Actuarial Outpost 1 = .999999.......?
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#21
06-15-2009, 09:49 AM
 Ginormous76 Member Join Date: Feb 2008 Posts: 23,654

Quote:
 Originally Posted by LifeAct I once had a calculator that would do 9/9 = .999999999999999999999 occasionally.
Me too, I call it "Microsoft Excel."
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#22
06-15-2009, 09:51 AM
 Ginormous76 Member Join Date: Feb 2008 Posts: 23,654

Quote:
 Originally Posted by 1695814 show me a finger between 2 and 4
Why do you want to see my ring finger?
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#23
06-15-2009, 09:58 AM
 campbell Mary Pat Campbell SOA AAA Join Date: Nov 2003 Location: NY Studying for duolingo and coursera Favorite beer: Murphy's Irish Stout Posts: 93,618 Blog Entries: 6

Quote:
 Originally Posted by Vorian Atreides But I'm more machine than human . . . how would I compare?
Well, we could order on height, but the ordering would change over time for some people.
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#24
06-15-2009, 10:07 AM
 spencerhs5 Member Join Date: Jan 2007 Posts: 1,533

I remember seeing a proof awhile back of 0= all real numbers, didnt remember it but I found it. I do remember doing a paper on this in high school and then when I didnt know an answer on an exam after that I would put zero. My teacher did not find it as amusing as I did.

http://uncyclopedia.wikia.com/wiki/A..._equal_to_zero

Let x=y
y^2=x*y
y^2-x^2=x*y-x^2
(y+x)(y-x)=x(y-x)
y+x=x
since x=y
y+x=y
x=0
#25
06-15-2009, 10:08 AM
 remilard Member Join Date: May 2005 Favorite beer: Talschänke Wöllnitzer Weißbier Posts: 9,546

Quote:
 Originally Posted by campbell Well, we could order on height, but the ordering would change over time for some people.
You can handle this with probability waves.
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#26
06-15-2009, 10:14 AM
 Buck Member CAS AAA Join Date: Mar 2005 Posts: 3,773

Question: Prove .9999999... = (.9999999...)^2 independently without referencing ".999999... =1".
#27
06-15-2009, 10:19 AM
 Griffin 1 Member Join Date: Sep 2001 Location: This is the worst kind of discrimination. The kind against me. Favorite beer: Grizzly. No, wait - polar bears. Posts: 10,444

Quote:
 Originally Posted by Buck Question: Prove .9999999... = (.9999999...)^2 independently without referencing ".999999... =1".
That's not a question.
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#28
06-15-2009, 10:24 AM
 campbell Mary Pat Campbell SOA AAA Join Date: Nov 2003 Location: NY Studying for duolingo and coursera Favorite beer: Murphy's Irish Stout Posts: 93,618 Blog Entries: 6

Quote:
 Originally Posted by spencerhs5 I remember seeing a proof awhile back of 0= all real numbers, didnt remember it but I found it. I do remember doing a paper on this in high school and then when I didnt know an answer on an exam after that I would put zero. My teacher did not find it as amusing as I did. http://uncyclopedia.wikia.com/wiki/A..._equal_to_zero

I've got a bunch of false proofs for your entertainment:
http://www.marypat.org/mathcamp/doc2000/false.pdf

I "prove" all numbers are equal, 2=1, $\pi = 2$, $\infty = -1$, and much much more. [None of these "proofs" are original to me. I just like collecting paradoxes/false proofs]
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#29
06-15-2009, 10:24 AM
 JUICE Member SOA AAA Join Date: Sep 2007 Favorite beer: Bell's Two Hearted Posts: 9,824

Quote:
 Originally Posted by Buck Question: Prove .9999999... = (.9999999...)^2 independently without referencing ".999999... =1".
This was problem 34 on Form 2 of exam C, may 09 sitting. Moderators!
#30
06-15-2009, 10:24 AM
 Kool-Aid Man Member Join Date: Jun 2009 Posts: 4,625

Quote:
 Originally Posted by Buck Question: Prove .9999999... = (.9999999...)^2 independently without referencing ".999999... =1".
I could do that if only I could prove that the sum of the infinite series N/(10^(N+1)) = 1/81, but I don't remember how to do infinite sums anymore.
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