let $x,y$ be any real numbers,define $*$,such $$x=(x*y)*y=y*(y*x)$$ fo any $a,b$,show that $$a*b=b*a$$
My try: $$x=(x*y)*y=y*(y*x)$$ then $$y=(y*x)*x=x*(x*y)$$ so $$y*x=x*(x*y)*x$$ and $$x*y=y*(y*x)*y$$ then I can't,Thank you
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Is your operation associative? – Phira Jan 01 '14 at 09:16
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With only above definition of * you can not prove commutativity, there is alot of counter example. – Farshad Nahangi Jan 01 '14 at 09:18
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My point is that you should not write $a\ast b\ast c$ without parentheses. – Phira Jan 01 '14 at 09:18
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@FarshadNahangi,why? – math110 Jan 01 '14 at 09:18
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forgive me @ china math I had mistake – Farshad Nahangi Jan 01 '14 at 09:25
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Here's the proof: http://www.proofwiki.org/wiki/Ordinal_Multiplication_is_Associative – Jan 01 '14 at 09:31
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Start from the result and do the only things you can (multiply by things that make something go away):
$\begin{align} a\ast b&=b\ast a\\ (a\ast b)\ast a &= b\\ (a\ast b)\ast((a\ast b)\ast a) &= (a\ast b) \ast b\\ a&=a \end{align}$
Now, read upwards to get a proof (you have to left-multiply by $a\ast b$ to get from line 3 to line 2, and right-multiply by $a$ to get from line 2 to line 1).
Phira
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2Note that it is really impossible not to find the solution with this approach: at the first step, all four possible multiplications are equally useful, at the second step, one of the possibilities takes you back to the beginning and the other one finishes the task. – Phira Jan 01 '14 at 09:25
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This proof cannot be accepted. How do you go back from line 3 to line 2, or from line 2 to line 1 without the cancelation rule? Unless you first prove the cancelation rule. – Farshad Nahangi Oct 30 '23 at 04:12