# If $xRy$ is equivalent to $xy=1$, what is $R^2$ equivalent to?

I don't understand why it is equivalent to $x=z$ and $x\neq0$, and not equivalent to:

$x=z$, $x\neq0$ and $y\neq0$. $y$ can't be equal to $0$ in order for $R^2$ to be true.

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You have asked multiple such questions. Are you just asking to answers for homework problems? – Jebruho Dec 6 '12 at 5:37
just some review questions – smiley Dec 6 '12 at 5:38
what is exactly your questions? – Mj125 Dec 6 '12 at 5:38
Also, it would almost certainly be better for you to provide some of your own work so that we can help you. – Jebruho Dec 6 '12 at 5:39
it's not really a homework question, it's just a theoretical question. – smiley Dec 6 '12 at 5:45

I assume that $R$ is a relation on the reals. By definition, $R$ is the set of all ordered pairs $\langle x,y\rangle$ of reals such that $xy=1$--that is, such that $x$ and $y$ are multiplicative inverses. Note that $x\in\Bbb R$ has a multiplicative inverse if and only if $x\neq 0$. Hence, $R$ is the set of all ordered pairs $\langle x,x^{-1}\rangle$ with $x\in\Bbb R$ and $x\neq 0$.

Now, by definition, $\langle x,z\rangle\in R^2$ if and only if there is some $y$ such that $\langle x,y\rangle,\langle y,z\rangle\in R$. Now, $\langle x,y\rangle\in R$ if and only if $x\neq 0$ and $y=x^{-1}$, and $\langle y,z\rangle\in R$ if and only if $z\neq 0$ and $y=z^{-1}$. Hence, $\langle x,z\rangle\in R^2$ if and only if $\langle x,x^{-1}\rangle,\langle z^{-1},z\rangle\in R$ and $x^{-1}=z^{-1}$, if and only if $x\neq 0$ and $x^{-1}=z^{-1}$, if and only if $x\neq 0$ and $x=z$.

We can't say that $x\:R^2\:z$ if and only if $x=z$ and $x\neq 0$ and $y\neq 0$--after all, where do you see $y$ in the expression $x\:R^2\:z$? I know it seems like we lose some information, but really we're just dropping some redundant information.

Let's look at it another way. Note that for each $x\in\Bbb R$, either $x=0$ or there is a unique $y\in\Bbb R$ such that $xy=1$. Thus, $R$ is not merely a relation, it is a function. When we're dealing with a relation $S$ and have $\langle a,b\rangle\in S$, we typically write $a\:S\:b$, but when $S$ happens to be a function, we instead tend to write $S(a)=b$. In this particular case, $R(x)=x^{-1}$ for all non-$0$ real $x$, which is a function into and onto the non-$0$ reals, and is in fact its own inverse function. Thus, $R^2=R\circ R$ is the identity function on the non-$0$ reals, meaning that for any $x,z\in\Bbb R$, we have $R^2(x)=z$ if and only if $x\neq 0$ (that is, $x$ is in the domain of $R$) and $x=z$ (since $R^2$ is the identity function and $R^2(x)=z$).

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there exist a x and y such that xy = 1 and a z and y such that zy = 1 if that is the case, then x=z and x =/= 0, and y =/= 0 – smiley Dec 6 '12 at 5:47
i don't understand why xy = 1 and yz = 1 is equivalent to x=/=0 and x=z, we kinda lose information – smiley Dec 6 '12 at 5:49
Let me try an alternate approach, which will hopefully elucidate things for you. – Cameron Buie Dec 6 '12 at 5:54

Just to be clear: the relation given means that $x R (R(z))$, i.e., as Cameron explained, there is $y$ such that $yRz$, and $xRy$. You are correct to notice that $y = \frac{1}{x}$ cannot be zero for this to occur; but you also notice that $x \neq 0 \iff \frac{1}{x} \neq 0$. Thus, since $x \neq 0 \implies R(x) \neq 0$ for any $y = R(x)$, there is no need to state this as a separate condition: it is implied by the first one.

(So no, you're not going crazy or anything. It is necessary.)

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