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Topic: (a*b)*c = a + b + c (Read 585 times) |
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Aryabhatta
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(a*b)*c = a + b + c
« on: Sep 17th, 2004, 5:39pm » |
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A problem from Russian Math Olmpiad 1998.
'*' is a binary operation on the set of real numbers such that:
(a*b)*c = a + b + c for all real a,b,c.
Show that * is same as '+'.
(+ is the usual addition operation)
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Grimbal
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Re: (a*b)*c = a + b + c
« Reply #1 on: Sep 18th, 2004, 8:23am » |
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[1] ((a*b)*c) = a+b+c
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Applying [1] in 2 ways on (((0*0)*a)*b) :
(((0*0)*a)*b) = (0+0+a)*b = a*b
and
(((0*0)*a)*b) = (0*0) + a + b
therefore
[2] a*b = (0*0) + a + b
Applying [2] with a=(0*0), b=0 :
(0*0)*0 = (0*0) + (0*0) + 0
but we know
(0*0)*0 = 0+0+0 = 0
therefore
(0*0) + (0*0) = 0
Assuming we are in a group where 1+1 is not 0, this implies
[4] (0*0) = 0
Combining [2] and [4] gives:
a*b = a+b
::
Note: if 1+1 = 0, a*b = a+b+1 satisfies [1]
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Aryabhatta
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Re: (a*b)*c = a + b + c
« Reply #2 on: Sep 20th, 2004, 10:09pm » |
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Well done Grimbal.
When I first saw this problem I was a bit surprised!
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Leo Broukhis
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Re: (a*b)*c = a + b + c
« Reply #3 on: Sep 21st, 2004, 12:27pm » |
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on Sep 18th, 2004, 8:23am, Grimbal wrote:[1] ((a*b)*c) = a+b+c
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Assuming we are in a group where 1+1 is not 0, this implies
[4] (0*0) = 0
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What if 2+2 is 0? 
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Grimbal
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Re: (a*b)*c = a + b + c
« Reply #4 on: Sep 21st, 2004, 2:32pm » |
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Hm... I said 'group' but I meant 'ring'. 
And when I mean 'ring' please understand 'division ring'. 
Whatever.... It works with real numbers.
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