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sleepymancer

Post Posted: Sun Dec 04, 2011 3:02 pm 


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Location: Graz, Austria

Ok, i know this isn't really speculation or directly Erfworld, but give me moment and I'll explain why its here (please move it to a better home if need be!)...
At some point on the boards in response to a recentish page (see, Erfworld link) somebody (probably Kreistor, but I can't actually recall!!) observed that infinity 1 still equals infinity. I agreed at the time, although I did find myself wondering if infinity minus infinity still equalled infinity or if it had to be zero. I digress. I have just moments ago found out the actual equation should read:
Infinity  1 = 999
My challenge to you all is to explain why this is correct (Obviously there is a trick in this, but I swear I am not making it up!)
Have fun!!
_________________ I tend to witter on, produce copious typos and run off on nonsensical tangents. If I've done this here, please forgive me I also get a bit obstinate and argumentative. If I'm not budging or understanding your counterargument call me on my manners







Thunder

Post Posted: Sun Dec 04, 2011 6:51 pm 


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infinity 1 = 999 > infinity 1 +1 = 999+ 1 > infinity = 1000.
thats not right so infinity  1 =/= 999
also if n = infinity, nn = n(11) = n(0) = 0
_________________ I live up in the land of ice and snow, where the hot springs blow, from the city which rhymes with fun. do you know from whence i hail?





MarbitChow

Post Posted: Mon Dec 05, 2011 12:09 am 


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Infinity is not a number, so standard arithmetical operations cannot be performed on it. Having said that, I'd like to see what 'proof' you offer; I assume it's similar to the proof that 2=1: Code: a = x [true for some a's and x's] a+a = a+x [add a to both sides] 2a = a+x [a+a = 2a] 2a2x = a+x2x [subtract 2x from both sides] 2(ax) = a+x2x [2a2x = 2(ax)] 2(ax) = ax [x2x = x] 2 = 1 [divide both sides by ax]





sleepymancer

Post Posted: Mon Dec 05, 2011 5:46 am 


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Joined: Thu Sep 16, 2010 9:14 am Posts: 106
Location: Graz, Austria

Nope, none of those and not the 1=2 trick either, I'm afraid! I'll give as a hint that the answer is historical rather than mathematical Hopefully, not too big of a hint though!
_________________ I tend to witter on, produce copious typos and run off on nonsensical tangents. If I've done this here, please forgive me I also get a bit obstinate and argumentative. If I'm not budging or understanding your counterargument call me on my manners





drachefly

Post Posted: Mon Dec 05, 2011 9:48 am 

Joined: Wed Jan 05, 2011 10:36 pm Posts: 1913

Welll. It depends what you mean by underlining. If it means 'repeat these digits', then it's pretty clear. The repeated digits were placed to the left of the decimal point.





MarbitChow

Post Posted: Mon Dec 05, 2011 10:37 am 


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Joined: Thu Apr 30, 2009 5:41 pm Posts: 2521

Ah, cute. The Roman Numeral representation of 1000 was CIƆ  which was reduced to ↀ and later transformed into infinity by John Wallis. So it's just a notational instead of a mathematical trick : "1000" equals "infinity" because they overloaded the symbol. http://en.wikipedia.org/wiki/Roman_nume ... ge_numbers(Google and Wikipedia has made us all geniuses.)





BLANDCorporatio

Post Posted: Mon Dec 05, 2011 10:49 am 



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Aww dang. Well that obviously was the right answer. In any case, great thread btw.
_________________ The whole point of this is lost if you keep it a secret.





sleepymancer

Post Posted: Mon Dec 05, 2011 11:59 am 


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Location: Graz, Austria

Yep, that was it!
Glad it caused such entertainment. embarrassed that it was one wiki, I had it mentioned to me in passing by a friend who is doing her PhD on philology and medieval manuscripts. I shall tease her now about where she gets her information from!
_________________ I tend to witter on, produce copious typos and run off on nonsensical tangents. If I've done this here, please forgive me I also get a bit obstinate and argumentative. If I'm not budging or understanding your counterargument call me on my manners





Thunder

Post Posted: Mon Dec 05, 2011 11:21 pm 


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Joined: Tue Dec 22, 2009 11:54 pm Posts: 243

MarbitChow wrote: Infinity is not a number, so standard arithmetical operations cannot be performed on it. Having said that, I'd like to see what 'proof' you offer; I assume it's similar to the proof that 2=1: Code: a = x [true for some a's and x's] a+a = a+x [add a to both sides] 2a = a+x [a+a = 2a] 2a2x = a+x2x [subtract 2x from both sides] 2(ax) = a+x2x [2a2x = 2(ax)] 2(ax) = ax [x2x = x] 2 = 1 [divide both sides by ax] i hate that proof so much, your not allowed to divide by 0 people "trick" or no
_________________ I live up in the land of ice and snow, where the hot springs blow, from the city which rhymes with fun. do you know from whence i hail?





AnubianDragon

Post Posted: Wed Jan 11, 2012 7:23 pm 

Joined: Sun Jan 08, 2012 12:26 am Posts: 8

Thunder wrote: MarbitChow wrote: Infinity is not a number, so standard arithmetical operations cannot be performed on it. Having said that, I'd like to see what 'proof' you offer; I assume it's similar to the proof that 2=1: Code: a = x [true for some a's and x's] a+a = a+x [add a to both sides] 2a = a+x [a+a = 2a] 2a2x = a+x2x [subtract 2x from both sides] 2(ax) = a+x2x [2a2x = 2(ax)] 2(ax) = ax [x2x = x] 2 = 1 [divide both sides by ax] i hate that proof so much, your not allowed to divide by 0 people "trick" or no Yep, this function's domain is all values where a =/= x. Since that was the original function, the whole proof is moot. As for "Infinity  Infinity = 0", such subtraction is incorrect (it is called an indeterminant form) because it is unclear which "Infinity" has a higher value. While we can subtract ordinary numerals from infinity (any real number), we cannot subtract Infinity from Infinity since it is unclear which number 'reaches' Infinity first. For example: Code: Limit_x>Infinity [2x  x] > 2*Infinity  Infinity > Infinity  Infinity (Indeterminant)
BUT! Technically 2x reaches infinity sooner than x does. So Infinity  Infinity in this case would be Infinity. You can evade these problems by simplifying before evaluating the limit.
Limit_x>Infinity [2x  x] > Limit_x>Infinity [x] > Infinity
It is the same type of error caused by attempting to divide 0 by 0 (0/0 should be 1, but it is also undefined, and it is also 0, and it is also infinity, so the whole thing is called "indeterminant")





