Question

pls solve this ASAP​

Answer 1

(a)  Let a relation $\sf{R:A\to A}$ defined as,

$\longrightarrow\sf{R=\{(a,\ b):|a-b|=3\}}$

which in roster form will be,

$\longrightarrow\sf{R=\{(1,\ 4),\ (4,\ 1)\}}$

We see that R is only symmetric here.

(b)  Let a relation $\sf{R:A\to A}$ defined as,

$\longrightarrow\sf{R=\{(a,\ b):0\leq b-a\leq 1,\ b\geq4\}}$

which in roster form will be,

$\longrightarrow\sf{R=\{(3,\ 4),\ (4,\ 4)\}}$

We see that R is transitive here. Also R is one among the smallest possible transitive relation since $\sf{n(R)=2.}$

(c)  We need no. of equivalence relations containing exactly 5 elements, out of which the 4 elements should be the following:

• $\sf{(1,\ 1)}$

• $\sf{(2,\ 2)}$

• $\sf{(3,\ 3)}$

• $\sf{(4,\ 4)}$

This means $\sf{(a,\ a)\in R\quad\forall\,a\in A.}$ Otherwise the relation won't be reflexive, thereby not being an equivalence relation.  

This implies the 5th element in the relation should be in the form $\sf{(a,\ b)}$ where $\sf{a\neq b}$ and $\sf{a,\ b\in A.}$

Now our relation is surely transitive, since elements $\sf{(a,\ a)}$ and $\sf{(a,\ b)}$ are enough for a relation to be transitive.

But since $\sf{(a,\ b)\in R}$ for $\sf{a,\ b\in A,\quad\!\!a\neq b,}$ it is necessary that $\sf{(b,\ a)\in R}$ otherwise the relation won't be symmetric. But we can't include more element in the relation because it already contained 5 elements.

For example, consider the relation on A,

$\longrightarrow\sf{R=\{(1,\ 1),\ (2,\ 2),\ (3,\ 3),\ (4,\ 4),\ (1,\ 2)\}}$

This relation is reflexive and transitive but not symmetric because $\sf{(2,\ 1)\notin R.}$

We face the same problem if we replace $\sf{(1,\ 2)}$ by any other possible element.

Hence no. of equivalence relations on A having exactly 5 elements is zero.