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Re: Finally the discussion is over: S = Lim S is a bad definition.
Posted:
Oct 6, 2017 5:58 AM


Den torsdag 5 oktober 2017 kl. 19:22:37 UTC+2 skrev John Gabriel: > On Thursday, 5 October 2017 09:44:43 UTC4, Markus Klyver wrote: > > Den onsdag 4 oktober 2017 kl. 11:27:43 UTC+2 skrev John Gabriel: > > > On Tuesday, 3 October 2017 20:32:47 UTC4, burs...@gmail.com wrote: > > > > Well its the mean value between 0.999... and 1.0. > > > > > > What are you saying you baboon, what are you mumbling about? You can only calculate an arithmetic mean between well formed numbers you moron! 0.999... is NOT a number of any kind. It represents a SERIES. If you define it stupidly as Euler did to be equal to its limit, then you are taking the mean of 1 and 1. There is nothing to do you HUGE APE!!!! > > > > > > Grow a brain. You sit pounding out your brain farts on sci.math. Get a job idiot!!!!! No one will hire you once they read the crap you write here. > > > > > > <6 year old rant> > > > > 0.999... is a short hand for an infinite sum (S = 0.9+0.09+0.009+...) which we define as the limit of a finite sum (Lim S = Lim {n \to \infty} 110^(n)) as the number of terms approaches infinity. > > Yes, S = Lim S. Clear as water. > > It is an illformed definition because S is NOT the same as Lim S. > > How about this definition: Mercedes = cheese cake. > > Do you think it is well formed? > > > > > Den torsdag 5 oktober 2017 kl. 08:01:05 UTC+2 skrev 7777777: > > > torstai 5. lokakuuta 2017 8.44.17 UTC+3 Zelos Malum kirjoitti: > > > > 0.999... must equal 1, there cannot be any other way for it to be. > > > > > > fail. > > > > > > in real numbers 0.(9) is not equal to 1. > > > > It indeed does. Or how do you define 0.(9)?
What's S and Lim S? You haven't actually defined what they mean. If S is the value of the series, then no, we don't define the value of the series as the limit of the value of the series.
Den torsdag 5 oktober 2017 kl. 20:46:07 UTC+2 skrev John Gabriel: > On Thursday, 5 October 2017 13:22:37 UTC4, John Gabriel wrote: > > On Thursday, 5 October 2017 09:44:43 UTC4, Markus Klyver wrote: > > > Den onsdag 4 oktober 2017 kl. 11:27:43 UTC+2 skrev John Gabriel: > > > > On Tuesday, 3 October 2017 20:32:47 UTC4, burs...@gmail.com wrote: > > > > > Well its the mean value between 0.999... and 1.0. > > > > > > > > What are you saying you baboon, what are you mumbling about? You can only calculate an arithmetic mean between well formed numbers you moron! 0.999... is NOT a number of any kind. It represents a SERIES. If you define it stupidly as Euler did to be equal to its limit, then you are taking the mean of 1 and 1. There is nothing to do you HUGE APE!!!! > > > > > > > > Grow a brain. You sit pounding out your brain farts on sci.math. Get a job idiot!!!!! No one will hire you once they read the crap you write here. > > > > > > > > <6 year old rant> > > > > > > 0.999... is a short hand for an infinite sum (S = 0.9+0.09+0.009+...) which we define as the limit of a finite sum (Lim S = Lim {n \to \infty} 110^(n)) as the number of terms approaches infinity. > > > > Yes, S = Lim S. Clear as water. > > > > It is an illformed definition because S is NOT the same as Lim S. > > > > How about this definition: Mercedes = cheese cake. > > > > Do you think it is well formed? > > > > > > > > Den torsdag 5 oktober 2017 kl. 08:01:05 UTC+2 skrev 7777777: > > > > torstai 5. lokakuuta 2017 8.44.17 UTC+3 Zelos Malum kirjoitti: > > > > > 0.999... must equal 1, there cannot be any other way for it to be. > > > > > > > > fail. > > > > > > > > in real numbers 0.(9) is not equal to 1. > > > > > > It indeed does. Or how do you define 0.(9)? > > Also, consider that you don't have a unique decimal representation of any number if you define the series to be the limit. For example, both of the following series have the same limit: > > 0.999... = 9/10 + 9/100 + ... > 0.875... = 1/2 + 1/4 + 1/8 + ... > > So if we define them both as 1, then 0.999... = 1 and 0.875... = 1. How do we identify what series we are dealing with if we only have the limit? There are innumerably many series with the same limit of 1. > > On the one hand you berate the limit and on the other hand you invoke it. What is more important, the series or its limit? You seem to think the limit does not matter and then you go right ahead and DEFINE the series to be equal to its limit?! Isn't that stupid? > > A series is NOT a LIMIT. In fact, no matter if you could hypothetically sum all the terms of 3/10 + 3/100 + ..., you would never arrive at 1/3 because 1/3 is not measurable in base 10. > > Do you understand that expression "measurable in base 10"? It means expressing any rational number using a given base. There is no fraction p/q such that q=10^n with n integer and p/q = 1/3. It's impossible my little stupid. There is a theorem stating this. You defining S = Lim S goes against the theorem. You just can't use illformed definitions. They break everything. > > Get it?
That is not how decimal expansions are defined. You can't just shop off a series and write 0.875... = 1/2 + 1/4 + 1/8 + ..., because you as the number of terms grows you will surpass 0.875 and the digits will not match. We use base10, and hence 0.875... = 8/10 + 7/100 + 5/1000 + ...
A series is defined as the limit of ITS PARTIAL SUMS. It is NOT defined as ITS OWN LIMIT. That's nonsense and an other strawman from your side. Put this madness down. You are either choosing not to listen to anything we explain, or you don't care. Summing "an infinite amount of terms" is defined as taking a limit.
The theorem you keep referring to only applies to finite decimal expansions.
Den torsdag 5 oktober 2017 kl. 22:52:59 UTC+2 skrev John Gabriel: > On Thursday, 5 October 2017 15:09:12 UTC4, burs...@gmail.com wrote: > > That ... > > > > You "decimal" means 10, so this a base 10 digit series. > > On the otherhand this one here is not a decimal, base 10, > > representation: > > > > 1/2 + 1/4 + 1/8 + ... = 1 > > 5/10 + 25/100 + 125/1000 + .... = 1 > > That's decimal you baboon!!!
But 25 > 10 and 125 > 100, so it's wrong.



