Chapter - 5, Functions

Section - 5.1 - Functions


Summary

  • Function : Suppose is a relation from to . Then is called a function from to if for every there is exactly one such that . In other words, is a function from to means:
    .
    To indicate that is a function from to , we will write .
  • Suppose and are functions from to . If , then . This theorem can be used in proving two functions are equal.
  • As can be seen from the definition of function, funtionsa are special kind of relations. Thus all the theorems of relations applies to functions also. Domain of the function from to must contain each element of the range as first coordinate of an ordered pair in . The elements of the range of will be the second coordinates of all the ordered pairs in . Thus range need not be all of . Also this second coordinate is also sometimes called as image of the first coordinate.
  • As noted earlier, since function is a relation, all definitions that apply to relation also aplly to function. Thus definition of composition of relation also apply here as follows:
    • Composition of function: Suppose and . Then , and for any , the value of at is given by the formula .

Soln1

(a) True

(b) False

(c) True


Soln2

(a) False. Because there is no image for point .

(b) is not a function. Because for eg: “ape” has images. is a function.

(c) True.


Soln3

(a) .

(b) .

(c) .


Soln4

(a) Capital of Italy = Rome.

(b) .

(c) .


Soln5

: It is an identity function on all countries.

: This function maps a city to another city such that is the capital of the country of of the city .


Soln6

.

.
.
.
.

Soln7

(a)

To prove is a function,we need to prove that for every there exists a unique such that .

Existence:

Suppose . Since , . Since , it follows that there exist an element such that . Thus . It follows that . Since and . It follows that .

Uniqueness:

Suppose and . Thus and . Since is a function, it follows that . Thus there exist only one element such that .

Suppose . Thus . Since , . Thus .

(b)

()Suppose . Suppose . Thus . It follows that . Thus .

()Suppose . Suppose . Since , there must exist an element such that . Thus . Since , it follows . Thus . Since and , it follows . Since and , if follows that . Thus . Since is arbitrary, it follows that .

(c)

We have and . By the definition of restiction, we have . Thus . Thus we have . Thus .


Soln8

To prove this we need to prove existence and uniqueness.

Existence:

Since is reflexive, transitive and symmetric, it is a equivalence relation. Also since for each value in there is a unique value \in A such that . Thus is a function and a equivalence relation.

Uniqueness:

Suppose there exists a function say , such that it is a equivalence relation. Suppose . Thus we have:

. iff
Since is symmetric,
. iff
Since is transitive,
. iff
Since is a function and and ,

Since are arbitrary, it follows that .

Thus is the only unique function.


Soln9

(a)

To prove that is a function from to , we need to prove that for all , there exist a image of in . Also we need to prove that for each , this image is unique. Thus we need to do existence and uniqueness proof.

Existence:

Suppose . Thus we have following cases:

  • Case 1: .
    Since and since is a function on . Thus there exist a such that . Thus . It follows that .

  • Case 2: . Since and since is a function on . Thus there exist a such that . Thus . It follows that .

Thus from both cases . Since is arbitrary, it follows that .

Uniqueness:

Suppose and such that and . Thus we have following possible cases:

  • Case 1: and .
    Since is a function, it follows that .

  • Case 2: and .
    Since is a function, it follows that .

  • Case 3: and .
    Thus and . But this is not possible since .

  • Case 4: and .
    Thus and . But this is not possible since .

Thus from all possible cases, . Thus there is only one image of in the set . Thus is a function.

(b)

()Suppose . Suppose . It follows that and . Since , it follows that . Also since , and , there exist an element such that . Since , it follows that . Since , it follows that . Thus . Since and , it follows that . Since are arbitrary, it follows .
Using similar argument it can be proved that . Thus we can conclude that .

()Suppose . To prove , we need to do existence and uniqueness proofs. Clearly in this case existence proof is exactly same as in part(a). Thus we shall only do the uniqueness proof:

Suppose and . Thus we have following possible cases:

  • Case 1: and .
    Since is a function, it follows that .

  • Case 2: and .
    Since is a function, it follows that .

  • Case 3: and .
    Thus . It follows that and . Since and
    , it follows that . Similarly since and
    , it follows that . Since , if follows that and both exist in and . Thus .

  • Case 4: and .
    This case is exactly same as Case 3, just need to swap and in the case 3.

Thus from all the possible cases .

Thus .


Soln10

(a)

Existence:

Suppose . Since , there must exist such that . Since is arbitrary, it follows that for all there exist an element such that .

Uniqueness:

Suppose and such that and . Since , it follows that there must exist an element such that . Thus and . But is a function, it follows that .

Thus we can conclude that is a function.

(b)



.
.
.

Clearly, and are functions but is not a function.


Soln11

(a)

Suppose is reflexive. Suppose is an arbitrary element. Since , thus . Since is relation on and is reflexive, it follows that . Thus . Thus is reflexive.

