Question

A relation $R$ is defined on set $A=-2,-1,0,1,2$ such that $R=\{(a,b):\sqrt{a^2}>a>\sqrt{b^2}\}$, then the number of ordered pairs in $R$ is

• A. $0$
• B. $5$
• C. $25$
• D. $10$

Answer 1

The correct option is A $0$

$A=-2,-1,0,1,2$

$R=\left\{\right(a,b):\sqrt{a^2}>a>\sqrt{b^2}\}$

Lets take $(-2,-1),\Rightarrow 2>-2\ngtr 1$

$(-2,0)\Rightarrow 2>-2\ngtr 0$

$(-1,0)\Rightarrow 1>-1\ngtr 0$

$(0,1)\Rightarrow 0\ngtr 0\ngtr 1$

$(2,1)\Rightarrow 2\ngtr 2>1$

Similarily, No ordered pairs satisfy the relation.

$\Rightarrow$ Number of ordered pairs $=0$.