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Let * be an operation on a set S where the operation * has commutative property but not associative property.

Is $((a*b)*c)*d = ((d*a)*b)*c$ true?

platoDev
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  • what does $\ast$ means? usual multiplication? convolution? – Marcos Mar 24 '22 at 13:36
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    What does $a\ast b\ast c\ast d$ mean without associativity? – ajr Mar 24 '22 at 13:37
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    "by associativity" "by commutativity"... What is $$ in the first place? If it is an arbitrary binary operation, then it need not necessarily be associative or commutative to begin with. If it is not associative, then what does $abcd$ mean to begin with? The notation is underspecified. If $$ was only defined as a binary operation, and not a quaternary operation... then $abcd$ is undefined and should have been written with enough parentheses to make it clear in what order it is being applied, e.g. as $((ab)c)*d$. – JMoravitz Mar 24 '22 at 13:39
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    If you do mean for $$ to be the usual multiplication... then what do you mean "without associative property"? The usual multiplication is* associative, you can't take that away from it. – JMoravitz Mar 24 '22 at 13:40
  • @JMoravitz $$ is the usual multiplication on real numbers. I was not trying to take associative property away from the usual multiplication. I was just wondering if $abcd$ = $(bcd)*a could be true if it were not to have associative property. Also, I will edit the question to clarify the order of the multiplications – platoDev Mar 24 '22 at 13:42
  • "Can it be true" is not the same thing as "Is it always true". Can it? Yes, the real multiplication is an example. Is it always? No, certainly not. – JMoravitz Mar 24 '22 at 13:48
  • @JMoravitz I edited the question to reflect your comment. – platoDev Mar 24 '22 at 13:53
  • @JMoravitz I will edit the question so that the operation is an arbitrary binary operation with commutative property but without associative property. – platoDev Mar 24 '22 at 13:56
  • Let $$ be a binary operation on ${0,1,2}$ such that $xy = \begin{cases}0&\text{if }x=2~\text{or }y=2\1&\text{otherwise}\end{cases}$. Then note that $xy$ is commutative but not associative. Here, let $a=b=c=1$ and $d=2$, you would have $((ab)c)d=((11)1)2=0$ but $((da)b)c=((21)1)*1=1$ – JMoravitz Mar 24 '22 at 13:58
  • How about this meaning: Take the usual list of axioms for the real numbers. https://en.wikipedia.org/wiki/Construction_of_the_real_numbers#Axioms Omit the associative law for multiplication. From that can we prove $((ab)c)d = ((da)b)c)$? – GEdgar Mar 24 '22 at 14:00
  • @JMoravitz thank you for your answer and your patience. – platoDev Mar 24 '22 at 14:01

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