Consider the following relations R=(x,y)|x,y are real numbers and x=wy for some rational n…

Question

Consider the following relations

$R={(x,y)|x,y$ are real numbers and $x=wy$ for some rational number $w}$;

$S={(\frac{m}{n},\frac{p}{q})|m,n,p$ and $q$ are integers such that $n,q
e 0$ and $qm=pm}$. Then

Accepted Answer

Let's evaluate each relation for the properties of an equivalence relation: reflexivity, symmetry, and transitivity.

Relation R : $R=(x, y) \mid x, y$ are real numbers and $x=w y$ for some rational number $w$.

Reflexivity : For all $x$ in $R$, $x = 1x$. Since 1 is a rational number, every element is related to itself.

Symmetry : For all $x, y$ in $R$, if $x = w y$ for some rational $w$, then $y = \frac{1}{w}x$. However, if $w = 0$, then $\frac{1}{w}$ is undefined, and therefore, $R$ doesn't satisfy symmetry.

Transitivity : If $x = w y$ and $y = v z$ for some rational numbers $w$ and $v$, then $x = (w v)z$. Since the product of rational numbers is rational, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.

Therefore, $R$ is not an equivalence relation on $R$ since it does not satisfy the symmetry property.

Relation S : $S=\left\{\left(\frac{m}{n}, \frac{p}{q}\right) \mid m, n, p\right.$ and $q$ are integers such that $n, q \neq 0$ and $q m=p m\}$

Reflexivity : For all $\frac{m}{n}$, $\frac{m}{n} = \frac{m}{n}$. Since $n \neq 0$ and $m = m$, every element is related to itself.

Symmetry : For all $\frac{m}{n}$, $\frac{p}{q}$, if $q m = p n$, then $n p = m n$. So if $\frac{m}{n}$ is related to $\frac{p}{q}$, then $\frac{p}{q}$ is related to $\frac{m}{n}$.

Transitivity :

$$
\[\begin{gathered} \frac{m}{n} R \frac{p}{q} \text { and } \frac{p}{q} R \frac{r}{s} \ \Rightarrow  m q=n p \text { and } p s=r q \ \Rightarrow  m q \cdot p s=n p \cdot r q \ \Rightarrow  \quad m s=n r \ \Rightarrow  \frac{m}{n}=\frac{r}{s} \Rightarrow \frac{m}{n} R \frac{r}{s} \end{gathered}$$
\]

So if $\frac{m}{n}$ is related to $\frac{p}{q}$ and $\frac{p}{q}$ is related to $\frac{r}{s}$, then $\frac{m}{n}$ is related to $\frac{r}{s}$. The relation $S$ is transitive.

Therefore, $S$ is an equivalence relation on the set of all fractions where denominator is not zero.

In conclusion, the correct answer is

Option C : $S$ is an equivalence relation but $R$ is not an equivalence relation.

Solutions · 1