defo m8.

wonder if fdawg will take it serious and do his patented wiki search!

http://en.wikipedia.org/wiki/Limit_of_a_sequence

GL HF

.

This question is about as old as time :)

little timmy was so proud of what he learned in pre-algebra that day...

And? Troll someone else.

haha so owned tho B>

SuPaRoWnEd. I cried myself to sleep last night :(

he's right, that's super basic stuff...

so?

super basic yet I understand none of it!11

And? I doubt I would have posted it here if I didn't find it funny / interesting :p

.999 != 1
.9 (continuous) == 1

That was what I meant by ~ though :P

~ is approximately equal to, e.g. 5.7 ~ 6

alright, my bad :(

I believe the proper way would be 0.99(9)

guess that would be 0.99 with an uncertain 3rd decimal :> as in error

I learned in primary school that's the way to show the recurring element when dealing with recurring decimal fractions.

nm,
lim_(m --> oo) sum_(n = 1)^m (9)/(10^n) = 1
PS
http://www.youtube.com/watch?v=2U3HMh1xBU0

http://www.youtube.com/img/down/how_it_works.jpg

HAHA, thats too bad =)

"==" != "="

hard to do 3 lines on top of each other ;) guess that's what Sujoy meant

I don't think that works out then.

agree. can i get my plus now?

who cares?

yeah, when you are infinitely close to 1, i can't see how you could claim different :p

when you are infinitey close to the speed of light you are going the speed of light!!!!!!!! and i am not a photon!!!!!!!! trust me!!!!!!!! im really not!!!!!!!!!!!!

It' wasn't exactly self-explanatory :(

retard?

how you figure?

shticks

?

???

1/3 + 2/3 = .999999... = 1.0, this is incorrect

1/3 + 2/3 = 3/3 = 1.0, this is correct

.333... + .666... = .999..., this is correct

the concept of infinity seems to have passed you by, friend.
the idea that the numbers are infinitely close to their whole counterparts, but never being whole is what you're missing in this logic problem. the number .999... is infinitely close to one, but it is never great enough to be one; it is .0000...(insert infinite zeros here)...001 away from the number one.

note: "..." in all references in this post equal infinite repetition

yep, he has a world of juicy infinity problems to explore :)

10x - x = 9.9999... - 0.9999... is wrong.

EDIT: Thought I would explain why it's wrong:
You either substract 'x' from both sides of the equation OR '0.99(9)' from both sides of the equation. You can't do it the way you did. Still that wouldn't change anything so it's useless. In this case you can either divide or multiply.

So no x does not equal 1.

lol =o)

To quote DJWheat "FUCK MY ASS".

don't see how this is wrong, it's taught like that at universities even?

I don't think he's realised what x has been equated too.

We already know what x is equated to so this entire 'problem' doesn't make sence. I was giving an example In case we didn't know the value of x. Blame sharakin for bringing those infinite decimals.

You completely misunderstood his equation. He equated x to 0.9999(9), so when he minused x from the left side he minused x from the right side. I guess you didn't pay attention in maths?

I understood what he did but it would make no sence to do that normally when you know the value of x.

Besides there's no way 0.99(9) equals 1. No matter what sort of calculations you make that prove it it just shows the limitations of those methods.

EDIT: I don't care what they teach at school about these things. It makes no sence to me hence my opinion.

EDIT2: I'm talking strictly about taking some recurring decimal fraction 'out of the sky' not converting it from normal fractions like: 1/3 = 0.33(3) and 2/3 = 0.66(6).

1/3 + 2/3 = 1
0.33(3) + 0.66(6) = 0.99(9) <- it should equal to a whole number tho but the method of writing decimal fractions has its limitations.

if you really knew the value of x you would agree that it equals 1. don't try to count the amount of 9 and think "so there's just 0,000000...0001 missing to 1", rather consider it as "no matter how much 9 I counted, there will always pop up another one". multiplied by 10 it changes nothing behind the "," as you just move it and there's still an infinite amount of 9 popping up at the end (infinite is infinite, there's no infinite +1 or -1). subtracting 0,999..999 you clean up all the decimal digits and what is left is 9x = 9,00000...000. so, x clearly equals 1.

Read EDIT2 paragraph and you'll see that I agree with you.

This is maths, for once opinions count for shit.

You're right.

