Chapter - 4, Relations

Section - 4.1 - Ordered Pairs and Cartesian Products


Summary

  • Ordered Pair:
    • Lets take an example. Suppose means “x is capital city of country y”. The pair is considered as an ordered pair since .
    • In an ordered pair , is called first coordinate and is called second coordinate.
  • Cartesian Product:
    Suppose and are sets. Then the Cartesian product of and , denoted , is the set of all ordered pairs in which the first coordinate is an element of A and the second is an element of B. In other words,
    .
  • Suppose and are sets, then:
    • .
    • .
    • .
    • .
    • .
  • Suppose and are sets. Then = iff either or .
  • Truth set of a statement:
    Suppose is a statement with two free variables in which ranges over a set and ranges over another set . Then is the set of all assignments to and that make sense in the statement . The truth set of is the subset of consisting of those assignments that make the statement come out true. Thus,
    truth set of

Soln1

(a)

(b)


Soln2

(a) .

(b) .


Soln3

It requires plotting points in the plane. So leaving it :)


Soln4

, , , and

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(a)


Thus, .

(b)


.

Thus,

(c)


Thus, .

(d)


Thus,

(e) and , thus .


Soln5

Proof for:

()Let be an arbitrary element of . Thus is an ordered pair with in first coordinate and second coordinate in . Thus can be either in or in . Thus we have following cases:

  • Case 1: :
    Thus we can say that . It follows that .

  • Case 2: :
    Thus we can say that . It follows that .

Thus from both cases, if , then .

()Suppose . Thus we have two cases:

  • Case 1: :
    Thus and . Since , it follows that . Now since and , it follows that .

  • Case 2: :
    Thus and . Since , it follows that . Now since and , it follows that .

Thus from both cases, if , then .

Thus from both directions .

Proof for:

()Let be an arbitrary element of . Thus and . It follows that , or . Similarly, since , it follows . Thus .

()Let be an arbitrary element of . Thus , it follows that . Similarly, since , it follows that . Thus we have .

Thus from both directions we have .


Soln6

It does not cover all the possible cases. Following cases are missed:

  • Case 1: .
  • Case 2: .

Soln7 .


Soln8

Theorem is correct.

A×(B\C) = (A×B)(A×C)

()Suppose . Thus and . It follows that , but . Thus .

()Suppose . Thus and . Since and , it follows that . Thus and , or . Thus we have .

Since is arbitrary, we can conclude that .


Soln9

()Suppose . Thus and . Since , it follows that and . Since , we can have following possible cases:

  • Case 1: :
    Since and , it follows that . Thus .

  • Case 2: :
    Since and , it follows that . Thus .

Thus from both cases, .

()Suppose . Thus we have following cases:

  • Case 1: .
    Here since and , it follows that . Also since , it follows that . Thus we have, and . It follows that .

  • Case 2: .
    Here since and , it follows that . Also since , it follows that . Thus we have, and . It follows that .

Thus from both cases, .

Thus from both directions, we have .


Soln10

Suppose . Suppose . It follows that . Thus there are following cases:

  • Case 1: :
    Since , it follows that .

  • Case 2: : Since , it follows that .

Thus either , or .


Soln11

(a)

Suppose . Thus there atleast exists an such that . Thus and . Since , it follows . Similarly, since , it follows . Thus . Since is arbitrary, it follows that .

(b)

∪p∈PCp =(∪i∈IAi)×(∪i∈IBi)

()Suppose . Thus exist in atleast one of the set, say such that . Thus and . It follows that and . Thus we have . Since is arbitrary, it follows that .

()Suppose . Thus such that and such that . Thus , or where . Since and , it follows that , or . Now, since and , it follows that . Since is arbitrary, it follows that .

Thus, from both directions we have: .


Soln12

The theorem and proof both are not correct. It does not consider if some of the sets or are empty.