Let $A=\{1,2,3\}, \,B=\{4,5,6,7\}$ and let $f=\{(1,4),\,(2,5),\,(3,6)\}$ be a function from $A$ to $B$. Show that $f$ is one-one.

(N/A) It is given that $A=\{1,2,3\}$ and $B=\{4,5,6,7\}$.
The function $f: A \rightarrow B$ is defined by the set of ordered pairs $f = \{(1,4), (2,5), (3,6)\}$.
From the definition of the function,we have:
$f(1) = 4$
$f(2) = 5$
$f(3) = 6$
$A$ function $f: A \rightarrow B$ is said to be one-one (injective) if distinct elements of $A$ have distinct images in $B$. That is,$f(x_1) = f(x_2) \implies x_1 = x_2$ for all $x_1, x_2 \in A$.
Here,the images of the distinct elements $1, 2, 3$ are $4, 5, 6$ respectively,which are all distinct.
Since $f(1) \neq f(2)$,$f(2) \neq f(3)$,and $f(1) \neq f(3)$,it follows that the function $f$ is one-one.