Gambitzoid, I will not waste bandwidth and pollute cyberspace by quoting your childishly naive reply to my previous post, which was obviously over your young head. Instead, I will post a suitably simplistic reply:
1) We cannot assume that subtraction applies to infinite decimals in the same way that we know it applies to finite decimals (Hence the original question of whether 0.9999999... = 1).
2) Point 1) above implies that we cannot assume that addition applies to infinite decimals in the same the same way that we know it applies to finite decimals.
3) Given that multiplication is no more than repeated addition, we cannot therefore assume that multipication applies to infinite decimals in the same the same way that we know it applies to finite decimals.
4) Therefore the said 'proof' is assuming something that it is trying to prove.
5) Therefore the 'proof' is bogus.
Which of the above steps do you disagree with, and why?
Originally posted by THUDandBLUNDERI disagree with the very first, but more specifically with the third. I mean, if multiplication is nonsense with an infinite number of decimals, how could you ever give meaning to things like 2*pi?
I will not waste bandwidth by quotng your childishly naive reply to my previous post, which was obviously over your young head.
Instead, I will post a suitably simplistic reply:
1) We cannot assume that subtraction applies to infinite decimals in the same way that we know it applies to finite decimals (Hence the original question of whether 0.9999 ...[text shortened]... ) Therefore the 'proof' is bogus.
Which of the above steps do you disagree with, and why?
So, all proofs are correct. 🙂
Originally posted by pidermanYou disagree with the first?
I disagree with the very first, but more specifically with the third. I mean, if multiplication is nonsense with an infinite number of decimals, how could you ever give meaning to things like 2*pi?
So, all proofs are correct. 🙂
Yet you fail to explain in what way you disagree. 🙄
You disagree with the third?
Multiplication [of infinite decimals] is nonsense?
I made no such claim.
If you cannot understand plain English then any further discussion is pointless.
adding and substracting with infinitely long numbers IS possible, only when you are very carefull.
9.999... - 0.999... is clearly 9, as for every 9 behind the decimalpoint of 9.999... gets removed by a 9 from 0.999...
This is in NO way similar to substracting infinity from infinity, wich is not defined. Infinity would slightly 'resemble' a number with infinite digits, non-zero, BEFORE the decimal point. Since this similarity has no value whatsoever, I will say nothing more about it.
Ok...sir Orfeo. You said yourself that 0.9 + 0.09 + 0.009 +... = 0.999... =X
You know that such a sum CAN be rewritten as a limit:
lim (1-1/10) + (1/10-1/100) + ... + (1/10^n-1 - 1/10^n) for n->inf
Now i'm going to do something wich is true for some reason (absolute convergence/whatever, i'm REALLY not in the mood to find that in my analasis for beginners book).
We change the order of the bracets. You see that only 1 - 1/10^n remains.
Despite this not-so-solid reasoning you cannot argue with the basics rules of adding and substracting.
Having said that I await what comments you have with my previous proofs 🙂
But see, when you say 9.999999 - 0.9999 is clearly 9 because EVERY 9 lines up, it sounds an awful lot like you're counting them, which you said I couldn't do...
I'm perfectly happy with everything you've said about the limit of the series. But can you point me to something that actually proves the sum REACHES that limit?
There was a post last page I think, where someone restated the question in terms of functions. Hang on while I go look.
Originally posted by SiskinAh, this is the post I was thinking of.
Series and functions can reach their limits.
if the lim as x->a of f(x) = f(a) then the function f is said to be continuous at a.
The problem being here, "a" is infinite, isn't it? Don't see how this helps, now that I think of it.
But the whole problem is, establishing whether that series reaches its limit rather than blithely assuming it. Don't tell me I'm going to have to rummage in my calculus notes!
Okay, I'm satisfied now. But not from what was up here, most of it missed some steps I needed to see.
What I found was this theorem (AND its proof):
The geometric series a + ar + ar^2 + ar^3 + ... + ar^n-1 + ...
converges and has the SUM S = a/1-r if |r|<1
So, for the sequence 0.9 + 0.09 + 0.09 + ...
You get a = 0.9 and r = 0.1
a/1-r = 0.9/(1-0.1) = 0.9/0.9 = 1
The sum of the series is 1. So 0.999... = 1.
IT'S TRUE!
Originally posted by orfeoa is not infinity. What was said here applies for every real number, as well as +/- infinity
Ah, this is the post I was thinking of.
The problem being here, "a" is infinite, isn't it? Don't see how this helps, now that I think of it.
But the whole problem is, establishing whether that series reaches its limit rather than blithely assuming it. Don't tell me I'm going to have to rummage in my calculus notes!
Every 9 cancels yes, but not because i can count them. It because there are countable infinite 9's behind the decimal point (infinite row of nines, but i can assign a number to every 9, so that if you point at one, i will say it's number, the set of counting numbers is countable infinite, the set of fractions is as well, but the set of reals is overcountable infinite). Being a countable infinite row, i can say that the first 9 cancels, and that every next 9 cancels as well. So the second cancels, the third...all of them cancel.
Limits are a bit hard to explain without offending mathematicians but i'll try it like this:
the limit of f(x) for x -> a, is what you need f(a) to be for the function to be continuous in a. This is more or less what limit means, even if a is infinity or not.
Originally posted by EinZweiDreiit's just two different ways of writign the same number, like sqrt(-1) and 'i'. and as for your 1=2 proof, it's cause 1*0=2*0 😉
I've seen a proof that proves that 2 is equal to 1, too. If one is clever and determined enough, they can nest a trick in a convincing-looking proof. Logically, .9~ cannot equal 1. According to the reflexive property, how can a number equal anything but itself?
Originally posted by geniusExactly. What we are saying is that .999~ is another way of writing the number 1. Same as 2/5 and 0.4 and 140/350 are all the same number.
it's just two different ways of writign the same number, like sqrt(-1) and 'i'. and as for your 1=2 proof, it's cause 1*0=2*0 😉
I'm no mathemetician, and you are all clearly well educated in this subject, but...
isn't this problem as simple as:
1 / 3 = 0.33333.....
0.333... X 3 = 0.9999.....
therefore 1 = 0.99999.....
it would certainly seem to this layman that any effort beyond this is completely wasted...but what do I know