be worrying at all???
Re: [Perm Unresolved] 0.999... = 1?
I sure hope you are being sarcastic...
Continuity at a point requires a function to be defined at the point...you can't say "The value is 0.5 because the function is continuous at that point!", this is obviously a circular argument...
I suggest you read here first (rules for continuity, plus example 1 in particular).
And 9/0 != inf.
Infinity is not a number...something/0 is undefined, definitely not equal to infinity. (just think, if it was, what would -9/0 be...negative infinity? what would 0/0 be then??? zero-infinity???)
Re: [Perm Unresolved] 0.999... = 1?
Quote:
Originally Posted by krtxmrtz
Of course, since 0.9r = 1 then it's equal to any fraction like A/A for all values of A as well.
Therefore the fraction a/a=0.9r=1, which means there is a fraction that completely describes 0.9r . . . . ? :confused:
Re: [Perm Unresolved] 0.999... = 1?
Quote:
Originally Posted by sql_lall
I sure hope you are being sarcastic...
Continuity at a point requires a function to be defined at the point...you can't say "The value is 0.5 because the function is continuous at that point!", this is obviously a circular argument...
I suggest you read
here first (rules for continuity, plus example 1 in particular).
Yes, I was wrong, this is what happens when you talk by heart after so many years without using the theory! My apologies. :blush: So the function value is not defined after all, I had simply forgotten about this specific type of (single-point) discontinuity.
Now I just can't remember whether this was relevant or not for the initial question of this thread... ???
Quote:
Originally Posted by sql_lall
And 9/0 != inf.
Infinity is not a number...something/0 is undefined, definitely not equal to infinity. (just think, if it was, what would -9/0 be...negative infinity? what would 0/0 be then??? zero-infinity???)
Whether infinity is or isn't a number I don't know for I'm not sure what the exact definition of number is in math. But 9/0 is infinity whatever infinity may be, and -9/0 is negative infinity, not undefined. I think we don't mean the same thing when we refer to "undefined". As for 0/0 THIS IS undefined as it can be equal to any number you like, i.e. 0/0 = x holds for any x. Here the equal sign must be interpreted meaning "compatible" in the sense that 0 = 0*x is true.
Re: [Perm Unresolved] 0.999... = 1?
Quote:
Originally Posted by yrwyddfa
Therefore the fraction a/a=0.9r=1, which means there is a fraction that completely describes 0.9r . . . . ? :confused:
What you won't find is a fraction A/B that produces 0.9r as you perform the division and keep finding more decimals for the quotient. The only fractions equal to 0.9r are those with the same numerator and denominator. If you like you can write 0.9r = 5/5 = 1 because 1 is just another way to write 0.9r.
Re: [Perm Unresolved] 0.999... = 1?
Quote:
Originally Posted by sql_lall
I suggest you read
here first (rules for continuity, plus example 1 in particular).
From what I've read on this site I see the function I chose has a "removable discontinuity".
Re: [solved?] 0.999... = 1? Yes.
Hi. I have been reading all the answers to this question. Let me try to convince ou that 0.999...=1
btw, the proof that you gave about the equality (excep the one involving division by 0) were right, but also the concerns of some sceptical people were with founded reason.
(i hold a phd in mathematics and i teach maths at the university. i have heard this question several times from my students as well).
Let's start fom the beginning of the story. you all know what are the natural numbers and the rational numbers (or fractions).
you can express a rational number in its fraction form (1/3) or in its decimal form (0.33333).
we still do not know what a real number is, and we DEFINE the real numbers as follow.
let a1, a2, a3, ... a sequence of rational numbers that converges in the Cauchy sense (that is, am-an ->0 when n, m goes to infinity) . then the LIMIT of this sequence, that we will cal A, belongs to the REAL LINE.
If the number A is not a rational number, it is called an irrational number.
For instance, this is the (ONLY) way to define numbers like sqr(2), pi, or the number e.
know, How to know if two REAL numbers are equal.
by definition, A=B if there exists a sequence a1, a2, a3 such that the limits of both sequences (again in the Cauchy sense) is at the same time A and B.
Now, let's construct the following convergente sequence
a1=0.9
a2=0.99
a3=0.999
the limit of this sequence is by definition, the number 0.99999...
but you can also see that 1 is the limit becase:
lim(1-an)=1-lim(an) when n -> infinity.
and 1-an=0.000..(n zeros)..1
lim(1-an)=0 (in the cauchy sense)
then 1-lin(an)=0, but lim(an)=0.9999999 therefore
1=0.999999999.... qed.
Remark 1. This is tru because we defined the real numbers the way we did, but if we define the real numbers in a different way (see the hatcher), then we will get different conclusion (the hatcher are NOT the real numbers, but other thing, related with nonstandard analysis, different axioms, different rules.
Remark 2. Some of you said that a serie or sequence or a limit is not a number, and the you were not accepting the right proofs. That's wrong. As you see, a real (irrational) number is exactly that, purely and simply a limit), but we are so used to them that it's easy to forget about this detail.
I hope that this clarify the question.
Victor.
Re: [Perm Unresolved] 0.999... = 1?
Quote:
Originally Posted by yrwyddfa
If 0.9r=1 then why do we have two symbols for it. I could understand it if one said 0.9r approximated 1, or 0.9r is so close to 1 it might as well be 1.
But why have multiple homogenous mathematical symbols for it? Why would anyone every use 0.9r when they can use 1 - if they are equal they are interchangeable. Why does 0.9r exist?
I believe thatits the other way around.
The assertion that .999... = 1 exists so that any series of equations, proofs, theorems, that end up creating the .999... series can then cite the rule .999...=1, and then use 1 from then on for .999...
Re: [Perm Unresolved] 0.999... = 1?
I have been thinking about this a lot the last few days and here is what I have come up with (And a little research uncovered a lot of insight too):
http://www.allmathematics.spiralmind...99EqualTo1.png
0.9999... = 1
Thus:
x = 0.9999...
10x = 9.9999...
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1.
That any good?
Cheers,
RyanJ
Re: [Perm Unresolved] 0.999... = 1?
I like the first one, but some people have problems with the second method.
They question whether 10x = 9.9999... is true since it isn't clear (unless it's been defined somewhere) that a new 9 is added to the end of the number when the multiplication is made.
So I don't think you can't prove this without using an infinite series as you did in the first method
