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       #Post#: 19497--------------------------------------------------
       An innocent-looking math problem
       By: Omni_Builder Date: January 15, 2015, 7:41 pm
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       So, I came across this very interesting problem:
       [hr]
       Consider a set "S" and an operation "&#9733;". For all elements
       "a" and "b" in S, you know the following:
       [list]
       [li](a &#9733; b) is also a member of S.[/li]
       [li]((a &#9733; b) &#9733; a) = b.[/li]
       [/list]
       Prove that, for all "a" and "b" in S, (a &#9733; (b &#9733; a))
       = b.
       [hr]
       It looks like there isn't much there, right? Well, it gets
       trickier when you realize all the things you don't yet know
       about S and &#9733;...
       This is a really neat problem, with an extremely satisfying
       solution. I couldn't solve it at first, only figuring it out
       after returning to the problem after several days. I'd be
       interested to see others' ideas.
       #Post#: 19812--------------------------------------------------
       Re: An innocent-looking math problem
       By: idodi Date: January 20, 2015, 8:15 am
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       Does this make sense?
       (A friend of mine came up with this theory.)
       [spoiler]
       Given that (a &#9733; b) &#9733; a = b
       Then the following is true: (one number &#9733; another number)
       &#9733; itself = the other number
       Therefore  (b &#9733; a) &#9733; b = a
       Substitute it into a &#9733; (b &#9733; a)
       Which would result in: ((b &#9733; a)&#9733; b) &#9733; (b
       &#9733; a)
       Based on the above principle, it must equal b.
       [/spoiler]
       #Post#: 19813--------------------------------------------------
       Re: An innocent-looking math problem
       By: Omni_Builder Date: January 20, 2015, 9:10 am
       ---------------------------------------------------------
       That's the solution I got as well!
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