Question

Determine whether each of these functions from Z to Z is one-to-one.

$$\begin{gathered} \left(a\right)f\left(n\right)=n-1 \\ \left(b\right)f\left(n\right)=n^2+1 \\ \left(c\right)f\left(n\right)=n^3 \\ \left(d\right)f\left(n\right)=\left[n/2\right] \end{gathered}$$

Short answer

1. One-to-one
2. Not one-to-one
3. One-to-one
4. Not one-to-one.

Step-by-step solution

Step: 1

a.$f\left(n\right)=n-1$

if$\begin{array}{l}n=a,f\left(n\right)=a-1\\ n=b,f\left(n\right)=b-1\end{array}$

Assume

$\begin{array}{l}f\left(a\right)=f\left(b\right)\\ a-1=b-1\\ a=b\end{array}$

The given function is one-to-one.

Step: 2

b.$f\left(n\right)=n^2+1$

if $\begin{array}{ll}n=a,f\left(n\right)=a^2+1& \\ n=b,f\left(n\right)=b^2+1& \end{array}$

Assume

$\begin{array}{l}f\left(a\right)=f\left(b\right)\\ a^2+1=b^2+1\\ a^2=b^2\\ \Rightarrow a=bora=-b\end{array}$

There are two possible outcomes.

Therefore the given function is not one-to-one.

c. $f\left(n\right)=n^3$

$\begin{array}{ll}n=a,f\left(n\right)=a^3& \\ n=b,f\left(n\right)=b^3& \end{array}$

Assume

$\begin{array}{l}f\left(a\right)=f\left(b\right)\\ a^3=b^3\\ \left(a-b\left)\right(a^2+ab+b^2\right)=0\\ \Rightarrow a=b\end{array}$

Therefore the given function is one-to-one.

Step: 3

d.$f\left(n\right)=[n/2]$

$\begin{array}{ll}n=a,f\left(n\right)=[a/2]& \\ n=b,f\left(n\right)=[b/2]& \end{array}$

Assume

role="math" localid="1668412125654" $\begin{array}{l}f\left(a\right)=f\left(b\right)\\ a=2,b=1\\ [a\mathit{/}2]=\left[\mathrm{b}l2\right]\\ 1=1\end{array}$

There are two possible outcomes.

Therefore the given function is not one-to-one