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x + y = x, solve for X (pg. 8)
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| DrUg_Tit0 |
| quote: | Originally posted by whiskers
i know you're smart and all, but this time you got it wrong.
1-0.999... = lim (1-0.999999999999999999999.....) = lim(0.000000000.....0001) = 0
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Yes, I admit you're right about this...I'm getting sleepy
1-0.999... = lim (1-0.999...)
But, since we've said that there's an infinite number of 9's then that 1 at the end is never going to show up, as there will be an infinite amount of zeros before it.
| quote: | | ***EDIT: caught you editing :whip: |
Yeah, sorry, but I didn't change any substance, just the form to make it look more complicated. :) |
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| Smeagol |
just so we all agree on the definition of limits:
lim(n-->inf)(An)=A means that for every e>0 there exists an N such that every An with n>N has |a-An|
So the limit is defenitely 1 (as proved on first page).
What has not been proved is that 0.999... should be interpreted as this limit of 0.9, 0.99, ...
That is a matter of definition of the representation of real numbers by infinite decimals. Can anyone (Drug_tito?) fill in on this?
but it 0.999..-1 couldn't reasonably be any other number that 0 and it would be a shame to let it pass outside the real numbers.
There exists no real number such as "almost zero". :) |
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| whiskers |
| quote: | Originally posted by DrUg_Tit0
Yes, I admit you're right about this...I'm getting sleepy
1-0.999... = lim (1-0.999...)
But, since we've said that there's an infinite number of 9's then that 1 at the end is never going to show up, as there will be an infinite amount of zeros before it.
Yeah, sorry, but I didn't change any substance, just the form to make it look more complicated. :) |
hehehe, i noticed, you used the sum symbol!
anyway, i find it kinda dumb to be arguing about this... in the end we're just trying to see who's gonna end up right to inflate his ego, but nothing will really be achieved... har har :p |
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| DrUg_Tit0 |
| quote: | Originally posted by whiskers
hehehe, i noticed, you used the sum symbol!
anyway, i find it kinda dumb to be arguing about this... in the end we're just trying to see who's gonna end up right to inflate his ego, but nothing will really be achieved... har har :p |
So you give up, eh? :)
But anyway, you're right, it's stupid to argue about this. I'll go sleep now. Good night. |
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| jdjd |
| quote: | Originally posted by DrUg_Tit0
Let's take the following sequence 0.9, 0.99, 0.999... Now, this sequence is converging and equal to 1 since for every e>0 exists a n0(e) from N so that for every n>n0 |an-1| |
nice hack job proof man! seems like you were in uni a looong time ago heh
its as easy as using the monotonic convergence theorem, and say that the sequence is non-decreasing and bounded above, so it converges to its least upper-bound 1. |
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| whiskers |
| quote: | Originally posted by Smeagol
just so we all agree on the definition of limits:
lim(n-->inf)(An)=A means that for every e>0 there exists an N such that every An with n>N has |a-An|
So the limit is defenitely 1 (as proved on first page).
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limits are asymptotes, for x->a f(x)=L means that as x APPROACHES a, f(x) APPROACHES L.
but the definition of a limit is a thing that i did not understand at all in 11th grade, even though my calc teacher explained it to use 3 or 4 times... no one understood it... and trust me, our teacher was REALLY good, the definition of a limit is the only thing i didn't understand the whole year with it's epsilons and deltas... |
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| jdjd |
| quote: | Originally posted by Smeagol
just so we all agree on the definition of limits:
lim(n-->inf)(An)=A means that for every e>0 there exists an N such that every An with n>N has |a-An|
So the limit is defenitely 1 (as proved on first page).
What has not been proved is that 0.999... should be interpreted as this limit of 0.9, 0.99, ...
That is a matter of definition of the representation of real numbers by infinite decimals. Can anyone (Drug_tito?) fill in on this?
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yes thats the definition and yes, according to a pure math prof, decimal expansions are a valid way of representing the real numbers |
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| astroboy |
| quote: | Originally posted by drizzt81
because the equation is only true if and only if y = 0
if y = 0, then x is element of the real numbers. |
So you're saying that if x was a non-real number like i, then y would not = 0 ? |
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| Resnick |
| quote: | Originally posted by whiskers
but the definition of a limit is a thing that i did not understand at all in 11th grade, even though my calc teacher explained it to use 3 or 4 times... no one understood it... and trust me, our teacher was REALLY good, the definition of a limit is the only thing i didn't understand the whole year with it's epsilons and deltas... |
what the? u learned delta epsilon proofs in grade 11? |
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| mezzir |
| quote: | | Suppose 0.999 ... = x. Then 10 x = 9.999 ... = 9 + x, so x = 1. |
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| whiskers |
| quote: | Originally posted by Resnick
what the? u learned delta epsilon proofs in grade 11? |
yes, it was an ap calc class and i took it my junior year because i skipped over a grade
good thing my teacher wasn't an ass and said that we should forget about the delta epsilon definition as it was irrelevant to us
and moncster LIES! |
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| mezzir |
| quote: | Originally posted by moncster
<---- 10th |
pics or stfu :happy2: |
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