# Relations and Inverse Relations

1. Mar 8, 2014

### Tsunoyukami

I am having difficulty understanding the following problem. I feel it should be very simple but am unsure how to interpret it.

A relation $R$ is defined on $N$ by $aRb$ if $\frac{a}{b} \in N$. For $c, d \in N$, under what conditions is $c R^{-1} d$? (Exercise 8.6 from Chartrand, Polemni & Zhang's Mathematical Proofs: A Transition to Advanced Mathematics; 3rd edition; page 210).

From page 192 of the same text, "Let $A$ and $B$ be two sets. By a relation $R$ from $A$ to $B$ we mean a subset of $A x B$. That is, $R$ is a set of ordered pairs, where the first coordinate of the pair belongs to $A$ and the second coordinate belongs to $B$. If $(a,b) \in R$, then we say that $a$ is related to $b$ by $R$ and write $a R b$. If $(a,b) \notin R$, then $a$ is not related to $b$ by $R$ and we write $a ~R b$." (Here I have used $~R$ to denote "not $R$" because I was unsure how to cross $R$ out as it appears in the text).

On the next page (193), "Let $R$ be a relation from $A$ to $B$. By the inverse relation of $R$ is meant the relation $R^{-1}$ from $B$ to $A$ defined by
$R^{-1}=\left\{ (b,a) : (a,b) \in R \right\}$."

Given these definitions I am having difficulty understanding what the notation means. For example, I understand what $a R b$ means, but I do not understand what $a R^{-1} b$ means. Because of the way $R^{-1}$ is defined (that is, $R^{-1}=\left\{ (b,a) : (a,b) \in R \right\}$) I am not sure whether this definition is precisely $a R^{-1} b$ or $b R^{-1} a$.

Any help with regard to this would be much appreciated. Thank you in advance!

2. Mar 8, 2014

### micromass

Staff Emeritus
Basically, we have that $aR^{-1}b$ if and only if $bRa$.

3. Mar 8, 2014

### Tsunoyukami

So then the solution to the problem is:

Because $c R^{-1} d$ if $d R C$, and by the way R is defined, $d R c$ if $\frac{d}{c} \in N$. So the condition for $c R^{-1} d$ is that $\frac{d}{c} \in N$.

4. Mar 8, 2014

### micromass

Staff Emeritus
Seems ok!

5. Mar 8, 2014

### HallsofIvy

Staff Emeritus
In particular,and this may be the point of the exercise, $aRb$ and $aR^{-1}b$ if and only if a/b= 1 or -1.