Question

Check the injectivity and surjectivity of the following functions: (i) f : N → N given by f ( x ) = x 2 (ii) f : Z → Z given by f ( x ) = x 2 (iii) f : R → R given by f ( x ) = x 2 (iv) f : N → N given by f ( x ) = x 3 (v) f : Z → Z given by f ( x ) = x 3

Answer 1

(i)

The given function f:N→N is defined by f(x)= x 2 .

Assume that for all x,y∈N,

f( x )=f( y ) ⇒ x 2 = y 2 ⇒x=y

Therefore, f is injective.

For 5∈N, there does not exist any natural number x such that,

f( x )= x 2 =5

Therefore, f is not surjective.

Hence, the given function is injective but not surjective.

(ii)

The given function f:Z→Z is defined by f(x)= x 2 .

It can be observed that,

f( −x )=f( x ) ∀x∈Z but −x≠x.

Therefore, f is not injective.

For −5∈Z, there does not exist any integer x such that ,

f( x )= x 2 =−5

Therefore, f is not surjective.

(iii)

The given function f:R→R is defined by f(x)= x 2 .

It can be observed that,

( −x ) 2 = ( x ) 2  ∀x∈R but −x≠x.

Therefore, f is not injective.

For −5∈R, there does not exist any real number x such that,

f( x )= x 2 =−5

Therefore, f is not surjective.

(iv)

The given function f:N→N is defined by f(x)= x 3 .

Let, f( x )=f( y )

⇒ x 2 = y 2 ⇒x=y

For all x,y∈N.

Therefore, f is injective.

For, 4∈N, there does not exist any real number x such that,

f( x )= x 3 =4

Therefore, f is not surjective.

(v)

The given function f:Z→Z is defined by f(x)= x 3 .

Let, f( x )=f( y )

⇒ x 3 = y 3 ⇒x=y

For all x,y∈Z

Therefore, f is injective.

For, 4∈Z, there does not exist any integer x such that,

f( x )= x 3 =4.

Therefore, f is not surjective.