Hey please help me answer this question: Given a set $A$ with at least 2 elements which on it the binary operation * is defined in that manner:

for every $a,b\in A, a*b=b$.

Check if the binary operation * is commutative, associative and idempotent.

  • $\begingroup$ We've learned so far elementary-set theory and very basic binary operations, I'm not yet familiar with axioms so please avoid using it in your answer. $\endgroup$ – PatentLobster Dec 17 '19 at 19:18
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    $\begingroup$ It should be easy to see it’s not commutative $\endgroup$ – J. W. Tanner Dec 17 '19 at 19:21
  • $\begingroup$ for me its still difficult since I struggle with wrapping my head around it $\endgroup$ – PatentLobster Dec 17 '19 at 19:28
  • $\begingroup$ " I struggle with wrapping my head around it ". That is why they gave you these exercises. But if we do them for you, will it help you wrap your head around it? $\endgroup$ – GEdgar Dec 17 '19 at 19:32
  • $\begingroup$ @GEdgar yes it will since I missed an important class and now Im behind the rest of the class, so now I have to learn it by myself, this is the most simple example from the tasks which I choose because I want to understand this simple stuff before I move any further and an example solution would be very useful for me to understand. $\endgroup$ – PatentLobster Dec 17 '19 at 19:42

An answer emphasizing substitutions

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    $\begingroup$ Thank you so much, No I understand it much better! $\endgroup$ – PatentLobster Dec 17 '19 at 23:11
  • $\begingroup$ You are welcome. $\endgroup$ – Saint James Dec 18 '19 at 14:55

Say $A$ has at least two different elements, name them $5$ and $2$. Then $$5*2 = 2\ne 5 = 2*5$$ so it is not commutative.

Since we have also: $$a*(b*c) = a*c = c$$ and $$(a*b)*c = b*c=c$$

we see it is associative. Also we have $b*b = b$ so ...

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  • $\begingroup$ Im sorry could you explain what: a∗(b∗c)=a∗c=c∧(a∗b)∗c=b∗c=c means? $\endgroup$ – PatentLobster Dec 17 '19 at 19:29
  • $\begingroup$ The confusing part is to write $\wedge$ instead of the word "and". Aqua was very lazy, avoiding 2 letters, but making it confusing. $\endgroup$ – GEdgar Dec 17 '19 at 19:31
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    $\begingroup$ Hello! This is a nitpick, but I think your answer would benefit from using the fact that $A$ has at least two elements to pick distinct $a, b$ from $A$ rather than using $2$ and $5$ in your counterexample (how do you know $2*5$ is even defined? let alone a counterexample to a property about elements of $A$) $\endgroup$ – Izaak van Dongen Dec 17 '19 at 19:32
  • $\begingroup$ Thank you @GEdgar but you are fast on words, wedge is 2 letters more than and :) $\endgroup$ – Aqua Dec 17 '19 at 19:33
  • $\begingroup$ @IzaakvanDongen thanks for exactnes $\endgroup$ – Aqua Dec 17 '19 at 19:35

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