I just realized that I answered to myself with my previous post: the limitations of converting and writing a decimal fraction. I approached this 'problem' as it would be something like '2 + 2 = 5'.

Up to this point I never really thought about recurring decimal fractions.

1 + 1 = 2 is correct

2 + 2 = 5

Did you do your homework yesterday spamerio?

i woz taut 2 + 2 = 5 must b tru

best nick ever

First and last time we agree on something :)

racist...

chaomi? :D

for extremely large values of 2 ?

always faszinating

why do you care who agrees with you?
I guess you still have the mentality of a 12 year old.

So's your face.

my face is unso-able!!!!!!!!!!!!!!!!!!! it is beyond repair!!!!!!!!!!!!!!!!!!!!!!!! because of the damage you have done to it with your words!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! i will forever reflect the feeling that i harbor inside which will always be anger and disgust with myself because of what sharakin has showed me!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! but if i were a normal person i would be amazed by the astounding mathematical postulations herein and my head would likely explode instead!!!!!!!!! so i am pretty thankful that sharakin has shown me that i am not a normal person so that i can keep my head!!!!!!!!!!!!!!!!!

lol :D

I can't believe that anyone is disputing this stuff... Not only is it fairly obviously correct, didn't your pre-algebra teachers go through this stuff with you? The second proof in particular is just a common way of converting infinite repeating decimals into fractions.

There i was thinking this was about the police

0.999... = 1 as there is no discrete number between 0.999... and 1 so they are considered the same.

discrete numbers to the rescue! :P

Person A runs at speed 5m/s and has a head start of 10m compared to Person B who runs at 10m/s. Obviously Person B is running faster and should overtake Person A. But Person B always has to run some time to reach the position Person A was at, while Person A ran away some more already. (Person A at 10m, Person B at 0. Person B runs to 10m but then A is already at 15m. Person B runs to 15m but by then Person A is already at 17,5m .. and so on).

How can Person B ever catch up with Person A then ???

Or a similar problem with less digits.
You can never arrive at your destination because you always have to walk half the distance first.

zzz

i didn't read it but it sounds like spiderman learned one of zeno's paradoxes!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! what a great day in fizzy fizziks!!!!

!!!!!!!!!!!!!!!! and
!!!!!!!!!!!!!!!!! imo

Those problems don't work becuase they don't take time into account - as you half the distance, you half the time, and obviously if you slow down time indefinately you'll never move anywhere.

ain't that the whole idea of convergence?

I don't know.

I'm just saying that the problem he stated isn't as odd as it first sounds, when you consider that speed is distance over time. When you half the distance travelled, you half the time taken aswell, so obviously continuing to half the distance will result in never reaching your destination.

the 2nd problem doesn't mention time at all :) and the first problem just tends to make people look helpless because they for sure know they can overtake a slower car with their car on the road, but can't express why it doesn't work here (well actually it does, it's just not so obvious)

Person A will take 6 seconds to move 30 meters
Person B will take 3 seconds to move 30 meters, however starts 2 seconds after person A.

Therefore at the 30 meter mark, person B will be 1 second behind person A

At the 40 meter mark,
Person A's time will be 8 seconds,
Person B's time will be 6 seconds,

Thus person B has over taken person A between the 30 and 40 meter mark.



don't think too much into it :P

not sure what you're talking about, is it a new problem or mine? :D

x1(t) = 10 + 5*t
x2(t) = 0 + 10*t

x1(t0) = x2(t0) :
10 + 5*t = 10*t
---> 5*t = 10, t = 2

x2(2)=10*2=20 there you go

don't think too much into it ;-)

B will arrive at the 30m mark in 5 seconds, 1 second before A

"Person B will take 3 seconds to move 30 meters, however starts 2 seconds after person A.

Therefore at the 30 meter mark, person B will be 1 second behind person A"

i'm pretty sure thats what I said.
Its not a new problem, just looking at it from a different perspective.

"Thus person B has over taken person A between the 30 and 40 meter mark."

Thus B has over taken A either at the 30m mark or before.

http://web.sdu.dk/mthom03/ehm.htm

sum(9/(10^k))=1 with k from 1 to infinite.
Yes I think everybody agrees, even if your notation is inappropriate.

e^(i*pi) + 1 = 0

So the two numbers are the same? o,o

rocket scientists rock?

WHYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

10x - x is 9x.

and 9*0.999... is not 9, but 8,99...(1)

rofl, made me laugh