Question

If $Z$ is the set of integers. Then, the relation $R=\left\{\left(a,b\right):1+ab>0\right\}$ on $Z$ is

• A. reflexive and transitive but not symmetric
• B. symmetric and transitive but not reflexive
• C. reflexive and symmetric but not transitive
• D. an equivalence relation

Answer 1

The correct option is C

reflexive and symmetric but not transitive

Explanation for the correct option.

Step 1. Check whether the relation is reflexive.

A relation $R$ in $Z$ is reflexive if every ordered pair $\left(a,a\right)$ where $a\in Z$ is in $R$.

It is given that relation $R$ is defined as: $R=\left\{\left(a,b\right):1+ab>0\right\}$.

Now, the ordered pair $\left(a,a\right)$ is in relation $R$ if $1+a·a>0$, so $1+a^2>0$ is always true

As square of an integer is always non-negative so $1+a^2>0$ is always true.

Hence, for every $a\in Z$ there is an ordered pair $\left(a,a\right)$ in relation $R$. So the relation is reflexive.

Step 2. Check whether the relation is symmetric.

A relation is symmetric if for every ordered pair $\left(a,b\right)$in the relation there is also an ordered pair $\left(b,a\right)$ in it.

If $\left(a,b\right)$ is in ordered pair satisfying the relation $R=\left\{\left(a,b\right):1+ab>0\right\}$ then $1+ab>0$.

Now it is known that multiplication is commutative so $ab=ba$.

And so if $1+ab>0$ is true then $1+ba>0$ is also true.

And thus $\left(b,a\right)$ is also in the relation $R$.

So there is an ordered pair $\left(b,a\right)$ in $R$ for every $\left(a,b\right)$ in $R$. So the relation is symmetric.

Step 3. Check whether the relation is transitive.

A relation is transitive if for every $\left(a,b\right),\left(b,c\right)$ there is an $\left(a,c\right)$ in $R$.

The ordered pair $\left(-1,0\right)$ is in relation $R$ because $1+\left(-1\right)\times 0>0$ is true.

The ordered pair $\left(0,2\right)$ is in relation $R$ because $1+0\times 2>0$ is true.

Now,

$$\begin{gathered} 1+\left(-1\right)\times 2>0 \\ \Rightarrow 1-2>0 \\ \Rightarrow -1>0 \end{gathered}$$

As $-1>0$ is a false statement so the inequality $1+\left(-1\right)\times 2>0$ is false and so the ordered pair $\left(-1,2\right)$ does not belong to relation $R$.

So relation $R$ contains $\left(-1,0\right)$ and $\left(0,2\right)$ but not $\left(-1,2\right)$. So relation $R$ is not transitive.

Thus, the relation $R$ defined as $R=\left\{\left(a,b\right):1+ab>0\right\}$ is reflexive and symmetric but not transitive.

Hence, the correct option is C.