Mathematics Revision Notes of Relations for NDA

Cartesian Products of Sets

-Let two non-empty sets `P` and `Q`. The cartesian product `P xx Q` is the set of all ordered pairs of elements from `P` and `Q`,
i.e., `P xx Q = { (p,q) : p ∈ P, q ∈ Q }`

- If either `P` or `Q` is the null set, then `P xx Q` will also be empty set.

Eg., : Consider the two sets `A= {a_1, a_2}` and `B = {b_1, b_2, b_3, b_4}`

`A xx B = {( a_1, b_1), (a_1, b_2), (a_1, b_3), (a_1, b_4), (a_2, b_1), (a_2, b_2),(a_2, b_3), (a_2, b_4)}`.
Q 2677612586

If `P = {a, b, c}` and `Q = {r}`, form the sets `P xx Q` and `Q xx P`. Are these two products equal?

Solution:

By the definition of the cartesian product,

`P xx Q = {(a, r), (b, r), (c, r)}` and `Q xx P = {(r, a), (r, b), (r, c)}`

Since, by the definition of equality of ordered pairs, the pair `(a, r)` is not equal to the pair `(r, a)`, we conclude that `P xx Q ≠ Q xx P.`

However, the number of elements in each set will be the same.

Relations

- A relation between two sets is a collection of ordered pairs containing one object from each set.

- If the object x is from the first set and the object y is from the second set, then the objects are said to be related if the ordered pair (x,y) is in the relation.

or

- A relation `R` from a non-empty set `A` to a non-empty set `B` is a subset of the cartesian product `A xx B`. The subset is derived by describing a relationship between the first element and the second element of the ordered pairs in `A xx B`. The second element is called the image of the first element.

`"Domain : -"` The set of all first elements of the ordered pairs in a relation `R` from a set `A` to a set `B` is called the domain of the relation `R`.

`"Codomain & Range :-"` The set of all second elements in a relation `R` from a set `A` to a set `B` is called the range of the relation R. The whole set B is called the codomain of the relation `R`. Note that range `⊆ `codomain.

Remarks :
(i) A relation may be represented algebraically either by the Roster method or by the Set-builder method.

(ii) An arrow diagram is a visual representation of a relation.




Q 2607612588

Let `A = {1, 2, 3, 4, 5, 6}`. Define a relation `R` from `A` to `A` by `R = {(x, y) : y = x + 1 }`

(i) Depict this relation using an arrow diagram.

(ii) Write down the domain, codomain and range of R.

Solution:

(i) By the definition of the relation, `R = {(1,2), (2,3), (3,4), (4,5), (5,6)}`.

The corresponding arrow diagram is shown in Fig

(ii) We can see that the domain `={1, 2, 3, 4, 5,}`

Similarly, the range `= {2, 3, 4, 5, 6}`

and the codomain `= {1, 2, 3, 4, 5, 6}`.

Types of Relations

`"Empty Relation :"` A relation `R` in a set `A` is called empty relation, if no element of `A` is related to any element of `A`, i.e., `R = φ ⊂ A xx A`.

`"Universal relation :"` A relation `R` in a set `A` is called universal relation, if each element of `A` is related to every element of `A`, i.e., `R = A xx A`.

`=>` Both the empty relation and the universal relation are some times called trivial relations.

Eg.: Let A be the set of all students of a boys school. Show that the relation `R` in A given by `R = {(a, b)` : a is sister of b} is the empty relation and `R′ = {(a, b)` : the difference between heights of a and b is less than 3 meters} is the universal relation.

Solution: Since the school is boys school, no student of the school can be sister of any student of the school. Hence, `R = φ`, showing that `R` is the empty relation. It is also obvious that the difference between heights of any two students of the school has to be
less than `3` meters. This shows that `R′ = A xx A` is the universal relation.


