Question

Q:1. Determine whether each set is a function from X = {1,2,3,4} to Y = {a,b,c,d}.
If it is a function, find its domain and range, draw its arrow diagram, and
determine if it is one-to-one, onto or both.
a) {(1,a,),(2,a),(3,c),(4,b)}
b) {(1,c),(2,a),(3,b),(4,c),(2,d)}
c) {(1,d),(2,d),(4,a)}

Q:2. List all possible functions from A to B, A = {a, b, c}, B = {0, 1}. Also indicate in
each case whether the function is one-to-one, is onto and one-to-one-onto.

Answer 1

Question 77237

1. Determine whether each set is a function from $X = \{1,2,3,4\}$ to $Y = \{a,b,c,d\}$. If it is a function, find its domain and range, draw its arrow diagram, and determine if it is one-to-one, onto or both.

a) $\{(1,a),(2,a),(3,c),(4,b)\}$

b) $\{(1,c),(2,a),(3,b),(4,c),(2,d)\}$

c) $\{(1,d),(2,d),(4,a)\}$

Solution

a) $f$ is function (each $x \in X$ has the only image $y \in Y$), not one-to-one (because $f(1) = f(2) = a$), not onto (because for $d \in Y$ there is not such $x \in X$ that $f(x) = d$). Domain $-X$, range $-\{a, b, c\}$.

b) $f$ is not function (because $f(2) = a$ and $f(2) = d$, i.e. for one $x \in X$ exist two different $y \in Y$).

c) $f$ is function (each $x \in X$ has the only image $y \in Y$), not one-to-one (because $f(1) = f(2) = d$), not onto (because for $b, c \in Y$ there is not such $x \in X$ that $f(x) = b$ and $f(x) = c$). Domain $-\{1, 2, 4\}$, range $-\{a, d\}$.

2. List all possible functions from $A$ to $B$, $A = \{a, b, c\}$, $B = \{0, 1\}$. Also indicate in each case whether the function is one-to-one, is onto and one-to-one-onto.

Solution

1) $f(a) = f(b) = f(c) = 0$ not one-to-one (all $x \in A$ have the same image), not onto (because for $1 \in B$ there is not such $x \in A$ that $f(x) = 1$);

2) $f(a) = f(b) = f(c) = 1$ not one-to-one (all $x \in A$ have the same image), not onto (because for $0 \in B$ there is not such $x \in A$ that $f(x) = 0$);

3) $f(a) = f(b) = 0, f(c) = 1$ not one-to-one (for different $a, b \in A$, $f(a) = f(b)$), onto (because for all $y \in B$ there exists such $x \in A$ that $f(x) = y$);

4) $f(a) = f(c) = 0, f(b) = 1$ not one-to-one (for different $a, c \in A$, $f(a) = f(c)$), onto (because for all $y \in B$ there exists such $x \in A$ that $f(x) = y$);

5) $f(b) = f(c) = 0, f(a) = 1$ not one-to-one (for different $c, b \in A$, $f(c) = f(b)$), onto (because for all $y \in B$ there exists such $x \in A$ that $f(x) = y$);

6) $f(a) = f(b) = 1, f(c) = 0$ not one-to-one (for different $a, b \in A$, $f(a) = f(b)$), onto (because for all $y \in B$ there exists such $x \in A$ that $f(x) = y$);

7) $f(a) = f(c) = 1, f(b) = 0$ not one-to-one (for different $a, c \in A$, $f(a) = f(c)$), onto (because for all $y \in B$ there exists such $x \in A$ that $f(x) = y$);

8) $f(c) = f(b) = 1, f(a) = 0$ not one-to-one (for different $ca, b \in A$, $f(c) = f(b)$), onto (because for all $y \in B$ there exists such $x \in A$ that $f(x) = y$);