- f : N → N given by f(x) = x2
- f : Z → Z given by f(x) = x2
- f : R → R given by f(x) = x2
- f : N → N given by f(x) = x3
- f : Z → Z given by f(x) = x3
Solution:
(i) f : N → N given by f(x) = x2
For x, y ∈ N => f(x) = f(y) which infers x2 = y2
x = y
Along these lines f is injective.
There are such quantities of co-area which have no picture in space N. Say, 3 ∈ N, yet there is no pre-picture in space of f. to such an extent that f(x) = x2 = 3. f isn’t surjective.
In this way, f is injective however not surjective.
(ii) Given, f : Z → Z given by f(x) = x2
Here, Z = {0, ±1, ±2, ±3, ±4, } f(- 1) = f(1) = 1
Yet, – 1 not equivalent to 1. f isn’t injective.
There are many quantities of co-area which have no picture in space Z.
For instance, – 3 ∈ co-space Z, however – 3 ∉ area Z f isn’t surjective.
In this way, f is neither injective nor surjective.
(iii) f : R → R given by f(x) = x2
f(- 1) = f(1) = 1
Yet, – 1 not equivalent to 1. f isn’t injective.
There are many quantities of co-area which have no picture in space R.
For instance, – 3 ∈ co-space R, however there doesn’t exist any x in area R where x2 = – 3 f isn’t surjective.
In this way, f is neither injective nor surjective.
(iv) f : N → N given by f(x) = x3
For x, y ∈ N => f(x) = f(y) which infers x3 = y3
x = y
In this manner f is injective.
There are many quantities of co-area which have no picture in space N.
For instance, 4 ∈ co-area N, yet there doesn’t exist any x in space N where x3 = 4. f isn’t surjective.
In this way, f is injective however not surjective.
(v) f : Z → Z given by f(x) = x3
For x, y ∈ Z => f(x) = f(y) which infers x3 = y3
x = y
Hence f is injective.
There are many quantities of co-area which have no picture in space Z.
For instance, 4 ∈ co-area N, however there doesn’t exist any x in space Z where x3 = 4. f isn’t surjective.
Thusly, f is injective yet not surjective.