Example

Problem 1: Consider a function defined as :

f : Z → Z b y f x = x 2 + 1 for all x ∈ Z f : Z → Z b y f x = x 2 + 1 for all x ∈ Z

Determine whether the function is an injection?

Solution : We consider two arbitrary elements of the domain set such that :

f x = f y f x = f y

We have deliberately considered a contradictory supposition with respect to the requirement of injectivity. If this supposition yields x = y, then the given function is an injection; otherwise not. Here,

⇒ x 2 + 1 = y 2 + 1 ⇒ x 2 + 1 = y 2 + 1

⇒ x 2 = y 2 ⇒ x 2 = y 2

⇒ x = ± y ⇒ x = ± y

This is not an unique solution. Here, “x” is not uniquely equal to “y”. We conclude that given function is not an injection. As a matter of fact, we can infer a check on our conclusion as,

⇒ f 1 = f - 1 = 2 ⇒ f 1 = f - 1 = 2

Thus, we see that two pre-images relate to one image, which is contradictory to the requirement of an injection.

Increasing and decreasing functions

The fact that function value is different for different arguments has an important bearing on the nature of injection plot. Consider two plots shown in the figure. In the plot shown on the left, a straight line parallel to x-axis intersects the plot only once. In the second plot, a line parallel to x-axis intersects the plot at two points for x = x 1 x = x 1 and x = x 2 x = x 2 . The function represented by second plot is not an injection as two values of arguments map to a single value of function – not two different values as required for an injection function.

Figure 2: Injection graph is either increasing or decreasing.

one-one function (injection)

It means that intersection plot intersects a line parallel to x-axis only once. This is possible only if the function is either (i) continuously increasing or (ii) continuously decreasing. Note the use of word “continuously”. An increasing plot, for example, if drops, then we can always find a line parallel ot x-axis, which intersects it at two points.

Hence, an injection graph is either an increasing or decreasing type. We can associate these characteristics with differential calculus. We can say that :

Either

d y d x > 0 for all x d y d x > 0 for all x

or,

d y d x < 0 for all x d y d x < 0 for all x

As a matter of fact the derivative can be equal to zero for certain values of "x" - not for an interval of "x". Thus, we can write the condition of increasing function : iif function is continuous and

d y d x ≥ 0 for all x ; equality holding for certain values of x d y d x ≥ 0 for all x ; equality holding for certain values of x

Similarly, we can write the condition of decreasing function : iif function is continuous and

d y d x ≤ 0 for all x ; equality holding for certain values of x d y d x ≤ 0 for all x ; equality holding for certain values of x

Example

Problem 2 : Consider a function defined as :

f : R → R b y f x = x 3 + 1 for all x ∈ R f : R → R b y f x = x 3 + 1 for all x ∈ R

Determine whether the function is a surjection?

Solution : We solve the rule for argument “x” as :

y = x 3 + 1 y = x 3 + 1

⇒ x = y − 1 1 / 3 ⇒ x = y − 1 1 / 3

We see that expression on the right hand side is a valid real expression for all values of “y” in "R" i.e co-domain. Hence, given function is an onto function or surjection.

Into function

Problem 3 : Consider a function defined as :

f : R → R b y f x = x 2 + 1 f o r a l l x ∈ R f : R → R b y f x = x 2 + 1 f o r a l l x ∈ R

Determine whether the function is an into function?

Solution : The rule of the function is :

y = x 2 + 1 y = x 2 + 1

The square of a real number is a positive number for all real number. Hence,

⇒ y = x 2 + 1 > 0 ⇒ y = x 2 + 1 > 0

It means that images are only the right half of the real number i.e. from zero to infinity. But, the co-domain of the function is “R”. It means that left half of the co-domain i.e. from negative infinity to less than zero has no image in “A”. Therefore, the given function is an into function.

One – one onto function (Bijection)

Problem 4 : Consider a function defined as :

f : A → B b y f x = x - 2 x - 3 f : A → B b y f x = x - 2 x - 3

Determine domain (A) and co-domain(B) of the function so that it is a bijection.

Solution : For determining domain of the function, we inspect the given rule,

f x = x − 2 x − 3 f x = x − 2 x − 3

We observe that the given rational function is defined for all values of “x” except for x = 3. Hence, its domain is :

Domain = R − { 3 } Domain = R − { 3 }

In order that the given function is a bijection, it should be both an injection and a surjection. For infectivity, we put :

f x = f y f x = f y

⇒ x - 2 x - 3 = y - 2 y - 3 ⇒ x - 2 x - 3 = y - 2 y - 3

⇒ x − 2 y − 3 = x − 3 y − 2 ⇒ x − 2 y − 3 = x − 3 y − 2

⇒ x y − 3 x − 2 y + 6 = x y − 2 x − 3 y + 6 ⇒ x y − 3 x − 2 y + 6 = x y − 2 x − 3 y + 6

⇒ − 3 x − 2 y = − 2 x − 3 y ⇒ − 3 x − 2 y = − 2 x − 3 y

⇒ x = y ⇒ x = y

Hence, function is an injection for the domain as determined above. Now, for surjection we solve the rule of the function for the argument (x),

⇒ y = x − 2 x − 3 ⇒ y = x − 2 x − 3

⇒ x y − 3 y = x − 2 ⇒ x y − 3 y = x − 2

⇒ x y − 1 = 3 y − 2 ⇒ x y − 1 = 3 y − 2

⇒ x = 3 y − 2 y - 1 ⇒ x = 3 y − 2 y - 1

This equation is valid for all real values of “y” except “1”. Hence, function is surjection for all real values of “y” except for “1”. Hence, co-domain for the function to be a surjection is :

Co-domain = R − { 1 } Co-domain = R − { 1 }

Thus, we conclude that the given function is bijection for the domain and co-domain as determined above.

Domain = R − { 3 } Domain = R − { 3 }

Co-domain = R − { 1 } Co-domain = R − { 1 }