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+--- Thread: Last post wins. (/showthread.php?tid=89)
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RE: Last post wins. - Kamikaze28 - 07-03-2012 05:03 PM
(07-03-2012 04:04 PM)knighthalo123 Wrote: Wrong you didnt subtract 2ab on the right side of the equation
He did subrtact 2ab, the sign flipped from '+' to '-'.
The mistake in this 'proof' is a hidden division by 0:
If a=b, then a^2-ab can be written as:
a^2-a*a
=a*a-a*a
=0
A division by 0 immediately destroys any meaning and validity of an equation.
RE: Last post wins. - knighthalo123 - 07-04-2012 12:46 AM
but still 2A^2-2ab = a^2-ab that doesnt look like a subtraction of 2ab...
RE: Last post wins. - aaronINdayton - 07-04-2012 01:03 AM
Wait, is that: (2A^2) - 2ab
OR is it: 2A^(2-2ab)?
RE: Last post wins. - Kamikaze28 - 07-04-2012 01:27 AM
I've reformatted this 'proof' to clarify a few things.
(07-03-2012 01:45 PM)Solan Wrote:HAHAHAHAHAHCode:
False proof that 2 = 1:
a = b | This is the premise
a^2 = ab | Multiply by a
2a^2 = a^2 + ab | Add a^2
2a^2 - 2ab = a^2 - ab | Subtract 2ab
2(a^2 - ab) = 1(a^2 - ab) | Factorise
2 = 1 | Divide by (a^2 - ab) <<
The line marked with << contains the fault of the proof: (a^2 - ab)=0, therefore the division is illegal.
The subtraction in line 4 is correct. The right hand side went from a^2 + ab to a^2 - ab.
RE: Last post wins. - aaronINdayton - 07-04-2012 01:44 AM
(07-04-2012 12:46 AM)knighthalo123 Wrote: but still 2A^2-2ab = a^2-ab that doesnt look like a subtraction of 2ab...
ab - 2ab = -ab
RE: Last post wins. - phrenikos - 07-04-2012 01:46 AM
I've got one for you non-stat majors (St. petersburg paradox):
Let's say I offer you a game: A coin is tossed at every stage. The pot starts at 1 dollar and is doubled every time a head appears. When a tail appears, the game ends and you win the money in the pot.
(i.e. You win 4 dollars if two heads appear, and then a tail appears after)
How much are you willing to pay for this game?
RE: Last post wins. - Kamikaze28 - 07-04-2012 02:15 AM
(07-04-2012 01:46 AM)phrenikos Wrote: I've got one for you non-stat majors (St. petersburg paradox):
Let's say I offer you a game: A coin is tossed at every stage. The pot starts at 1 dollar and is doubled every time a head appears. When a tail appears, the game ends and you win the money in the pot.
(i.e. You win 4 dollars if two heads appear, and then a tail appears after)
How much are you willing to pay for this game?
That's clever - the computer science part of me calculated the expected winnings and therefore wants to play at any cost, because the expected winnings is infinite. My gut however wouldn't pay more than $8. And in reality, I don't gamble for money and wouldn't play at all.
RE: Last post wins. - Solan - 07-04-2012 09:52 AM
haha kamikaze you got me... yeah i was really confused by that at first, just i realized later that a=b... so a^2-ab = 0 indeed.
also to phrenikos... what would be the difference in how much i'd pay for the game? does the pot start at how much i pay or something? o jjust 1 dollar every challenge?
RE: Last post wins. - knighthalo123 - 07-04-2012 03:50 PM
guys do you know five = nine?
RE: Last post wins. - QuantumApocalypse - 07-04-2012 08:21 PM
0.9 recurring = 1
