Q.Let $R$ be a relation on the set $A$ of ordered pairs of positive integers defined by $(x, y) \, R \, (u, v)$ if and only if $xv = yu$. Show that $R$ is an equivalence relation.

An equivalence relation must be reflexive, symmetric and transitive; $R=\{(a,b):a-b\in\mathbb Z\}$ on $\mathbb Z$ satisfies all three, so it is an equivalence relation. (OR: $f(x)=2x-3$ is one-one and onto on $\mathbb R$, hence invertible, with $f^{-1}(x)=(x+3)/2$.)

Definition. A relation $R$ on a set $A$ is called an equivalence relation if it is:

1. Reflexive: $(a,a)\in R$ for every $a\in A$.
2. Symmetric: $(a,b)\in R \Rightarrow (b,a)\in R$.
3. Transitive: $(a,b)\in R$ and $(b,c)\in R \Rightarrow (a,c)\in R$.

Checking $R=\{(a,b): a-b\in\mathbb{Z}\}$ on $\mathbb{Z}$:

Reflexive: For any $a\in\mathbb{Z}$, $a-a=0\in\mathbb{Z}$. So $(a,a)\in R$ for all $a$. Reflexive ✓.

Symmetric: If $(a,b)\in R$, then $a-b\in\mathbb{Z}$. Then $b-a = -(a-b)$, and the negative of an integer is an integer, so $b-a\in\mathbb{Z}$, i.e. $(b,a)\in R$. Symmetric ✓.

Transitive: If $(a,b)\in R$ and $(b,c)\in R$, then $a-b\in\mathbb{Z}$ and $b-c\in\mathbb{Z}$. Then

$a-c = (a-b)+(b-c),$

which is a sum of two integers, hence an integer. So $(a,c)\in R$. Transitive ✓.

Since $R$ is reflexive, symmetric and transitive, $R$ is an equivalence relation.

OR: $f:\mathbb{R}\to\mathbb{R}$, $f(x)=2x-3$ — show invertible and find $f^{-1}$.

A function is invertible iff it is bijective (one-one and onto).

One-one: Suppose $f(x_1)=f(x_2)$. Then $2x_1-3=2x_2-3 \Rightarrow 2x_1=2x_2 \Rightarrow x_1=x_2$. So $f$ is one-one.

Onto: Let $y\in\mathbb{R}$ be any real number. We want $x$ with $f(x)=y$, i.e. $2x-3=y \Rightarrow x=\dfrac{y+3}{2}$, which is a real number for every real $y$. So every $y\in\mathbb{R}$ has a pre-image; $f$ is onto.

Since $f$ is one-one and onto, $f$ is bijective, hence invertible.

Finding the inverse: Let $y=2x-3$. Solve for $x$:

$x = \frac{y+3}{2}.$

Interchanging the roles (writing the inverse as a function of $x$):

$f^{-1}(x) = \frac{x+3}{2}.$

$R$ is reflexive, symmetric and transitive (an equivalence relation); elements related to $1$ are those differing from $1$ by a multiple of $4$.

Reflexive: For any $a\in A$, $|a-a|=0=4\times0$, a multiple of $4$. So $(a,a)\in R$ for all $a$. Reflexive ✓.

Symmetric: If $(a,b)\in R$, then $|a-b|$ is a multiple of $4$. Since $|a-b|=|b-a|$, $|b-a|$ is also a multiple of $4$, so $(b,a)\in R$. Symmetric ✓.

Transitive: If $(a,b)\in R$ and $(b,c)\in R$, then $a-b=4k$ and $b-c=4m$ for integers $k,m$. Adding, $a-c=(a-b)+(b-c)=4(k+m)$, a multiple of $4$, so $|a-c|$ is a multiple of $4$ and $(a,c)\in R$. Transitive ✓.

Since $R$ is reflexive, symmetric and transitive, it is an equivalence relation.

Elements related to 1: we need $x\in A=\{0,1,\dots,12\}$ with $|x-1|$ a multiple of $4$: $x-1\in\{0,4,8,-4,\dots\}$, i.e. $x\in\{1,5,9,13,-3,\dots\}$. Within $[0,12]$, this gives $x=1,5,9$.

$\{x: (x,1)\in R\} = \{1,5,9\}$.

OR: $f:\mathbb{R}-\{3\}\to\mathbb{R}-\{1\}$, $f(x)=\dfrac{x-2}{x-3}$.

One-one: Let $f(x_1)=f(x_2)$: $\dfrac{x_1-2}{x_1-3}=\dfrac{x_2-2}{x_2-3}$.

Cross-multiplying: $(x_1-2)(x_2-3)=(x_2-2)(x_1-3)$

$x_1x_2-3x_1-2x_2+6 = x_1x_2-3x_2-2x_1+6$

$-3x_1-2x_2=-3x_2-2x_1 \Rightarrow -x_1=-x_2 \Rightarrow x_1=x_2$.

So $f$ is one-one.

Onto: For any $y\in\mathbb{R}-\{1\}$, we solve $y=\dfrac{x-2}{x-3}$ for $x$:

$y(x-3)=x-2 \Rightarrow yx-3y=x-2 \Rightarrow x(y-1)=3y-2 \Rightarrow x=\dfrac{3y-2}{y-1}$, well-defined since $y\ne1$.

Also $x\ne3$ always (setting $x=3$ gives $3y-2=3y-3$, i.e. $-2=-3$, impossible), so $x\in\mathbb{R}-\{3\}$ and $f(x)=y$. So $f$ is onto.

Since $f$ is one-one and onto, $f$ is bijective.

Parallelism is reflexive, symmetric and transitive, so it is an equivalence relation; the equivalence class of $y=2x+4$ is all lines $y=2x+c$.

Let $L$ be the set of all lines in the $XY$-plane and $R = \{(l_1, l_2) : l_1 \parallel l_2\}$. (Take a line to be parallel to itself.)

Reflexive: every line is parallel to itself, so $(l, l) \in R$ for all $l \in L$.

Symmetric: if $l_1 \parallel l_2$, then $l_2 \parallel l_1$; so $(l_1,l_2)\in R \Rightarrow (l_2,l_1)\in R$.

Transitive: if $l_1 \parallel l_2$ and $l_2 \parallel l_3$, they all share the same direction, so $l_1 \parallel l_3$; hence $(l_1,l_2),(l_2,l_3)\in R \Rightarrow (l_1,l_3)\in R$.

Since $R$ is reflexive, symmetric and transitive, it is an equivalence relation.

Class of $y = 2x + 4$: two lines are related iff they have the same slope. The line $y = 2x+4$ has slope $2$, so the lines related to it are exactly those of the form $y = 2x + c$ for any real $c$.