08 Dec 04
Originally posted by GambitzoidI believe this was already brought to KINGSI's attention, and I'm sure he has modified his reasoning accordingly; no need to shout.
that was the most horrible logic i ever saw, you suck at math
.333333333333 = y (ok, fine)
then 3.33333333333 = 10y (still fine, okay)
and 3 = 3y WTF ARE YOU TALKING ABOUT, YOU JUST SAID y=.333...???
so 3y= 1!!!!! WHAT ARE YOU SMOKING????
Clearly .333333 = 1 ACTUALLY THEY ARENT EQUAL AT ALL!!!!
and yes .999999 repeating does equal 1
15 Dec 04
Originally posted by opsoccergurl11The number of 9s needed to adequately represent multiplying x by 10 is NOT the same as the number of 9s needed to adequately represent subtraction of x.
well, by algebra,
if .999999=x, then
9.999999=10x
and 9=9x after you subtract the two equations
when you divide by 9, you get 1=x
so, i just stated x=.99999999 and x=1 by just modifying the original equation
15 Dec 04
1 edit
While I'm at it, converging to 1 and equalling 1 are not, and never will be, the same thing. Limits are approached, but never reached.
I've seen at LEAST half a dozen "proofs" throwing around equals signs when they're not warranted. The limit of an infinite series may equal 1, or whatever number, but that is not the same as being able to say the series itself equals that number. God knows my maths lecturer jumped on anyone who didn't say "THE LIMIT OF... equals".
15 Dec 04
1 edit
Originally posted by Siskin[Ignore this one please, pressed the wrong button!] 😳
Theorem: Between any two rational numbers there are infinitely many more rational numbers.
Proof: Let a and b be different rational numbers, with a<b. Then (a+b)/2 is also a rational number, and between a and b.
Suppose that (strictly) ...[text shortened]... aren't any rational numbers between them, they must be the same.
15 Dec 04
Originally posted by orfeoWhat are you doing talking about a number of 9's? The row of 9's is INFINITE, wich means that there are more 9's then any number you can think of. Infinity is NOT a number!
The number of 9s needed to adequately represent multiplying x by 10 is NOT the same as the number of 9s needed to adequately represent subtraction of x.
Multiplying x = 0.999... by 10 changes NOTHING about the row of 9's behind the decimal point. There still is an INFINITE row of 9's there.
So stating 10x-x = 9 is absolutely true!
x = lim 1 - 1/10^n for n->infinity THIS STATEMENT IS TRUE!
lim 1- 1/10^n = 1- lim 1/10^n for n->infinity THIS STATEMENT IS TRUE!
lim 1/10^n = 0 for n->infinity THIS STATEMENT IS TRUE!
Putting everything in a row we get x=1
15 Dec 04
Originally posted by GambitzoidHOORAY!
here what i think is the absolute final proof...
,99... can be rewritten as:
[9/(10^1) + 9/(10^2) + 9/(10^3) + ... + 9/(10^n-1) + 9/10^n]
where n APPROACHES infinity.
The LIMIT of .999... as n approaches infinity is 1.
HOWEVER!!! n never can actually reach infinity because a number n+1 can always be found... so while .999 becomes infinitesimally close to 1, it never can actually equal 1.
15 Dec 04
1 edit
Originally posted by TheMaster37Yes, because you defined x as the limit. 0.99999 recurring isn't the limit. 1 is. 0.9999 recurring is the ACTUAL SUM.
What are you doing talking about a number of 9's? The row of 9's is INFINITE, wich means that there are more 9's then any number you can think of. Infinity is NOT a number!
Multiplying x = 0.999... by 10 changes NOTHING about the ...[text shortened]... THIS STATEMENT IS TRUE!
Putting everything in a row we get x=1
15 Dec 04
let me refrase it then:
0.999... = lim 1 - 1/10^n for n->infinity what is it that you don't like about this? This is a simple and true statement.
If you still do not believe that 0.999... = 1 I suggest you read several of my previous post in this thread. I've given a simple proof, only requiring the rules of calculation in Standard Analasis.
15 Dec 04
Originally posted by TheMaster37No it isn't. Replace the lim with a sum (sigma).
let me refrase it then:
0.999... = lim 1 - 1/10^n for n->infinity what is it that you don't like about this? This is a simple and true statement.
If you still do not believe that 0.999... = 1 I suggest you read several of my previous post in this thread. I've given a simple proof, only requiring the rules of calculation in Standard Analasis.
Some of your other posts are interesting and I will have to think about them, but there are plenty of equations on here indicating that 0.999... is the SUM of an infinite series, and it's quite easy to understand when you start adding terms - 0.9 + 0.09 + 0.009 etc.. But you cannot tell me, no matter how hard you try, that is the same as the LIMIT of the series. Series approach but DO NOT REACH their limits.
Other issue: My very post was wrong in its expression, I agree. You're right, multiplying an infinite number of 9s after the decimal point won't affect how many 9s there are. But at the same time, how do you justify subtracting infinite 9s from infinite 9s and ending up with nothing after the decimal point?
In other words I'm not convinced that infinity minus infinity equals zero, why doesn't it equal infinity?
16 Dec 04
Originally posted by orfeoI'm confused - are you saying that there's something deeply wrong with calling 0.999... a number? If you'd rather not represent real numbers by rational Cauchy sequences, which 0.999... is an example of, that's fair enough, but how would you rather represent them? Real numbers are a construct like any other, so we can pick any representation we like that has the required algebraic and analytical properties. Here's a Dedekind cut version:
Yes, because you defined x as the limit. 0.99999 recurring isn't the limit. 1 is. 0.9999 recurring is the ACTUAL SUM.
Let X = {x in Q: x < 1 - 10^-n for some n}
Let Y = {x in Q: x < 1}
Then 'the limit of' 0.999... is (X,Q\X) and 1 is (Y,Q\Y). Now our question 'does 0.999... = 1?' becomes 'does X = Y?' It really doesn't make a lot of difference, though.
16 Dec 04
Originally posted by orfeoactually, there's no hooray because i made a mistake... orfeo is wrong
HOORAY!
i said that .999.... can be represented as
[9/(10^1) + 9/(10^2) + 9/(10^3) + ... + 9/(10^n-1) + 9/10^n]
^ ^ ^
^ ^ ^
^ ^ ^
^ ^ ^
| | |
THAT IS WRONG
because there is another +9/10^(n+1) and then add 9/10^(n+2) ad infinitum. .99999... is not the funtion i described above but the limit of the function above as n->inf which is 1.
.9999...=1
Look in any textbook!!!!
17 Dec 04
Originally posted by THUDandBLUNDERwhat is X*10?
If it is to be proven that 0.9999... = 1, don't you think it is rather begging the question to blithely assume that
0.99999... = x implies 9.99999... = 10x?
10X, wow; just amazing!
now, what is .999999999... * 10?
thats right, 9.999999....
so where is the assumption?