Classify the following functions as injection, surjection or bijection :

Solution:

(i) $f: \mathbf{N} \rightarrow \mathbf{N}$, given by $f(x)=x^{2}$

Injection test:

Let $x$ and $y$ be any two elements in the domain $(\mathbf{N})$, such that $f(x)=f(y)$.

$f(x)=f(y)$

$x^{2}=y^{2}$

$x=y$               (We do not get $\pm$ because $x$ and $y$ are in $\mathbf{N}$ )

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

f(x) = y

$x^{2}=y$

$x=\sqrt{y}$, which may not be in $\mathbf{N}$.

For example, if $y=3$,

$x=\sqrt{3}$ is not in $\mathbf{N}$.

So, f is not a surjection.

So, f is not a bijection.

(ii) $f: \mathbf{Z} \rightarrow \mathbf{Z}$, given by $f(x)=x^{2}$

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

$f(x)=f(y)$

$x^{2}=y^{2}$

$x=\pm y$

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

$f(x)=y$

$x^{2}=y$

$x=\pm \sqrt{y}$ which may not be in Z.

For example, if $y=3$,

$x=\pm \sqrt{3}$ is not in $\mathbf{Z}$.

So, f is not a surjection.

So, f is not a bijection.

(iii) $f: \mathbf{N} \rightarrow \mathbf{N}$, given by $f(x)=x^{3}$

Injection test:

Let x and y be any two elements in the domain (N), such that f(x) = f(y).

$f(x)=f(y)$

$x^{3}=y^{3}$

$x=y$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (N), such that f(x) = y for some element x in N (domain).

$f(x)=y$

$x^{3}=y$

$x=\sqrt[3]{y}$ which may not be in $\mathbf{N}$.

For example, if $y=3$,

$x=\sqrt[3]{3}$ is not in $\mathbf{N}$

So, f is not a surjection and  f is not a bijection.

(iv) $f: \mathbf{Z} \rightarrow \mathbf{Z}$, given by $f(x)=x^{3}$

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y)

$f(x)=f(y)$

$x^{3}=y^{3}$

$x=y$

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

$f(x)=y$

$x^{3}=y$

$x=\sqrt[3]{y}$ which may not be in $\mathbf{Z}$.

For example, if $y=3$,

$x=\sqrt[3]{3}$ is not in $\mathbf{Z}$.

So, f is not a surjection and f is not a bijection.

(v) $f: \mathbf{R} \rightarrow \mathbf{R}$, defined by $f(x)=|x|$

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y)

$f(x)=f(y)$

$|x|=|y|$

$x=\pm y$

So, f is not an injection .

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

$f(x)=y$

$|x|=y$

$x=\pm y \in \mathbf{Z}$

$|x|=y$

$x=\pm y \in \mathbf{Z}$

(vi) $f: Z \rightarrow Z$, defined by $f(x)=x^{2}+x$

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

$f(x)=f(y)$

$x^{2}+x=y^{2}+y$

Here, we cannot say that $x=y$.

For example, $x=2$ and $y=-3$

Then,

$x^{2}+x=2^{2}+2=6$

$y^{2}+y=(-3)^{2}-3=6$

So, we have two numbers 2 and $-3$ in the domain $\mathbf{Z}$ whose image is same as 6 .

So, $f$ is not an injection.

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

$f(x)=y$

$x^{2}+x=y$

Here, we cannot say $x \in \mathbf{Z}$.

For example, $y=-4$.

$x^{2}+x=-4$

$x^{2}+x+4=0$

$x=\frac{-1 \pm \sqrt{-15}}{2}=\frac{-1 \pm i \sqrt{15}}{2}$ which is not in $\mathbf{Z} .$

So, f is not a surjection and  f is not a bijection.

(vii) $f: \mathbf{Z} \rightarrow \mathbf{Z}$, defined by $f(x)=x-5$

Injection test:

Let x and y be any two elements in the domain (Z), such that f(x) = f(y).

$f(x)=f(y)$

$x-5=y-5$

$x=y$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).

$f(x)=y$

$x-5=y$

$x=y+5$, which is in $\mathbf{Z}$.

