Chapter - 5, Functions

Section - 5.3 - Inverses of Functions


Summary

  • Suppose . If is one-to-one and onto, then .
  • Suppose is a function from to , and suppose that is a function from to . Then and .
  • Suppose . Then,
    • If there is a function such that then is one-to-one.
    • If there is a function such that then is onto.
  • Suppose . Then the following statements are equivalent.
    • is one-to-one and onto.
    • .
    • There is a function such that and .
  • Suppose , , , and . Then .

Soln1

Required function .


Soln2

Required function .


Soln3

Suppose . Thus we have:
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Thus we have .

Also, we have .
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Thus we have .

Since and , it follows that is one-to-one and onto and .


Soln4

Suppose . Clearly is defined for all values of (cube root of negative number is defined). Thus .

Now, .
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Thus we have .

Also, .
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Thus we have .

Since and , it follows that is one-to-one and onto and .


Soln5

Suppose . Clearly .

We have . Thus .

Also, we have . Thus .

Since and , it follows that is one-to-one and onto and .


Soln6

Suppose is a function from to . Clearly for any value of . Thus we can say that , where and .

Now, .
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Thus we have .

Similarly it can be shown that, .

Since and , it follows that is one-to-one and onto and .

Thus (a) and (b) both are solved.


Soln7

(a)

To prove , we need to prove that . Suppose . Thus . Since is arbitrary, .

(b)

Clearly and . Thus .


Soln8

(a)

Suppose is an arbitrary element of . Let . Then , so and therefore .
Thus, .
Since was arbitrary, we have shown that , so .

(b)

Suppose . Since , it follows . Since , . Thus (using the first half of the proof as required in the solution). It follows .


Soln9

Suppose . Suppose such that . Suppose . Thus . It follows that . Thus There exist an element such that and . Thus there exist an element such that . Since is arbitrary, it follows that . Thus is onto.


Soln10

()Suppose is an arbitrary pair in . Since , it follows . Since , it follows . Thus for some element , and . Since is a function and , it follows . Thus . It follows . Thus .

() Suppose is an arbitrary pair in . Thus and . Since , it follows . Thus for some element , and . Since is a function and , it follows . Thus . Thus .

Since and , it follows that .


Soln11

(a)

Since and and , it follows that is onto by the theorem(3.1 in above summary). It is given that is one-to-one. Thus is one-to-one and onto. Thus .

Thus, .

(b)

Since and and , it follows that is one-to-one by the theorem(3.2 in above summary). It is given that is onto. Thus is one-to-one and onto. Thus .

Thus, .

(c)

Since and and , it follows from theorem(3rd theorem form above summary) that is onto and is one-to-one.

Proof for is not onto:

Suppose is onto. Thus is one-to-one and onto. Thus by theorem, . Applying on both sides, . Since is one-to-one and onto, . Thus . But it contradicts with the given that . Thus we can conclude that is not onto.

Proof for is not one-to-one:

Suppose is one-to-one. Thus is one-to-one and onto. Thus by theorem, . Thus . Since is one-to-one and onto, . Thus . But it contradicts with the given that . Thus we can conclude that is not one-to-one.


Soln12

Suppose is one-to-one. Suppose . Clearly . Now we shall prove . Thus we shall do it by proving existence and uniqueness:

Existence:

Suppose . Thus . It follows that there exists such that . Thus . Thus there exist such that . Since is arbitrary, it follows .

Uniqueness:

Suppose . Suppose . Thus and . Since is one-to-one, it follows . Thus there exists a unique element such that .

Thus .


Soln13

(a)

We have . It follows that . Clearly we have is compatible with , since . Thus it follows directly from Ex18 from sec-5.1, that is a function such that .

(b)

Since , it follows from ex16 from sec5.2 that is one-to-one.

Proof for is onto. Suppose . Since is onto, it follows that there exists an such that . Since and is an equivalence relation, it follows . Thus . Since is arbitrary, it follows that for every , there exists an element . Thus is onto.

(c)

Since is one-to-one and onto, it follows that exists. Suppose . Thus . It follows there exist such that and . Thus . Suppose . Thus , or . Since , it follows . Thus . Since , it follows . Since , it follows .

(d)

It seems like there is a little typo in the question, instead of , it seems it actually should be .

Suppose .

() Suppose . Suppose . Thus there must exist an element such that , or . Thus , which is equivalent to . Since , it follows , or . Thus from part(c) of this solution it follows that . Since , it follows . Since is arbitrary, it follows .

() Suppose . Suppose is an arbitrary element. Thus there must exist an element such that , or . Thus . By part(c) it follows that . Since , it follows . Thus . Since is arbitrary, it follows that .


Soln14

(a)

Suppose . Thus . It follows that there exist an element such that . Thus we have . Since is arbitrary, it follows that for all .

(b)

We know from Sec5.1 Ex-7, . Since , it follows . Also from part(a) of this solution we have . Thus is one-to-one and onto and .


Soln15

Suppose . Thus clearly . Now with , it follows from Soln14(b) that . Since , it follows .


Soln16

(a)

We have . Suppose . Thus we have . Thus . It is a quadratic equation, thus we have . Clearly . Thus .

(b)

Suppose such that where and . Thus we have . Thus .
Similarly . Thus . Thus . Now from Soln14(b) we have .


Soln17

We shall prove by proving that and , where is set of all partial orders and is set of all strict partial order.

Proof of :

Suppose . Thus is strict partial order. Thus and are disjoint. It follows . Since and are disjoint, it follows . Thus .

Proof of :

Suppose . Thus is partial order. Thus . It follows . Since , it follows . Thus .

Thus is one-to-one and onto and .


Soln18

(a)

To prove is an equivalence relation, we need to prove that is reflexive, symmetric and transitive.

Reflexive:

Suppose . Suppose . Thus , or . It follows . Thus is reflexive.

Symmetric:

Suppose such that . Thus there exist an such that . It follows (by applying h on both sides). Similarly taking value corresponding to on both sides, . Thus . Thus is symmetric.

Transitive:

Suppose such that and . Thus for some , we have and . Thus we have . Thus . It follows is transitive.

Thus is an equivalence relation.

(b)

Suppose . Thus there exist an such that . Thus we have . Thus . Thus .

(c)

Suppose and . Thus there exist an such that . Thus . Applying on both sides, . Thus . Thus also has a fixed point = .