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Clearly S is positive. Now multiply each side by 2:
But that's just the same as S - 1.
And if 2S = S - 1, then S = -1.
So -1 is positive.
(Found at Futility Closet)
If you enjoy such things, there are 17 other mathematics posts in TYWKIWDBI. Click here.
There's definitely a problem here (especially since it violates the definition of a positive number). Granted, I'm a physics major, so the purer proofs are usually a bit beyond me, but take the following definitions of the series:S=Sum(2^n) from 0 to Inf.2S=Sum(2n)-1 from 0 to Inf.Take the first terms, though.2S-S=2-1=1First two terms:2S-S=(2+4)-(1+2)=3First 3 terms:2S-S=(2+4+8)-(1+2+4)=7First 4:2S-S=(2+4+8+16)-(1+2+4+8)=15So it's clear that 2S-S diverges and will not converge to any value, including -1. I'm sure that there's some fancy theorem that covers this, but I'll just stick to this.
I'm an English major, so I'm on even shakier ground than you are, but let me try to address your argument.As you point out, we're dealing with infinitely long series.You've demonstrated that the series 2S and S diverge by subtracting one from another. Instead of doing that, create a third series - the S-1 series - by subtracting 1 from the S series:S = 1+2+4+8+16... Then S-1 = 2+4+8+16...Note the 2S series is the same:2S = 2+4+8+16...So 2S = S-1, and simpliflyingS = -1which is negative. But the original postulate was that it was positive, being composed only of the sum of positive numbers.I think all that this means is that logic doesn't apply to infinite series. Using infinite series, you can prove that 4 = 3 (see this link: http://tywkiwdbi.blogspot.com/2007/12/proof-that-4-equals-3.html)
It's simple. Based upon the first result, S is equal to infinity, and therefore is not a rational number, and cannot be treated as such. But, it's still fun to mess with stuff like this.
Aha, I think Mike has the official mathematical reasoning behind it. I can sleep tonight.
So what you're saying is that S = summation of 2^n from 0 to some termLet's assume that this term is not infinity. Which would mean that S = 2^(n+1) - 1Now manipulating this equation for both cases 2S = 2^(n+2) - 2 S - 1 = 2^(n+1)Now, lets find some terms that will show that 2S and S - 1 are equal 2^(n+2) - 2 = 2^(n+1) => 2^n = 1The only possible term that will prove this is 0. Therefore, your proof of -1 being a positive number is disproven, due to the fact that the term is obviously not 0, but closer to infinity. I hope that makes sense? Just another look at how this isn't possible.
For me the simplest "disproof" is not a mathematical one, but a logical one - that two times infinity is nonsense.
but S isn't positive is it now