In the previous section, we saw some miscellaneous examples. In this section, we will see a few more miscellaneous examples.
Solved example 17.47
Let A = {1,2,3}. Then show that the number of relations containing (1,2) and (2,3) which are reflexive and transitive but not symmetric is three.
Solution:
1. The fig.17.12 below shows all the 9 possible pairs.
 |
| Fig.17.12 |
2. We need to pick suitable pairs from the above nine and form various relations.
• The relations thus formed should:
♦ be reflexive
♦ be transitive
♦ contain (1,2) and (2,3)
♦ not be symmetric.
3. The pairs in the diagonal yellow cells are to be included because, the relation should be reflexive.
4. The relation should be transitive. Also (1,2) and (2,3) must be present. So (1,3) also must be present. Thus the three red cells will be present.
5. The relation should not be symmetric. So all green cells must be avoided. If we include the green cells, then:
(1,2) will get the symmetric pair (2,1)
(1,3) will get the symmetric pair (3,1)
(2,3) will get the symmetric pair (3,2)
6. So the smallest relation possible is obtained by including yellow and red.
We get: R1 = {(1,1),(2,2),(3,3),(1,2),(1,3),(2,3)}
7. In (5), we wrote that, all greens must be avoided.
In fact, it is true that, we must not include the three greens together. But we can pick (2,1) and (3,2) one by one.
8. Picking (2,1) and adding it to R1, we get:
R2 = {(1,1),(2,2),(3,3),(1,2),(1,3),(2,3),(2,1)}
9. Picking (3,2) and adding it to R1, we get:
R3 = {(1,1),(2,2),(3,3),(1,2),(1,3),(2,3),(3,2)}
10. We cannot add (3,1) to R1. This is because, it will group with (1,2). So we will be forced to add (3,2) to maintain transitivity. This will give:
R4 = {(1,1),(2,2),(3,3),(1,2),(1,3),(2,3),(3,1),(3,2)}
11. Now, (2,3) will group with (3,1) and force us to include (2,1)
This will give:
R5 = {(1,1),(2,2),(3,3),(1,2),(1,3),(2,3),(3,1),(3,2),(2,1)}
This contain all greens.
So R4 and R5 are not possible.
12. In (10), we wrote that, the green (3,1) must never be added.
Can we add the remaining two greens together?
We get: R6q = {(1,1),(2,2),(3,3),(1,2),(1,3),(2,3),(2,1),(3,2)}
Here, (3,2) will group with (2,1) and force us to add (3,1).
So R6q is also not possible.
13. Thus the only possible relations are: R1, R2 and R3.
We can write:
The number of relations which satisfy the given conditions is three.
Solved example 17.48
Show that the number of equivalence relation in the set {1,2,3} containing (1,2) and (2,1) is two.
Solution:
1. The fig.17.13 below shows all the 9 possible pairs.
 |
| Fig.17.13 |
2. We need to pick suitable pairs from the above nine and form various relations.
• The relations thus formed should:
♦ be equivalence relations
♦ contain (1,2) and (2,1).
3. The pairs in the diagonal yellow cells are to be included because, the relation should be reflexive.
4. The pairs (1,2) and (2,1) in the red cells are to be included because, it is a given condition.
5. So the smallest relation is R1 = {(1,1), (2,2),(3,3),(1,2),(2,1)}
This is an equivalence relation.
6. Now there are four pairs remaining. If we add any one of those remaining pairs, we will be forced to add all of them.
For example, if we add (1,3), we get the group: (2,1), (1,3). Then we have to add (2,3) to make it transitive.
As a consequence, we are forced to add (3,2) to make it symmetric.
Also, when (1,3) is added, we have to add (3,1)
7. When the remaining four pairs are added, we get the universal relation.
It is an equivalence relation.
8. So there are two relations which satisfy the given conditions.
R1 and the universal relation.
Solved example 17.49
Show that the number of binary operations on {1,2} having 1 as identity and having 2 as the inverse of 2 is exactly one.
Solution:
1. A binary operation is a function.
• In our present case, the domain is {1,2} × {1,2}.
♦ That means, the domain is {(1,1),(1,2),(2,1),(2,2)}
♦ That means, the only possible input pairs are (1,1),(1,2),(2,1) and (2,2)
• The range is {1,2}
♦ That means, the output will be either 1 or 2
2. Let us see what happens when each of the pairs mentioned in (1) are given as input.
(i) First we input (1,1)
• So the operation is 1∗1
• Given that, 1 is the identity (e).
• Recall that, if e is the identity, then:
a∗e = a = e∗a for all a ∈{1,2}
• So we can write:
1∗1 = 1 = 1∗1 (Here we apply e to the element 1)
• That means:
For the input (1,1), the output is 1
(ii) Next we input (1,2)
• So the operation is 1∗2
Here also, since ‘1’ is the identity, we can write:
2∗1 = 1 = 1∗2 (Here we apply e to the element 2)
• That means:
For the input (1,2), the output is 1
(iii) Next we input (2,1)
• So the operation is 2∗1
We already saw that, for this operation, the result is 1
• That means:
For the input (2,1), the output is 1
(iv) Finally, we input (2,2)
So the operation is 2∗2
•
Here we cannot use the property of identity because, identity is ‘1’. This operation does not involve ‘1’.
•
So we use the property of inverse.
• Given that, 2 is the inverse of 2.
• Recall that, if b is the inverse of a, then:
a∗b = e = b∗a for all a ∈{1,2}
• So we can write:
2∗2 = 1 = 2∗2 (Here we apply the inverse ‘2’ to the element 2)
• That means:
For the input (2,2), the output is 1
3. Now consider the given operation:
The operation has ‘1’ as identity and ‘2’ as the inverse of 2.
