Re: Oops I broke the laws of math.
round one
fight
round one
fight
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Re: Oops I broke the laws of math.
all this math
its making my head hurt
**** you guys
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Re: Oops I broke the laws of math.
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Re: Oops I broke the laws of math.
Fine. I guess there's nothing to argue because there's a series that represents .999... that converges to 1?
But what about the proofs in the OP? I don't think anybody in the thread definitively posted answers to them.
Proof 1:
You're taking the positive square root of the left side and the negative square root of the right side.
Proof 2:
You're dividing by zero.
Proof 3:
Those properties of square roots only work for positive numbers.
Don't know enough about series to answer the rest
. I need more maths classes.
I'm retired
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Re: Oops I broke the laws of math.
4:
There's one "-1" missing in the end.
5:
This one is just stupid, x is not a function so you can't take it's derivative.
6:
They assume x * x is x+x+x+x+... which is false if x is negative. I don't think this is why the proof fails but I don't care. I think it's the same as proof 5.
7:
The sum S is infinite, it doesn't make any sesne to say S-A is 4S because nothing can be the 4 times more infinite than something else.Comment
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Re: Oops I broke the laws of math.
What's two plus two divided by two?
I AM ASHAMED OF MY STUPIDITY AND MUST COMMIT SEPPUKU. FORGIVE ME, FFR.
Ninjas NEVER die. Now on a 30 kill streak.Comment
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Re: Oops I broke the laws of math.
either 2 or 3.
Duh. Haven't really learned about series and such yet."AW MAN, WE SUCK AGAIN!"
- Rob Schneider, The Waterboy (1998)Comment
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Re: Oops I broke the laws of math.
IEAT USES SUPER - MATHRYUKEN
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Firstly, remember that we can't really interpret 0.000...1 as a finite decimal value, but rather, the point that by definition bounds the limit of 1/n as n gets infinitely large. This point is of course 0 (which will be subsequently proven).
For any two values x1 and x2, x1/n can be made arbitrarily close to x2/n by taking n to be a sufficiently large number. Thus,
lim (n→∞) x1/n = lim (n→∞) x2/n, for all x
Since x is a constant, we can extend the notion and say that
lim (n→∞) x1/n = lim (n→∞) x2/n for all x =>
x1[lim (n→∞) 1/n] = x2[lim (n→∞) 1/n] for all x
Thus, taking x2 to be 0,
x1[lim (n→∞) 1/n] = x2[lim (n→∞) 1/n] =>
x1[lim (n→∞) 1/n] = 0[lim (n→∞) 1/n] =>
x1[lim (n→∞) 1/n] = 0 =>
lim (n→∞) x1/n = 0
And that's why lim (n→∞) x/n = 0 for all x. Proceeding with the proof...
Taking x1 to be 1,
lim (n→∞) x1/n = 0 =>
lim (n→∞) 1/n = 0 =>
From the aforementioned interpretation, 0.000…1 is equivalent to lim (n→∞) 1/n. Thus,
lim (n→∞) 1/n = 0 =>
0.000...1 = 0
And finally getting back to the problem at hand...
Since lim (n→∞) 1/n = 0.000...1 = 0 = lim (n→0) n,
15.68 - lim (n→∞) 1/n = 15.68 - lim (n→0) n =>
15.68 - 0.000...1 = 15.68 - 0 =>
15.67999... = 15.68
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IT'S SUPER EFFECTIVE!Comment
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