(b)

Suppose is symmetric. Suppose . Thus . Since is symmetric, it follows that . Thus . Thus symmetric.

(c)

Suppose is transitive. Suppose such that and . Thus and . Since is transitive, it follows that . Thus . Thus is transitive.


Soln12

(a) False.

Counter example:

.

Clearly , because there is no element in such that .

(b) True.

Suppose is symmetric. Suppose . Thus there must exist some such that and there must exist some such that and . Since is symmetric, it follows that . Since and , it follows that . Thus is symmetric.

(c) False.

.
.
.
.

Clearly is transitive. Clearly , because . Similarly , since . But because .


Soln13

(a)

True. Suppose is reflexive. Suppose . Suppose . Since is a function, will be a unique value. Since , and is reflexive, it follows that . Since is arbitrary, . Thus . Thus is reflexive.

(b)

True. Suppose is symmetric. Suppose . Thus . Since is transitive, it follows that . Thus . Thus is symmetric.

(c)

True. Suppose is transitive. Suppose and . Thus and Thus . Since is transitive, it follows that . Thus . Thus is transitive.


Soln14

(a)

Suppose . Thus . Also, . Thus . Since is arbitrary, it follows (using theorem for that talks about two functions are equals if for all values of , corresponding images are also equal).

(b)

Suppose . Since is arbitrary, can be any function. Suppose is a constant function such that . Suppose is an arbitrary element. Thus . Since , it follows . It follows that . Since is a function, is a unique value. Since is arbitrary, it follows that . Thus is a constant function.


Soln15

(a)

Clearly for , we have . Thus .

(b)

Reflexive:

Suppose . Suppose is any real number, clearly for all . Thus . Thus is reflexive.

Transitive

Suppose and . Since , it follows that there is an element such that . Similarly since , it follows that there is an element such that .

Now there are two cases, or . If , then suppose is any arbitrary number. Then and . Thus .
Similarly if , and suppose is any arbitrary number. Then also we have .
Thus in either case there exist a number such that . Thus is transitive.

Symmetric:

Suppose . Thus there is an element such that . Or we can also say that . It follows that . Thus is symmetric.


Soln16

(a)

Suppose . Then for any ,

.

Thus .

We will prove it by contradiction. Suppose . Then for some and , we have . Thus . Suppose is an arbitrary number such that . Thus . Also, since ,we have . Thus . But . Thus we have a contradiction. Thus .

(b)

If a relation is preorder, it means it is reflexive and transitive. Thus we will prove these two properties to prove the relation as preorder.

Suppose . Thus for and , we have . Thus . Thus is reflexive.

Suppose and . Thus there exist such that and . Suppose is the maximum of . Suppose , it follows that . Since is arbitrary, we have , where . Thus . Thus is transitive.

To prove that is not partial order, we shall prove that is not antisymmetric. We will take a counter example to prove is not symmetric. Suppose and . Then and . Thus and but . Thus is not antisymmetric.

(c)

Suppose and . Thus there exist such that and . Suppose is maximum of . Suppose is any arbitrary number. Thus we have and . Now . Thus by triangular inequality, we have .
Since and , it follows that . Thus , where . Thus .


Soln17

(a)

Suppose . Since , . Thus . Thus is reflexive.

Suppose . Thus . Or we can also say . Thus . Thus is symmetric.

Suppose and . Thus . Since , it follows . Thus is transitive.

Since is reflexive, symmetric and transitive, it follows that is an equivalence relation.

(b)

Suppose is an equivalence relation on . Suppose and .

()Suppose . Since is an equivalence relation, it follows that . Thus . Thus .

()suppose . Since , it follows that . Thus . Thus .

Thus from both directions, we have .


Soln18

(a)

Suppose is compatible with .

Suppose .

Existence:

Suppose and such that . Thus there exist an such that and . Since and , it follows that . Thus . Or . Thus such a relation exists.

Uniqueness:

Suppose there is another relation . Suppose such that , and . Suppose . Thus . Since is compatible with , it follows that . Thus there exists a such that and . It follows that (where h is the function we defined). Or . Since are arbitrary, it follows that .

Note: Not confident about the solution.

(b)

Suppose and . Suppose . Since , . Similarly, since , . Since , it follows that . Since is a function, it follows . Thus . Thus if , then . Since are arbitrary, it follows that . Thus is compatible with .


Soln19

(a)

Suppose , then . Thus . Thus , or . Thus . Or we can also say that if then .

Now consider the function .

Now we will prove that is the unique function to satisfy .

Existence:

Suppose . Thus there exists an element and such that . Since is an equivalence relation, it follows that and . Thus , or .

Uniqueness:

Suppose is a function such that . Suppose . Thus . We proved above that if then . Thus . It is also clear that , or . Thus there exist a such . Thus . Since is arbitrary, it follows that .

(b)

Here it will be enough to show just one counter example. Suppose . Clearly . If is a function, it follows that, . Now and . Thus . But this is not correct because . Thus is not a function.