Infinity minus one equals ?

[sleepymancer]

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!!

[Thunder]

infinity -1 = 999 -> infinity -1 +1 = 999+ 1 -> infinity = 1000.

thats not right so infinity - 1 =/= 999

also if n = infinity, n-n = n(1-1) = n(0) = 0

[MarbitChow]

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: Select all

     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]
2a-2x = a+x-2x        [subtract 2x from both sides]
2(a-x) = a+x-2x       [2a-2x = 2(a-x)]
2(a-x) = a-x          [x-2x = -x]
     2 = 1            [divide both sides by a-x]

[sleepymancer]

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!

[drachefly]

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]

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]

Aww dang. Well that obviously was the right answer. In any case, great thread btw.

[sleepymancer]

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!

[Thunder]

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: Select all

     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]
2a-2x = a+x-2x        [subtract 2x from both sides]
2(a-x) = a+x-2x       [2a-2x = 2(a-x)]
2(a-x) = a-x          [x-2x = -x]
     2 = 1            [divide both sides by a-x]

i hate that proof so much, your not allowed to divide by 0 people "trick" or no

[AnubianDragon]

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: Select all

     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]
2a-2x = a+x-2x        [subtract 2x from both sides]
2(a-x) = a+x-2x       [2a-2x = 2(a-x)]
2(a-x) = a-x          [x-2x = -x]
     2 = 1            [divide both sides by a-x]

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: Select all

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")