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This is Exercise 4 from page 28 of Analysis I by Amann and Escher. I searched and found two questions that appear to concern this same exercise. However, the first one is cryptic to me and the second one doesn't answer the second part of the question. Therefore I would like to solicit feedback on the correctness and clarity of my attempt.

Exercise:

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My attempt:

Consider elements $x$ and $r$ from $X$, $x \neq r$. We see that $x \circledast x = r$ from (ii), and that $x \circledast r = x$ from (i).

If the operation $\circledast$ were commutative, then we would have $r \circledast x = x \circledast r = x$. Using (ii), we see that

$$(r \circledast x) \circledast (x \circledast r) = x \circledast x = r,$$

which implies that $x = r$. Since this contradicts our assumption that $x$ and $r$ are distinct, we conclude that the operation $\circledast$ is not commutative.

To show that there is no identity element, assume that there were one, and call it $e$. Using (i), we see that $e \circledast r = e $, but because $e$ is the identity, it is also true that $e \circledast r = r$, so we see that $e = r$ and therefore $r$ is the identity element. But we showed above that $r$ doesn't commute with $x$, so $r$ is not the identity, and therefore $\circledast$ has no identity element.

Edit:

I have one more question. I looked at the Wikipedia page for "Anticommutative property", and it says that an anticommutative operation is one for which reversing the order of the two arguments produces the inverse. That definition looks similar to the one copied above; is it exactly the same?

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1 Answers1

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... However, the first one is cryptic to me ...

Not more then yours I guess. Maybe it's confusing when using same variable name in different manner. To clearify the arguments of the first proof, let us rewrite the predications of the defintion:

(i) Existence of right identity: $\exists w\in X:u\circledast w=u$ for all $u\in X$

(ii) $u\circledast v=w\iff(u\circledast v)\circledast(v\circledast u)=w\iff u=v$ for all $u,v\in X$

Back to the the first proof, the author starts to prove if there exists a two-sided identity, there for he considers three assumptions:

By formula (i), one may infer there exists $r_0$ (where we assotiated $r_0$ with $w$) with: $$ x\circledast r_0=x\hspace{1cm} \forall\ x.\ \ (1)$$

For same $r_0$ he supposes an addtional proposition: $$r_0 \circledast x=x \hspace{1cm} \forall\ x.\ \ (2).$$

And from (ii) he concludes: $$x\circledast x=r_0\hspace{1cm} \forall\ x.\ \ (3)$$

Apply (1) and (2) within (3), one my infer $$(x\circledast r_0)\circledast (r_0\circledast x)=r_0\hspace{1cm} \forall\ x.\ \ (4)$$

If we compare (4) with (ii) we found a candidate for (ii) by associating $r_0$ with v and x with u (remember $r_0$ is already our candidate for $w$): $$(u\circledast v)\circledast(v\circledast u)=r_0\iff u=v$$

And can conclude $x=r_0$. But now if our set X has more than one element, then $r_0$ should be $r_0\neq x$ for almost all. So one of our assumptions must be wrong, because we can conclude $x=r_0$ and $r_0\neq x$ at the same time. And (2) is the only proposition that could be wrong, because the rest of our assuptions was given by default and should be true. Hence we can conclude $\neg(r_0 \circledast x=x)$ or $r_0 \circledast x\neq x$. In conclusion, if $r_0$ is right-sided identity, one may infer that $r_0 \circledast x\neq x$ is true, that means $r_0$ is not left-sided identity at the same time.

That the operation is not commutative, which means $\neg(u\circledast v = v\circledast u)$ is true for all u,v $\in X$, follows by counterexample. We already sorted out that $x\circledast r_0=x$ and $r_0\circledast x\neq x$ is true for an arbitrary x, then $x\circledast r_0 \neq r_0\circledast x$ must be true as well. So $u\circledast v = v\circledast u$ can't be true in general $_\square$

and the second one doesn't answer the second part of the question. Therefore I would like to solicit feedback on the correctness and clarity of my attempt.

Your proof is close to the structure of the second one but you make unnecessary or wrong assuptions and using variable names in unclear manner:

Consider elements $x$ and $r$ from $X$... and that $x\circledast r = x$ from (i).

That means you suppose x and r are arbitrary elements and you assume for those arbitrary elements, one may infer $x \circledast r = x$ from (i). But this is wrong, (i) only ensures that there exists minimum one $r$ for all $x$. Moreover you dont need this assumption for the proof.

Suppose an additional proposition that operation is commutative: $$x\circledast y=y\circledast x\hspace{1cm} \forall\ x,y.\ \ (1)$$

From (ii) we know there exists an $r_0\in X$, where: $$u=v\implies u\circledast v=r_0.\hspace{1cm} \forall\ u,v.\ \ (2)$$

For same $r_0$ we know from (ii): $$(u\circledast v)\circledast (v\circledast u)=r_0\implies u=v.\hspace{1cm} \forall\ u,v.\ \ (3)$$

Assume x,y are arbitrary elements of X and: $$x\neq y\hspace{1cm}\ \ (4)$$

Now assign $u:=x\circledast y$ and $v:=y\circledast x$, then with (1) and (2) one may infer $(x\circledast y)\circledast (y\circledast x)=r_0$ is true. With the truth of $(x\circledast y)\circledast (y\circledast x)=r_0$ for an arbitrary x and y we reassign u:=x and v:=y. Then with (3) one may infer that $x=y$ is true. Now we have the same situation as in the first proof, that we run into a contradiction, which means $x\neq y$ and $x=y$. Thats why we can conclude that the additional assumption (1) must be wrong or in orther words $\neg(x\circledast y=y\circledast x)$ is true.

To show that there is no identity element, we know from (i) that $x\circledast r_0=x$. Now suppose $r_0$ is left-sided as well, that means $r_0\circledast x=x$. Then one may infer $r_0\circledast x=x\circledast r_0$. But before we proved also that $x\circledast y\neq y\circledast x$ is true for an arbitrary x and y, particulary its true for $\neg (r_0\circledast x=x\circledast r_0)$, so we run into contradiction again which implies that $r_0\circledast x\neq x$ is true $_\square$

Hölderlin
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