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*is just a sad Algebra II student*
*tries to solve problem*
*head explodes*
I agree, but your proof is wrong. You need to use the same number of nines as length of number you are dividing into (27/99, not 27/999).Quote:
Originally Posted by Moo Moo the Ner Cow
0.27 = 27/99
0.1 = 1/9
therefore
0.9 = 9/9 = 1
Yeah, nit-pick detail. :rolleyes2
I agree :DQuote:
Originally Posted by Moo Moo the Ner Cow
my math teacher had an interesting proof based upon sums of infinite series... I'll see if I can reproduce it.
.999999999..... = 9/10 + 9/100 + 9/1000 +.... = .9 * (1/10)^(n-1)
agree so far?
If you see that so far, you can take the infinite sum as X of that series. The infinite sum of any geometric series [an = a1*r^(n-1)] is a1/(1-r).
Thus, looking at the series, we get .9 / (1 - 1/10) or .9/.9 or 1.
If you want to know where that infinite sum formula comes from, I can post that, too....
I already did. Give me a number other than 1 that when subtracted from 1 will result in 0. Unless someone can answer this than .999... is equal to and less than 1. :eek:Quote:
Originally Posted by Arche
EDIT: there, that looks better
The problem is.... .999999 isn't a number in it's own right.... it's another way of saying 1. We claim that 100/100 is one.... 1 - (100/100) = 1... so why can't .99999.... also be 1?
Beside the fact that the proof is pretty watertight
Doesn't 1-(100/100)=0?Quote:
Originally Posted by bennator
whoops... that too.... that's what I get for posting too quickly
"Give me a number other than 1 that when subtracted from 1 will result in 0. Unless someone can answer this than .999... is equal to and less than 1. "
Why? I might as well say "unless 2 is green, 2 does not exist"
Is it safe to say that .999... + lim (1/x), x --> infinty = 1 ?
Yeah, I'd find some maths quite refreshing given that my degree now consists of absorbing vast quantities of facts, with the occaisional stoichiometrical calculation thrown in.
oh, and by the way, now that I've started to do calculus, math is quite possibly the most amazing thing ever.
no, but I think the answer to your question is "the smallest number greater than one" :D
But I think that if we apply the same logic that gives us that .999... =1, then lim (1/x), x--> infinity = 0. I think, anyway. :mog:
No.Quote:
Originally Posted by Doomgaze
My formula is 1 - x = 0, where x = any number other than 1. Your answer so far has been .999..., but 1 - .999... != 0.
Basically I'm asking you to prove that 1 - .999... = 0, because if 1 - .999... is equal to zero, then 1 is equal to .999...
EDIT: Bah, even my fiance's against me, I hate arguing math with someone that has a Master's in math. Anyway, just to add to the ".999... = 1" agrument, what is 1/3 + 1/3 + 1/3? Yeah, I think you can follow this train of thought out. it's an easier proof than most of what I've seen.