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June 10th, 2013, 02:42 AM   #11
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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! + ......
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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
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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...

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June 11th, 2013, 07:04 AM   #14
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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
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June 11th, 2013, 07:22 AM   #15
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Re: impossible (prove that 0 = 1)

so -1= 1. Add 1 and divide by 2 to get 0 = 1.
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June 11th, 2013, 07:24 AM   #16
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Re: impossible (prove that 0 = 1)

0! =1!, then 0=1
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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. does NOT have a proper monotonic inverse.
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