Question

Question: Determine whether each of these functions from Z to Z is one-to-one. a) $\mathit{f}\mathbf{\left(}\mathit{n}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathit{n}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}$b) $\mathit{f}\mathbf{\left(}\mathit{n}\mathbf{\right)}\mathbf{=}{\mathit{n}}^{\mathbf{2}}\mathbf{+}\mathbf{1}$ c) $\mathit{f}\mathbf{\left(}\mathit{n}\mathbf{\right)}\mathbf{=}{\mathit{n}}^{\mathbf{3}}$ d) $\mathit{f}\mathbf{\left(}\mathit{n}\mathbf{\right)}\mathbf{=}\lceil n/2\rceil$

Short answer

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

if

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

Assume

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

The given function is one-to-one.

Step: 2

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

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

Assume

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

There are two possible outcomes.

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

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

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

Assume

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

Therefore the given function is one-to-one.

Step: 3

d) $f\left(n\right)=⌈n/2⌉$

$$\begin{gathered} n=a,f\left(n\right)=⌈a/2⌉ \\ n=b,f\left(n\right)=⌈b/2⌉ \end{gathered}$$

Assume

$$\begin{gathered} f\left(a\right)=f\left(b\right) \\ a=2,b=1 \\ ⌈a/2⌉=⌈b/2⌉ \\ 1=1 \end{gathered}$$

There are two possible outcomes.

Therefore the given function is not one-to-one