Are the following set of ordered pairs functions? If so examine whether the mapping is injective or surjective.
(i) $\{(x, y): x$ is a person, $y$ is the mother of $x\}$.
(ii) $\{(a, b): a$ is a person, $b$ is an ancestor of $a\}$.
(i) Given set of ordered pair is $\{(x, y): x$ is a person, $y$ is the mother of $x\}$.
It represent a function. Here, the image of distinct elements of $x$ under $f$ are not distinct, so it is not a injective but it is a surjective.
(ii) Set of ordered pairs $=\{(a, b)$ : $a$ is a person, $b$ is an ancestor of $a\}$
Here, each element of domain does not have a unique image. So, it does not represent function.
If the mappings $f$ and $g$ are given by $f=\{(1,2),(3,5),(4,1)\}$ and $g=\{(2,3),(5,1),(1,3)\}$, write $fog$.
$$\begin{aligned} \text{Given that,}\quad f & =\{(1,2),(3,5),(4,1)\} \\ \text{and}\quad g & =\{(2,3),(5,1),(1,3)\} \\ \text{Now,}\quad f \circ g(2) & =f\{g(2)\}=f(3)=5 \\ f \circ g(5) & =f\{g(5)\}=f(1)=2 \\ f \circ g(1) & =f\{g(1)\}=f(3)=5 \\ f \circ g & =\{(2,5),(5,2),(1,5)\} \end{aligned}$$
Let $C$ be the set of complex numbers. Prove that the mapping $f: C \rightarrow R$ given by $f(z)=|z|, \forall z \in C$, is neither one-one nor onto.
$$\begin{aligned} \text{The mapping}\quad & f: C \rightarrow R \\ \text{Given, }\quad & f(z)=|z|, \forall z \in C \\ & f(1)=|1|=1 \\ & f(-1)=|-1|=1 \\ & f(1)=f(-1) \\ \text{But }\quad & 1 \neq-1 \end{aligned}$$
So, $f(z)$ is not one-one. Also, $f(z)$ is not onto as there is no pre-image for any negative element of $R$ under the mapping $f(z)$.
Let the function $f: R \rightarrow R$ be defined by $f(x)=\cos x, \forall x \in R$. Show that $f$ is neither one-one nor onto.
Given function, $f(x)=\cos x, \forall x \in R$
Now, $$f\left(\frac{\pi}{2}\right)=\cos \frac{\pi}{2}=0$$
$$\begin{array}{lc} \Rightarrow & f\left(\frac{-\pi}{2}\right)=\cos \frac{\pi}{2}=0 \\ \Rightarrow & f\left(\frac{\pi}{2}\right)=f\left(\frac{-\pi}{2}\right) \\ \text { But } & \frac{\pi}{2} \neq \frac{-\pi}{2} \end{array}$$
So, $f(x)$ is not one-one.
Now, $f(x)=\cos x, \forall x \in R$ is not onto as there is no pre-image for any real number. Which does not belonging to the intervals $[-1,1]$, the range of $\cos x$.
Let $X=\{1,2,3\}$ and $Y=\{4,5\}$. Find whether the following subsets of $X \times Y$ are functions from $X$ to $Y$ or not.
(i) $f=\{(1,4),(1,5),(2,4),(3,5)\}$
(ii) $g=\{(1,4),(2,4),(3,4)\}$
(iii) $h=\{(1,4),(2,5),(3,5)\}$
(iv) $k=\{(1,4),(2,5)\}$
Given that,
$$\begin{aligned} X & =\{1,2,3\} \text { and } Y=\{4,5\} \\ X \times Y & =\{(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)\} \end{aligned}$$
(i) $f=\{(1,4),(1,5),(2,4),(3,5)\}$
$f$ is not a function because $f$ has not unique image.
(ii) $g=\{(1,4),(2,4),(3,4)\}$
Since, $g$ is a function as each element of the domain has unique image.
(iii) $h=\{(1,4),(2,5),(3,5)\}$
It is clear that $h$ is a function.
(iv) $k=\{(1,4),(2,5)\}$
$k$ is not a function as 3 has not any image under the mapping.