Question

For each function below, determine whether or not the function is injective and whether or not the function is surjective. (You do not need to justify your answer). (a) f:R + R given by f(x) = x2 (b) f:N + N given by f(n) = n2 (c) f: Zx Z → Z given by f(n, k) = n +k

Answer 1

a.Neither injective nor surjective

b.Injective but not surjective

c.Surjective but not injective

Explanation:

Injective function:It is also called injective function.If f(x)=f(y)

Then, x=y

Surjective function:It is also called Surjective function.

If function is onto function then Range of function=Co-domain of function

a.We are given that

`f:R\rightarrow R`

`f(x)=x^2`

It is not injective because

f(1)=1 and f(-1)=1

Two elements have same image.

If function is one-to-one then every element have different image.

Function is not surjective because negative elements have not pre-image in R

Therefore, Co-domain not equal to range.

Given function neither injevtive nor surjective.

b.
`f:N\rightarrow N`

`f(n)=n^2`

If
`f(n_1)=f(n_2)`

`n^2_1=n^2_2`

`n_1=n_2`

Because N={1,2,3,...}

Hence, function is Injective.

2,3,4,.. have no pre- image in N

Therefore, function is not surjective

Because Range not equal to co-domain.

Hence, given function is injective but not Surjective.

c.
`f:Z* Z \rightarrow Z`

f(n,k)=n+k

It is not injective because

f(1,2)=1+2=3

f(2,1)=2+1=3

Hence, by definition of one-one function it is not injective.

For every element belongs to Z we can find pre- image in
`Z* Z`

Hence, function is surjective.