Question

Determine whether each of the following functions from Z to Z is one to one and onto.

a.f(n)=n-1

b.f(n)=[n/2]

C.f(n)=n²+1

Answer 1

a)

$$f(n)=n-1, n\in \Z$$

Let $f(n_1)=f(n_2), n_1, n_2\in \Z.$ It means that

$$n_1-1=n_2-1$$

$$n_1=n_2$$

The function $f(n)=n-1$ is one-to-one from $\Z$ to $\Z.$

Let $y=n-1,y\in \Z.$ Then

$$n=y+1$$

We see that $\exist n\in \Z\ \forall y\in \Z.$

The function $f(n)=n-1$ is onto from $\Z$ to $\Z.$

The function $f(n)=n-1$ is one-to-one and onto from $\Z$ to $\Z.$

(b)

$$f(1)=\lceil1/2\rceil=1$$

$$f(2)=\lceil2/2\rceil=1$$

$$1\not=2, \ but\ f(1)=1=f(2)$$

The function $f(n)=\lceil n/2\rceil$ is not one-to-one from $\Z$ to $\Z.$

Let $y=k,y\in \Z.$ Take $n=2k, n\in \Z.$ Then

$$\lceil n/2\rceil=\lceil 2k/2\rceil=\lceil k\rceil=k$$

We see that $\forall y\in \Z\ \exist n\in\Z$ such that $\lceil n/2\rceil=y.$

The function $f(n)=\lceil n/2\rceil$ is onto from $\Z$ to $\Z.$

The function $f(n)=\lceil n/2\rceil$ is onto but is not one-to-one from $\Z$ to $\Z.$

(c)

$$f(1)=(1)^2+1=2$$

$$f(-1)=(-1)^2+1=2$$

$$-1\not=1, \ but\ f(-1)=2=f(1)$$

The function $f(n)=n^2+1$ is not one-to-one from $\Z$ to $\Z.$

$\forall n\in \Z$ the value of $f(n)=n^2+1$ is non negative integer value.

Let $y=-1, y\in \Z.$

We cannot find $n\in \Z$ such that $f(n)=n^2+1=-1.$

The function $f(n)=n^2+1$ is not onto from $\Z$ to $\Z.$

The function $f(n)=n^2+1$ is not ontoand is not one-to-one from $\Z$ to $\Z.$