Exercise 1.2.8

Question

Here are two important definitions related to a function $f:$ $A ightarrow B .$ The function $f$ is \emph{one-to-one} $(1-1)$ if $a_{1} 
eq a_{2}$ in $A$ implies that $f\left(a_{1}ight) 
eq$ $f\left(a_{2}ight)$ in $B$. The function $f$ is \emph{onto} if, given any $b \in B$, it is possible to find an element $a \in A$ for which $f(a)=b$
  Give an example of each or state that the request is impossible:
  \begin{enumerate}[label=(\alph*)] 
  \item $f: \mathbf{N} ightarrow \mathbf{N}$ that is $1-1$ but not onto.
  \item $f: \mathbf{N} ightarrow \mathbf{N}$ that is onto but not $1-1$.
  \item $f: \mathbf{N} ightarrow \mathbf{Z}$ that is $1-1$ and onto.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Let $f(n) = n + 1$ does not have a solution to $f(a) = 1$
  \item Let $f(1) = 1$ and $f(n) = n - 1$ for $n > 1$
  \item Let $f(n) = n/2$ for even $n$, and $f(n) = -(n+1)/2$ for odd $n$.
   \end{enumerate}

Exercise 1.2.8

Answers

Comments

Exercise 1.2.8

Answers

Comments

Add answer