•
We will denote this operation as '∗'. We used this operation and found out the outputs for all possible inputs.
4. Our next task is to prove that, ∗ is unique. It can be done in steps.
(i) In step (2), we saw the inputs and the corresponding outputs. We can write them as a set:
∗ = {[(1,1),1], [(1,2),1], [(2,1),1], [(2,2),1]}
(ii) Suppose that, ∗' is another operation having 1 as identity and having 2 as the inverse of 2.
•
Here also, we will get the same set. That means:
∗' = {[(1,1),1], [(1,2),1], [(2,1),1], [(2,2),1]}
•
The same set is obtained because, all possible inputs are already present in ∗.
(iii) Since both sets are same, we can write: ∗ = ∗'.
Therefore, ∗is unique.
Solved example 17.50
Consider the identity function IN : N → N defined as IN(x) = x ∀ x ∈ N.
Show that although IN is onto but IN + IN : N → N defined as
(IN + IN ) (x) = IN (x) + IN (x) = x + x = 2x is not onto.
Solution:
1. Consider the function
IN : N → N defined as IN(x) = x ∀ x ∈ N.
•
Let y be any element in the codomain N
Then we can write: IN(x) = x = y
•
So it is clear that:
♦ If we want any ‘y’ (in the codomain N) to be an image,
♦ we need to pick an ‘x’ (from the domain N) in such a way that,
♦ that ‘x’ is equal to 'y'.
• In this way, all y in the codomain can become an image. So this function is an onto function.
2. Consider the function
IN + IN : N → N defined as
(IN + IN ) (x) = IN (x) + IN (x) = x + x = 2x.
•
Let y be any element in the codomain N
Then we can write: (IN + IN )(x) = 2x = y
• So it is clear that:
♦ If we want any ‘y’ (in the codomain N) to be an image,
♦ we need to pick an ‘x’ (from the domain N) in such a way that,
♦ that ‘x’ is equal to 'y/2'.
•
Suppose that, y = 3. Then y/2 = 1.5.
•
This 1.5 is not present in the domain N. So all elements in the codomain N cannot become images. That means, this function is not an onto function.
Solved example 17.51
Consider a function $f:~\left[0,\frac{\pi}{2} \right] \to R$ given by f(x) = sin x and another function $g:~\left[0,\frac{\pi}{2} \right] \to R$ given by g(x) = cos x. Show that f and g are one-one. But (f+g) is not one-one.
Solution:
1. Both f and g have the same domain.
•
We can use any real number from 0 to $\frac{\pi}{2}$ (both included) as the input.
2. First consider f.
•
In a one-one function, if f(x1) is to be equal to f(x2), then x1 must be equal to x2. This
condition can be used to prove that, a given function is a one-one function.
•
In our present case, suppose that, f(x1) is equal to f(x2). Then we can write:
$\begin{array}{ll}{} &{f(x_1)} & {~=~} &{f(x_2)} &{} \\
{\Rightarrow} &{\sin x_1} & {~=~} &{\sin x_2} &{} \\
{\Rightarrow} &{x_1} & {~=~} &{x_2} &{} \\
\end{array}$
• So f is a one-one function.
◼ Let us see an example:
•
We know that $\frac{\pi}{6}$ lies between 0 and $\frac{\pi}{2}$.
•
If we use $\frac{\pi}{6}$ as the input, we will get:
output = $f \left(\frac{\pi}{6} \right) ~=~\sin \left(\frac{\pi}{6} \right)~=~\frac{1}{2}$
•
Only the input $\frac{\pi}{6}$ will give $\frac{1}{2}$ as the output. We will never find another input value (from 0 to $\frac{\pi}{2}$) which will give $\frac{1}{2}$ as the output.
3. Next, consider g.
•
In a one-one function, if g(x1) is to be equal to g(x2), then x1 must be equal to x2. This
condition can be used to prove that, a given function is a one-one function.
•
In our present case, suppose that, g(x1) is equal to g(x2). Then we can write:
$\begin{array}{ll}{} &{g(x_1)} & {~=~} &{g(x_2)} &{} \\
{\Rightarrow} &{\cos x_1} & {~=~} &{\cos x_2} &{} \\
{\Rightarrow} &{x_1} & {~=~} &{x_2} &{} \\
\end{array}$
• So g is a one-one function.
◼ Let us see an example:
•
We know that $\frac{\pi}{6}$ lies between 0 and $\frac{\pi}{2}$.
•
If we use $\frac{\pi}{6}$ as the input, we will get:
output = $g \left(\frac{\pi}{6} \right) ~=~\cos \left(\frac{\pi}{6} \right)~=~\frac{\sqrt{3}}{2}$
•
Only the input $\frac{\pi}{6}$ will give $\frac{\sqrt{3}}{2}$ as the output.
We will never find another input value (from 0 to $\frac{\pi}{2}$) which
will give $\frac{\sqrt{3}}{2}$ as the output.
4. Finally, we consider (f+g).
•
We have: (f+g)(x) = sin x + cos x
•
We want to prove that, (f+g) is not a one-one function.
•
Let us use 0 as the input. We get:
(f+g)(0) = sin 0 + cos 0 = (0+1) = 1
•
Let us use $\frac{\pi}{2}$ as the input. We get:
$(f+g) \left(\frac{\pi}{2} \right)~=~\sin \left(\frac{\pi}{2} \right)~+~ \cos \left(\frac{\pi}{2} \right)~=~ (1+0)~=~1$
•
So ‘1’ is the image of both 0 and $\frac{\pi}{2}$
Therefore, (f+g) is not a one-one function.
The link below gives a few more solved examples
In the next chapter, we will see inverse trigonometric functions.
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