In the previous section, we saw properties of composite functions. In this section, we will see invertible functions.
First we will see a solved example.
Solved example 17.22
Let f: {1,2,3} → {a,b,c}
be one-one and onto function given by f(1) = a, f(2) = b and f(3) =
c. Show that there exists a function g: {a, b, c} → {1, 2, 3}such that
g∘f = IX and f∘g = IY, where, X = {1, 2, 3} and Y = {a, b, c}.
Solution:
1. We have two sets:
• X = {1,2,3}
• Y = {a,b,c}
2. We have a function which is one-one and onto:
• f: X→Y defined by:
♦ f(1) = a
♦ f(2) = b
♦ f(3) = c
3. We can think about a new function in the reverse order. This function is also one-one and onto:
• g: Y→X defined by:
♦ g(a) = 1
♦ g(b) = 2
♦ g(c) = 3
4. Now we can calculate (g∘f)
• Recall that, (g∘f) means, output of f is used as the input of g.
♦ (g∘f)(1) = g(f(1)) = g(a) = 1
♦ (g∘f)(2) = g(f(2)) = g(b) = 2
♦ (g∘f)(3) = g(f(3)) = g(c) = 3
• We see that:
Whatever input we give for the function (g∘f), the output will be the same. So (g∘f) is an identity function.
• But we must specify the domain for the identity function. We see that, the input values for (g∘f) are taken from the set X. So the domain is X. That means, it is an identity function on set X.
• We can write: (g∘f) = IX.
5. Similarly, we can calculate (f∘g)
• Recall that, (f∘g) means, output of g is used as the input of f.
♦ (f∘g)(a) = f(g(a)) = f(1) = a
♦ (f∘g)(b) = f(g(b)) = f(2) = b
♦ (f∘g)(c) = f(g(c)) = f(3) = c
• We see that:
Whatever input we give for the function (f∘g), the output will be the same. So (f∘g) is an identity function.
• But we must specify the domain for the identity function. We see that, the input values for (f∘g) are taken from the set Y. So the domain is Y. That means, it is an identity function on set Y.
• We can write: (f∘g) = IY.
The result obtained from the above solved example, can be applied in general. It can be written in 2 steps:
1. We have a function f: X→Y
• The function f, is one-one and onto.
2. Then there exists another function g: Y→X such that:
♦ (g∘f) = IX
♦ (f∘g) = IY
The converse is also true. It can be written in 2 steps:
1. There exists two functions: f: X→Y and g: Y→X
• The two functions together satisfy two conditions:
♦ (g∘f) = IX
♦ (f∘g) = IY
2. Then f is both one-one and onto.
Now we can write about invertible function. It can be written in 4 steps:
1. We have a function f: X→Y
2. There exists another function g: Y→X such that:
♦ (g∘f) = IX
♦ (f∘g) = IY
3. Then f is an invertible function.
4. The function g is called inverse of f.
Inverse of f is denoted as: f-1.
5. Note the subscripts X and Y in step(2)
♦ X is the domain of the original function f.
♦ Y is the domain of the inverse function g.
Checking whether a function is invertible or not
This can be written in 4 steps:
1. Consider the condition:
♦ (g∘f) = IX
♦ (f∘g) = IY
2. This condition can be satisfied only if f is one-one and onto.
3. So we can write:
• If f is invertible, then it will be one-one and onto.
• Conversely, if f is one-one and onto, then it will be invertible.
4. This information can be used to prove a function to be invertible.
• All we need to do is: prove that, it is one-one and onto.
• This is especially helpful in situations where we do not need to find f-1.
Solved example 17.23
Let f: N → Y be a function defined as f(x) = 4x + 3, where, Y = {y ∈ N : y = 4x + 3 for some x ∈ N}.
Show that f is invertible. Find the inverse.
Solution:
1. Based on the given data, we can write 5 points:
(i) f is a set. It contains ordered pairs of the form (x,y).
(ii) The domain of f is N. So x can be natural numbers only.
(iii) For every x, the corresponding y can be calculated using the equation y = 4x + 3
(iv) Since y is calculated using this equation, it will also be a natural number.
(v) The y values thus calculated, forms the set Y. Set Y is the codomain.
• The codomain Y will not contain natural numbers like 1, 2, 8, 9 etc., This is because, these numbers do not satisfy the equation y = 4x + 3.
• Since the codomain in this case contain only the outputs, we can say that, codomain is same as the range.
2.Now we have the details about f. So we can write the reverse of f. We will call it g.
g: Y→N
• The details about g can be written in 5 steps:
(i) g is a set. It contains ordered pairs of the form (y,x)
(ii) The domain of g is set Y. The set Y is already formed when function f is defined.
(iii) Each element y in Y was derived from an element x in N.
• So when we work in reverse from Y to N, we can write:
Every element y in Y will have a corresponding element x in N.
(iv) y was derived using the equation: y = 4x + 3
• So when working in reverse, the x corresponding to y can be obtained as: $x = \frac{y-3}{4}$
(v) Based on this, we can define g as follows:
g: Y→N is defined as: $x~=~g(y)~=~ \frac{y-3}{4}$
• The inputs for g are taken from Y. The outputs will be present in N.
3. Now we have both f and g. We can calculate (g∘f)
• (g∘f) means, output of f is used as the input of g.
• The two functions are: f: N→Y and g: Y→N
• We know that, (g∘f) directly connects the domain of f to the codomain of g. So (g∘f) is from N to N.
• We denoted the inputs from N as x. So we are calculating (g∘f)(x).
