2
$\begingroup$

A binary operation $\circledast$ on a set $X$ is called anticommutative if

  1. $\exists r\in X: x\circledast r = x,\;\; x\in X$ and
  2. $x\circledast y=r\Leftrightarrow (x\circledast y)\circledast(y\circledast x)=r\Leftrightarrow x=y$

I have to prove that an anticommutative bin. operation on $X$ is not commutative and that there is no identity element $e\in X$, if $X$ has more than two elements.

I proved is as follows: Let $\circledast$ be an anticommutative binary operation on a set $X$, which contains two distinct elements $x$ and $y$ with $x\circledast y=y\circledast x$. Because of 2., it follows that $(x\circledast y)\circledast(y\circledast x)=r$, which, since $\circledast$ is anticommutative, is equivalent to $x=y$, contradicting our assumption that $x$ and $y$ are distinct elements.

END OF PROOF

I think that this proves that any anticommutative binary operation on a set that contains at least two elements is not commutative and does not have an identity element $e$. However, it bothers me that the exercise says I should prove the statement for anticom. bin. operations on sets of at least three elements. Is my proof incomplete?

$\endgroup$
1
  • 1
    $\begingroup$ I suspect that more than two elements is simply an error for at least two elements. $\endgroup$ Apr 20, 2016 at 22:34

1 Answer 1

1
$\begingroup$

Your proof looks fine. In this question and this book they say exactly the same things except the condition is that $X$ has more than one element. It looks like perhaps you've got an earlier edition or an incorrect copy of the exercise?

$\endgroup$
3
  • $\begingroup$ Are you sure that the binary operation you described is anticommutative? I thought that from $r\circledast a=a=a\circledast r$ it followed that $(r\circledast a)\circledast(a\circledast r)=r$, which is equivalent to $r=a$; this, to me, looks like a contradiction. Am I wrong? $\endgroup$
    – user312277
    Apr 20, 2016 at 20:59
  • $\begingroup$ @N.Pich: Sorry, of course you're right. My answer was contradictory, since I said your proof was right but then didn't notice that the operation I wrote was commutative, which it can't be. $\endgroup$
    – joriki
    Apr 20, 2016 at 21:37
  • $\begingroup$ @N.Pich: I've corrected the answer. $\endgroup$
    – joriki
    Apr 20, 2016 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy