Question

If $X=\{1,2,3,4,5\}$ and $Y=\{1,3,5,7,9\}$, determine which of the following sets represent a relation and also a mapping?

• A. $R_1=\left\{\right(x,y)$:$y=x+2,x\in Y,y\in Y\}$
• B. $R_2=\left\{\right(1,1),(1,3),(3,5),(4,7),(5,9\left)\right\}$
• C. $R_3=\left\{\right(1,1),(2,3),(3,5),(3,7),(5,7\left)\right\}$
• D. $R_4=\left\{\right(1,3),(2,5),(4,7),(5,9),(3,1\left)\right\}$

Answer 1

The correct option is D $R_4=\left\{\right(1,3),(2,5),(4,7),(5,9),(3,1\left)\right\}$

Here we will check all options one by one:

Option A:

$R_1=\left\{\right(x,y):y=x+2,x\in X,y\in Y\}$

$⟹R_1=\left\{\right(1,3),(2,4),(3,5),(4,6),(5,7\left)\right\}$

Since $4$ amd $6$ are the images of $2$ and $4$ respectively but $4$ and $6$ do not belong to $Y$

$\therefore (2,4),(4,6)\notin X\times Y$

Hence $R_1$ is not a relation as well as not a mapping.

Option B:
$R_2$: It is a relation but not a mapping because the element $1$ has two different images.

Option C:

$R_2$: It is a relation but not a mapping because the element $3$ has two different images.

Option D:
$R_4$: It is both a relation and a mapping because every element in $X$ is mapped to the elements in $Y$. Also, every element of $X$ has a one and only one image in $Y$ and every element in $Y$ has its pre-image in $X$. Hence, it is also one-one and onto mapping and hence it is a bijection.