So, f is a surjection and f is a bijection.

(viii) $f: \mathbf{R} \rightarrow \mathbf{R}$, defined by $f(x)=\sin x$

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

$f(x)=f(y)$

$\sin x=\sin y$

Here, $x$ may not be equal to $y$ because $\sin 0=\sin \pi$.

So, 0 and $\pi$ have the same image 0 .

So, $f$ is not an injection.

Surjection test:

Range of $f=[-1,1]$

Co-domain of $f=\mathbf{R}$

Both are not same.

So, f is not a surjection and f is not a bijection.

(ix) $f: \mathbf{R} \rightarrow \mathbf{R}$, defined by $f(x)=x^{3}+1$

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

$f(x)=f(y)$

$x^{3}+1=y^{3}+1$

$x^{3}=y^{3}$

$x=y$

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

$f(x)=y$

$x^{3}+1=y$

$x=\sqrt[3]{y-1} \in \mathbf{R}$

So, f is a surjection.

So, f is a bijection.

(x) $f: \mathbf{R} \rightarrow \mathbf{R}$, defined by $f(x)=x^{3}-x$

Injection test:

Let x and y be any two elements in the domain (R), such that f(x) = f(y).

$f(x)=f(y)$

$x^{3}-x=y^{3}-y$

Here, we cannot say $x=y$.

For example, $x=1$ and $y=-1$

$x^{3}-x=1-1=0$

$y^{3}-y=(-1)^{3}-(-1)-1+1=0$

So, 1 and $-1$ have the same image 0 .

So, f is not an injection.

Surjection test:

Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain).

$f(x)=y$

$x^{3}-x=y$

By observation we can say that there exist some $x$ in $\mathbf{R}$, such that $x^{3}-\mathrm{x}=\mathrm{y}$.

So, $f$ is a surjection and $f$ is not a bijection.

(xi) $f: \mathbf{R} \rightarrow \mathbf{R}$, defined by $f(x)=\sin ^{2} x+\cos ^{2} x$

$f(x)=\sin ^{2} x+\cos ^{2} x=1$

So, for all elements in the domain, the image is 1.

So, f is not an injection.

Range of f = {1}

Co-domain of f = R

Both are not same.

So, f is not a surjection and  f is not a bijection.

(xii) $f: \mathbf{Q}-\{3\} \rightarrow \mathbf{Q}$, defined by $f(x)=\frac{2 x+3}{x-3}$

Injection test:

Let x and y be any two elements in the domain (Q − {3}), such that f(x) = f(y).

$f(x)=f(y)$

$\frac{2 x+3}{x-3}=\frac{2 y+3}{y-3}$

$(2 x+3)(y-3)=(2 y+3)(x-3)$

$2 x y-6 x+3 y-9=2 x y-6 y+3 x-9$

$9 x=9 y$

$x=y$

So, f is an injection.

Surjection test:

Let y be any element in the co-domain (Q − {3}), such that f(x) = y for some element x in Q (domain).

$f(x)=y$

$\frac{2 x+3}{x-3}=y$

$2 x+3=x y-3 y$

$2 x-x y=-3 y-3$

$x(2-y)=-3(y+1)$

$x=\frac{3(y+1)}{y-2}$, which is not defined at $y=2$

So, f is not a surjection and f is not a bijection.

(xiii) $f: \mathbf{Q} \rightarrow \mathbf{Q}$, defined by $f(x)=x^{3}+1$

Injection test:

Let x and y be any two elements in the domain (Q), such that f(x) = f(y).

$f(x)=f(y)$

$x^{3}+1=y^{3}+1$

$x^{3}=y^{3}$

$x=y$

So, f is an injection .

Surjection test:

Let y be any element in the co-domain (Q), such that f(x) = y for some element x in Q (domain).

$f(x)=y$

$x^{3}+1=y$

$x=\sqrt[3]{y-1}$, which may not be in $\mathbf{Q} .$

For example, if $y=8$,

$x^{3}+1=8$

$x^{3}=7$

$x=\sqrt[3]{7}$, which is not in $\mathbf{Q}$.

So, f is not a surjection and