• We get:
$$\begin{array}{ll}{} &{(g \circ f)(x)} & {~=~} &{g(f(x))} &{} \\
{} &{} & {~=~} &{g(4x+3)} &{} \\
{} &{} & {~=~} &{\frac{[(4x+3)-3]}{4}} &{} \\
{} &{} & {~=~} &{\frac{4x}{4}} &{} \\
{} &{} & {~=~} &{x} &{} \\
\end{array}
$$
• So whatever input x we give, the output will be the same x. That means, (g∘f) is an identity function.
• The inputs are taken from N. So we can write:
(g∘f) = IN.
4. Similarly, we can calculate (f∘g)
• (f∘g) means, output of g is used as the input of f.
• The two functions are: g: Y→N and f: N→Y
• We know that, (f∘g) directly connects the domain of g to the codomain of f. So (f∘g) is from Y to Y.
• We denoted the inputs from Y as y. So we are calculating (f∘g)(y).
• We get:
$$\begin{array}{ll}{} &{(f \circ g)(y)} & {~=~} &{f(g(y))} &{} \\
{} &{} & {~=~} &{f \left(\frac{y-3}{4} \right)} &{} \\
{} &{} & {~=~} &{f \left(4 \times \frac{y-3}{4} + 3 \right)} &{} \\
{} &{} & {~=~} &{f \left(y-3 + 3\right)} &{} \\
{} &{} & {~=~} &{y} &{} \\
\end{array}
$$
• So whatever input y we give, the output will be the same y. That means, (f∘g) is an identity function.
• The inputs for (f∘g) are taken from Y. So we can write:
(f∘g) = IY.
5. Let us write a summary:
• We are given a function f: N → Y
• We wrote the reverse function g: Y → N
• From (3), we got: (g∘f) = IX
• From (4), we got: (f∘g) = IY
• So f is invertible.
• Also, the inverse of f = f-1 = g
• That means:
The inverse of $f(x) = 4x + 3$ is:
$g(y) = \frac{y-3}{4}$
Solved example 17.24
Let Y = {n2 : n∈N} ⊂ N. Consider f: N→Y as f(n) = n2. Show that f is invertible. Find the inverse of f.
Solution:
1. Based on the given data, we can write 5 points:
(i) f is a set. It contains ordered pairs of the form (x,y).
(ii) The domain of f is N. So x can be natural numbers only.
(iii) For every x, the corresponding y can be calculated using the equation y = x2
(iv) Since y is calculated using this equation, it will also be a natural number.
(v) The y values thus calculated, forms the set Y. Set Y is the codomain.
•
The codomain Y will not contain natural numbers like 2, 5, 7, 8 etc.,
This is because, these numbers do not satisfy the equation y = x2, where x is a natural number.
• Since the codomain in this case contain only the outputs, we can say that, codomain is same as the range.
2. Now we have the details about f. So we can write the reverse of f. We will call it g.
g: Y→N
• The details about g can be written in 5 steps:
(i) g is a set. It contains ordered pairs of the form (y,x)
(ii) The domain of g is set Y. The set Y is already formed when function f is defined.
(iii) Each element y in Y was derived from an element x in N.
• So when we work in reverse from Y to N, we can write:
Every element y in Y will have a corresponding element x in N.
(iv) y was derived using the equation: y = x2
• So when working in reverse, the x corresponding to y can be obtained as: $x = \sqrt{y}$
(v) Based on this, we can define g as follows:
g: Y→N is defined as: $x~=~g(y)~=~ \sqrt{y}$
• The inputs for g are taken from Y. The outputs will be present in N.
3. Now we have both f and g. We can calculate (g∘f)
• (g∘f) means, output of f is used as the input of g.
• The two functions are: f: N→Y and g: Y→N
• We know that, (g∘f) directly connects the domain of f to the codomain of g. So (g∘f) is from N to N.
• We denoted the inputs from N as x. So we are calculating (g∘f)(x).
• We get:
$$\begin{array}{ll}{} &{(g \circ f)(x)} & {~=~} &{g(f(x))} &{} \\
{} &{} & {~=~} &{g(x^2)} &{} \\
{} &{} & {~=~} &{\sqrt{x^2}} &{} \\
{} &{} & {~=~} &{x} &{} \\
\end{array}
$$
• So whatever input x we give, the output will be the same x. That means, (g∘f) is an identity function.
• The inputs are taken from N. So we can write:
(g∘f) = IN.
4. Similarly, we can calculate (f∘g)
• (f∘g) means, output of g is used as the input of f.
• The two functions are: g: Y→N and f: N→Y
• We know that, (f∘g) directly connects the domain of g to the codomain of f. So (f∘g) is from Y to Y.
• We denoted the inputs from Y as y. So we are calculating (f∘g)(y).
• We get:
$$\begin{array}{ll}{} &{(f \circ g)(y)} & {~=~} &{f(g(y))} &{} \\
{} &{} & {~=~} &{f \left(\sqrt{y}\right)} &{} \\
{} &{} & {~=~} &{y} &{} \\
\end{array}
$$
• So whatever input y we give, the output will be the same y. That means, (f∘g) is an identity function.
• The inputs for (f∘g) are taken from Y. So we can write:
(f∘g) = IY.
5. Let us write a summary:
• We are given a function f: N → Y
• We wrote the reverse function g: Y → N
• From (3), we got: (g∘f) = IX
• From (4), we got: (f∘g) = IY
• So f is invertible.
• Also, the inverse of f = f-1 = g
• That means:
The inverse of $f(x) = x^2$ is:
$g(y) = \sqrt{y}$
In the next section, we will see a few more solved examples.
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