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# Two is equal to One

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Post Voting Period
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after 8 votes the winner is...
bencbartlett
 Voting Style: Open Point System: 7 Point Started: 9/2/2012 Category: Science Updated: 5 years ago Status: Post Voting Period Viewed: 1,696 times Debate No: 25411
Debate Rounds (3)

 Pro In R2 I will present a proof that 2 = 1.Please don't worry. The proof is flawed. In R2 it will be task of Con to spot the flaw. If she is able to do this, she wins the debate. If she is unable to spot the flaw in R2, I win this debate.If Con is unable to spot the flaw in R2 (or provides an errenous explanation), I will provide the correct explanation in R3. Con need not do anything in R1 and R3.I will be using Taylor series expansion and logarithms. Nothing in much detail. But you need to be comfortable with basics of these two concepts and series calcuations to accept this debate.Hoping to get a sincere opponent. Report this Argument Con I accept this challenge - it'll be nice to actually be able to argue mathematics on here for once, since I've tried starting quite a few math debates before. I'll post my solution as soon as I find an error in whatever proof you offer, and I commit to working this problem alone. Good luck!Report this Argument Pro Thanking bencbartlett for accepting this debate. I agree that it feels nice to debate mathematics.I am numbering the steps in order to assist discussion.Taylor expansion of ln (1+x) is:1-> ln (1+x) = x - x^2/2 + x^3/3 - x^4/4 + x^5/5 -+-Putting x = 1.2-> ln (1 + 1) = ln (2) = 1 - 1/2 + 1/3 - 1/4 + 1/5 -+-Rearranging the terms3-> ln (2) = (1 + 1/3 + 1/5 +++) - (1/2 + 1/4 + 1/6 +++)4-> Let P = 1 + 1/3 + 1/5 +++5-> Let Q = 1/2 + 1/4 + 1/6 +++From steps 3, 4 and 56-> ln (2) = P - QConsider Q7-> Q = 1/2 + 1/4 + 1/6 + 1/8 + 1/10 +++Taking 1/2 as common from LHS.8-> Q = 1/2 [1 + 1/2 + 1/3 + 1/4 + 1/5 +++]Rearranging terms9-> Q = 1/2 [ (1 + 1/3 + 1/5 +++) + (1/2 + 1/4 + 1/6 +++) ]Subsituting series in step 9 with help of step 4 and 5,10-> Q = 1/2 [ P + Q]11-> 2Q = P + Q12-> Q = PSubstituting Q in step 6.13-> ln (2) = P - P14-> ln (2) = 0Taking antilogs on both sides15-> 2 = 1Best of Luck to my esteemed opponent.Report this Argument Con My opponent’s argument breaks down at step 2. The Taylor expansion for ln(x) has a radius of convergence of 1, around the point x = 1, such that 0
5 comments have been posted on this debate. Showing 1 through 5 records.
Posted by baggins 5 years ago
Changed the time limit.
Posted by Ore_Ele 5 years ago
There is not a problem to working alone.
Posted by baggins 5 years ago
In 72 hours, you will be able to ask people in mathematics department in some university and get response from them. If you commit to working alone, I will change time limit to 72 hours.

Any flaw which invalidates the proof is sufficient.
Posted by Ore_Ele 5 years ago
Also, I'd take this but I need the 72 hours per round.
Posted by Ore_Ele 5 years ago
So we only need to point out a mathematical flaw, or does it have to be the specific one that you are looking for?
8 votes have been placed for this debate. Showing 1 through 8 records.