`"(i) Reflexive Relation : "` : A relation in a set A is called reflexive relation if `(a,a) ∈ R` for every element `a ∈ A`. if `(a, a) ∈ R`, for every `a ∈ A`,

`"(ii) Symmetric Relation :"` A relation in a set `A` is called if `(a,b) ∈ R` the `(b,a) ∈ R` for all `a,b ∈ A`
if `(a_1, a_2) ∈ R` implies that `(a_2, a_1) ∈ R`, for all `a_1, a_2 ∈ A`.

`"(iii) Transitive Relation : "` A relation R on a set A is called transitive if whenever (a, b) is in R and (b, c) is in R, then (a, c) is in R. if `(a_1, a_2) ∈ R` and `(a_2, a_3) ∈ R` implies that `(a_1, a_3) ∈ R`, for all `a_1, a_2,a_3 ∈ A.`
Q 2636867772

Let L be the set of all lines in a plane and `R` be the relation in `L `defined as `R = {(L_1, L_2) : L_1` is perpendicular to `L_2`}. Show that `R` is symmetric but neither reflexive nor transitive.

Solution:

`R` is not reflexive, as a line `L_1` can not be perpendicular to itself, i.e., `(L_1, L_1)∉ R`. `R` is symmetric as (L1, L2) ∈ R

`⇒ L_1` is perpendicular to `L_2`

`⇒ L_2` is perpendicular to `L_1`

`⇒ (L_2, L_1) ∈ R`.

`R` is not transitive. Indeed, if `L_1` is perpendicular to `L_2` and `L_2` is perpendicular to `L_3`, then `L_1` can never be perpendicular to `L_3`. In fact, `L_1` is parallel to `L_3`, i.e., `(L_1, L_2) ∈ R, (L_2, L_3) ∈ R` but `(L_1, L_3) ∉ R`.
Q 2626291171

Show that the relation `R` in the set `{1, 2, 3}` given by `R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}` is reflexive but neither symmetric nor transitive.

Solution:

`R` is reflexive, since `(1, 1), (2, 2)` and `(3, 3)` lie in `R`. Also, `R` is not symmetric, as `(1, 2) ∈ R` but `(2, 1) ∉ R`. Similarly, `R` is not transitive, as `(1, 2) ∈ R` and `(2, 3) ∈ R` but `(1, 3) ∉ R`

Equivalence relation

- A relation `R` in a set `A` is said to be an equivalence relation, If `R` is reflexive, symmetric and transitive.

Eg : Let `T` be the set of all triangles in a plane with `R` a relation in `T` given by `R = {(T_1, T_2) : T_1` is congruent to `T_2`}. Show that `R` is an equivalence relation.
Solution : `R` is reflexive, since every triangle is congruent to itself. Further, `(T_1, T_2) ∈ R ⇒ T_1` is congruent to `T_2 ⇒ T_2` is congruent to `T_1 ⇒ (T_2, T_1) ∈ R`. Hence, `R` is symmetric. Moreover, `(T_1, T_2), (T_2, T_3) ∈ R ⇒ T_1` is congruent to `T_2` and `T_2` is congruent to `T_3⇒ T_1` is congruent to `T_3⇒ (T_1, T_3) ∈ R`. Therefore, `R` is an equivalence relation.

Q 2656291174

Show that the relation `R` in the set `Z` of integers given by `R = {(a, b) : 2` divides `a – b`} is an equivalence relation.

Solution:

`R` is reflexive, as `2` divides `(a – a)` for all ` a ∈ Z`. Further, if `(a, b) ∈ R`, then `2` divides `a – b`. Therefore, `2` divides `b – a`. Hence, `(b, a) ∈ R`, which shows that `R` is symmetric. Similarly, if `(a, b) ∈ R` and ` (b, c) ∈ R`, then `a – b` and `b – c` are divisible by
`2`. Now, `a – c = (a – b) + (b – c)` is even (Why?).

So, `(a – c)` is divisible by `2`. This shows that `R` is transitive. Thus, `R` is an equivalence relation in `Z`.

 
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