My Math Forum impossible (prove that 0 = 1)

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 June 10th, 2013, 02:42 AM #11 Newbie   Joined: Jan 2010 Posts: 20 Thanks: 0 Re: impossible (prove that 0 = 1) Hoempa you can't divide by 0. Its not defined. Soroban,In third step,?-a * ?-b is not equal to ?ab. Hence, ?1 can't be split into ?-1 * ?-1. KyVancchay, I didn't get how you simplify the indefinite. Soroban,your Maclaurin series is wrong. ln(1+x) = x - x^2/2! +x^3/3! -x^4/4! + ......
June 10th, 2013, 02:51 AM   #12
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Re: impossible (prove that 0 = 1)

Quote:
 Originally Posted by Denis If a hockey game ends 0-0, then overtime is played until 1-0; the loser gets 1 point in the standings, so 0 = 1

The winner get's 2 points in the standings so 1 = 2 :P

June 10th, 2013, 02:58 AM   #13
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Re: impossible (prove that 0 = 1)

Quote:
Originally Posted by mathbalarka
Quote:
 Originally Posted by agentredlum 0 = 1 In the trivial ring.
Best answer of all, you rule
Only after you my liege...

 June 11th, 2013, 07:04 AM #14 Member   Joined: May 2013 From: Phnom Penh, Cambodia Posts: 77 Thanks: 0 Re: impossible (prove that 0 = 1) Another way to solve that 1=0. Suppose that we have a=b, then a^2 =ab a^2 -b^2 =ab -b^2 (a-b)(a+b)=b(a-b) a+b=b 2b=b b=0 b=0*b 1=0
 June 11th, 2013, 07:22 AM #15 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: impossible (prove that 0 = 1) $\sqrt{1}=-1 \vee 1$ so -1= 1. Add 1 and divide by 2 to get 0 = 1.
 June 11th, 2013, 07:24 AM #16 Member   Joined: May 2013 From: Phnom Penh, Cambodia Posts: 77 Thanks: 0 Re: impossible (prove that 0 = 1) 0! =1!, then 0=1
June 11th, 2013, 10:26 PM   #17
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Re: impossible (prove that 0 = 1)

Quote:
 Originally Posted by KyVanchhay 0! =1!, then 0=1
This is surely a bad reasoning. $\Gamma$ does NOT have a proper monotonic inverse.

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