# Binary operation

Let binary operation $\circ$ on set $X$ be function $\circ : X \times X \rightarrow X$.

Binary operation on set X is :

• unitary if for some element $1 \in X$ and any $x \in X$ we've got $(1 \circ x) = x = (x \circ 1)$

• alternate if for any $x,y \in X$ we've got $x \circ y = y \circ x$.

Show that:

1) $x \circ (y \circ x) = y \implies (x \circ y) \circ x = y$

2) Let $\circ$ and $\star$ be binary operation on $X$. $\circ$ and $\star$ are unitary and for any $a,b,c,d \in X$ we've got $(a \circ b) \star (c \circ d) = (a \star c) \circ (b \star d)$. Show that $\star$ and $\circ$ are alternate and identical.

Thanks for help.

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FYI, "alternate" is more commonly called "commutative" or "abelian". –  Chris Eagle Jun 30 '13 at 19:06
Thanks :D I used google translate :) –  JohnCina51 Jun 30 '13 at 19:28
What have you tried? –  Vectk Jun 30 '13 at 19:33

Here’s an explicit, fully-defined example showing that (1) is false in general.

$$\begin{array}{c|cc} \circ&a&b\\ \hline a&b&b\\ b&a&b \end{array}$$

Then

$$a\circ(b\circ a)=a\circ a=b\;,$$

but

$$(a\circ b)\circ a=b\circ a=a\;.$$

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Number 1 is false. Let $X = \{1, 2, 3, 4, 5\}$

We can have $1 \circ (2 \circ 1) = 2$ with $2 \circ 1 = 3$, but $(1 \circ 2) \circ 1 = 5$ with $1 \circ 2 = 5$.

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How do you know the values of the result of operating by $\circ$? –  triomphe Jun 30 '13 at 19:05
@MLT The user didn't mention anything about the binary operation. I'm defining it however I want. –  user84510 Jun 30 '13 at 19:14
@MLT You cannot prove this implication without additional assumptions on $\circ$. I have defined a binary operation that doesn't satisfy it. –  user84510 Jun 30 '13 at 19:25
Well, you haven't defined it, but you started defining it. :) –  Thomas Andrews Jul 23 '13 at 20:48

2) Let $a=c=1$ then, $$b\star c=c \circ b \qquad(1)$$ Then you can use this identity (1) to change order as $$(b \star a) \star (d \star c) = (b \star d) \star (a \star c).$$ Then let $b=c=1$ so, $$a \star d = d \star a,$$ which implies that $\star$ is alternative. You can prove for $\circ$ similarly. Now that $\star$ is alternative we can rewrite (1) as $$c\star b=c \circ b,$$which proves that $\star$ and $\circ$ are identical.

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Be careful - there might be different $1$ elements for $\star$ and $\circ$." –  Thomas Andrews Jul 23 '13 at 20:41
Which $1$ do you mean when you set $a=c=1$? –  Thomas Andrews Jul 23 '13 at 20:47