Question

If $A=\{1,2,3,4\},$ then which among the following relations are not functions from $A$ to itself

• A. $f_1=\left\{\right(x,y)|y=x+1,\forall x,y\in A\}$
• B. $f_2=\left\{\right(x,y)|x+y>4,\forall x,y\in A\}$
• C. $f_3=\left\{\right(x,y)|y<x,\forall x,y\in A\}$
• D. $f_4=\left\{\right(x,y)|x+y=5,\forall x,y\in A\}$
  

Answer 1

Answer: (A), (B), (C)

$f_3=\left\{\right(x,y)|y<x,\forall x,y\in A\}$

$f_1=\left\{\right(x,y)|y=x+1,\forall x,y\in A\}$
$f_1=\left\{\right(1,2),(2,3),(3,4\left)\right\}$
Since element $4$ is mapped to $5$ according to relation but $5\notin A,$ it is not a function.

$f_2=\left\{\right(x,y)|x+y>4,\forall x,y\in A\}$
$f_2=\left\{\right(1,4),(2,3),(2,4),(3,3),(3,4),(4,4),(4,1),(3,2\left)\right(4,2),(4,3\left)\right\}$
Since for $x=2,$ there are two mappings. So, it is not a function.

$f_3=\left\{\right(x,y)|y<x,\forall x,y\in A\}$
$f_3=\left\{\right(2,1),(3,1),(3,2)\cdots \}$
Since for $x=3,$ there are two mappings. so, it is not a function.

$f_4=\left\{\right(x,y)|x+y=5,\forall x,y\in A\}$
$f_4=\left\{\right(1,4),(2,3),(3,2),(4,1\left)\right\}$
Since, Domain $=\{1,2,3,4\}$
Since, every element in $A$ is associated to a unique element in $A.$
Therefore, it is a function from $A$ to itself.