CS 390 Solutions to Unit 3 Exercises

pp. 32 - 36

1.22 (a) It is one-to-one but not onto unless $a = 0$. For if $x_{1} + a = x_{2} + a$, i.e. $f_{a}(x_{1}) = f_{a}(x_{2})$, then $x_{1} = x_{2}$. Hence by the definition of one-to-one function =, $f_{a}$ is one-to-one.
If $a \neq 0$, then there is no number that is mapped to 0 by $f_{a}$. In fact there is no nonnegative real number which are mapped by $f_{a}$ to a nonnegative real number less than $a$. Hence the range of $f_{a}$ is the set of real numbers not less than (i.e. greater than or equal to) $a$.

1.57 (a) $f(S \cup T) \subseteq f(S) \cup f(T)$.
Proof: Let $y$ be an element of $f(S \cup T)$. Then there is an element $x$ in $S \cup T$ such that $f(x) = y$. Thus $x \in S$ or $x \in T$.
If $x \in S$, then $f(x) \in f(S)$, and if $x \in T$, then $f(x) \in f(T)$. Hence $y \in f(S) \cup f(T)$.

1.26(b) Since $-1$ is not in ${\cal R}^{+}$, $f(x)$ is defined for every element of the domain and its value is between $0$ and $1$ since $1/(1+x) \leq 1$ for all nonnegative number $x$. Also for every $x$, $1/(1+x)$ is unique. Thus $f$ is a function from ${\cal R}^{+}$ to $\{y \mid 0 < y \leq 1 \}$.
$f$ is onto. For let $y$ be an arbitrary numnber in $\{y \mid 0 < y \leq 1 \}$. Then from $1/(1+x) = y$, $x = 1/y - 1$ is obtained. This $x$ is unique for a given $y$ and it is mapped to $y$ by $f$. Hence $f$ is onto.
$f$ is 1-1. For if $1/(1+x_{1}) = 1/(1+x_{2})$, then $(1+x_{1}) = (1+x_{2})$. Hence $x_{1} = x_{2}$.
Thus $f$ is a 1-1 onto function, that is, it is a bijection.

1.27 Let us first prove that $f^{-1}$ is a bijection.
$f^{-1}$ is onto. For for every $x \in A$, there exists a $y \in B$ such that $y = f(x)$ since $f$ is a function from $A$ to $B$. Hence $x = f^{-1}(y)$, that is, every element $x$ of $A$ is mapped to from an element of $B$ by $f^{-1}$. Hence $f^{-1}$ is onto.
$f^{-1}$ is 1-1. For if $f^{-1} (y_{1})$ = $f^{-1} (y_{2})$, then $y_{1} = f(x)$ and $y_{2} = f(x)$, where $f^{-1} (y_{1}) = x$. Since $f$ is a function, this means that $y_{1} = y_{2}$. Hence $f^{-1}$ is 1-1.
Thus $f^{-1}$ is a bijection.
Now for $(f^{-1})^{-1}$. For any $x \in A$ and any $y \in B$, $(f^{-1})^{-1}(x)$ = $y$ iff $f^{-1}(y) = x$. Also $y = f(x)$ iff $f^{-1}(y) = x$. Hence $(f^{-1})^{-1}(x)$ = $y$ iff $y = f(x)$. Thus for every $x \in A$, $(f^{-1})^{-1}$ and $f$ take the same value. Hence they are equal.