Question

Determine whether each of these functions from Z to Z is one-to-one. a) f(n) = n - 1 b) f(n) = n2 + 1 c) f(n) = nº d) f(n) = [n/2]

Answer 1

Answer:  The correct option is

(a) f(n) = n - 1.

Step-by-step explanation:  We are given to determine whether the given functions are one-to-one or not.

We know that a function y = f(x) is one-to-one if and only if

f(x) = f(y)  ⇔  x = y.

That is, any two distinct elements cannot have the same image.

(a) The given function is

$f(n)=n-1.$

Let us consider that

$f(n_1)=f(n_2)\\\\\Rightarrow n_1-1=n_2-1\\\\\Rightarrow n_1=n_2.$

Similarly,

$n_1=n_2\\\\\Rightarrow n_1-1=n_2-1\\\\\Rightarrow f(n_1)=f(n_2).$

So, this function is one-to-one.

(b) The given function is

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

Let us consider that

$f(n_1)=f(n_2)\\\\\Rightarrow n_1^2+1=n_2^2+1\\\\\Rightarrow n_1^2=n_2^2\\\\\Rightarrow n_1=\pm n_2.$

That is, there may be two unequal elements having same image.

For example, f(-1)=(-1)²+1=1+1=2,  f(1)=(1)²+1=1+1=2.

It implies that f(-1)=f(1) but 1 ≠ -1.

So, the given function is not one-to-one.

(c) The given function is

$f(n)=n^0.$

Here, the image of all the elements is 1.

For example, $f(2)=2^0=1,~~f(3)=3^0=1.$

f(2)=f(3)  but  2≠3.

So, more than one element is having the same image and so the function cannot be one-to-one.

(d) The given function is

$f(n)=\left[\dfrac{n}{2}\right].$

Here, we see that

$f(2)=\left[\dfrac{2}{2}\right]=[1]=1,\\\\\\f(3)=\left[\dfrac{3}{2}\right]=[1.5]=1.$

So, f(2)=f(3) but 2≠3.

So, the given function is not one-to-one.

Thus, the correct option is (a).