If R be a relation defined as a R b ⇔ | a | < b, then R is(a) Reflex

1.	If R be a relation defined as a R b ⇔ \| a \| < b, then R is(a) Reflexive only (b) Symmetric only (c) Transitive only (d) Reflexive and transitive but not symmetric
Answer» (c) Transitive only • There exists a real number –2, such that \| –2 \| is not less than –2 as 2 \(\not\leq\)- 2. Thus for all negative, real numbers \(x\), \| \(x\) \| \(\not\leq\) \(x\). Hence (\(x\), \(x\)) ∉ R V real numbers. Hence R is not reflexive. • R is not symmetric since there exist real numbers –2 and 3 such that \| –2 \| ≤ 3 but \| 3 \| \(\not\leq\) - 2, i.e., (–2, 3) ∈ R \(\not\Rightarrow\) (3, –2) ∈ R. • R is transitive since V real numbers a, b, c (a, b) ∈ R, (b, c) ∈ R ⇒ \| a \| ≤ b and \| b \| ≤ c ⇒ \| a \| ≤ b ≤ \| b \| ≤ c (∵ For all \(x\), \| \(x\) \| ≤ \(x\)) ⇒ \| a \| ≤ c ⇒ (a, c) ∈ R

If R be a relation defined as a R b ⇔ | a | < b, then R is(a) Reflexive only (b) Symmetric only (c) Transitive only (d) Reflexive and transitive but not symmetric

Answer»

(c) Transitive only

• There exists a real number –2, such that | –2 | is not less than –2 as 2 \(\not\leq\)- 2.

Thus for all negative, real numbers \(x\), | \(x\) | \(\not\leq\) \(x\).

Hence (\(x\), \(x\)) ∉ R V real numbers.

Hence R is not reflexive.

• R is not symmetric since there exist real numbers –2 and 3 such that | –2 | ≤ 3 but | 3 | \(\not\leq\) - 2,

i.e., (–2, 3) ∈ R \(\not\Rightarrow\) (3, –2) ∈ R.

• R is transitive since V real numbers a, b, c

(a, b) ∈ R, (b, c) ∈ R ⇒ | a | ≤ b and | b | ≤ c

⇒ | a | ≤ b ≤ | b | ≤ c

(∵ For all \(x\), | \(x\) | ≤ \(x\))

⇒ | a | ≤ c

⇒ (a, c) ∈ R

Discussion

No Comment Found

Related InterviewSolutions

In fig. shows a relationship between the sets P and Q. Write this relation in :a. Set builder form b. Roster form c. What is its domain and range?
Let A = {1, 2, 3, 4, 5, 6}. Let R be a relation on A defined by R = {(a, b): a, b A, b is exactly divisible by a} i. Write R in roster form ii. Find the domain of R iii. Find the range of R.
Let `A = {1, 2, 3, 4, 6}`. Let R be the relation on A defined by `{(adot b): a , b in A , b`is exactly divisible by a}.(i) Write R in roster form(ii) Find the domain of R(iii) Find the range of R.
Let `A={1,2,3,5}" and "B={4,6,9}.` Define a relation from A to B, given by `R={(a,b):a inA,b inB" and "(a-b)" is odd"}.` (i) Write R in roster form. (ii) Find dom (R) and range (R).
Show that the relation R on the set Z of integers, given by R = {(a, b) : 2 divides a – b}, is an equivalence relation.
Let W denote the words in the English dictionary. Let the relation R be defined by R = {(x, y) ∈ W × W : the words x and y have at least one letter in common}. Then R is (a) Reflexive and transitive, not symmetric (b) Reflexive and symmetric and not transitive (c) Symmetric and transitive, not reflexive (d) Reflexive, symmetric and transitive
Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, the minimum number of ordered pairs which can be added to R to make it an equivalence relation is(a) 2 (b) 3 (c) 5 (d) 7
If R be a relation defined as a R b ⇔ | a | < b, then R is(a) Reflexive only (b) Symmetric only (c) Transitive only (d) Reflexive and transitive but not symmetric
Let I be the set of integers and R be a relation on I defined by R = {(x, y) : (x – y) is divisible by 11, x, y ∈ I}. Then R is(a) An equivalence relation (b) Symmetric only (c) Reflexive only (d) Transitive only
Let Z be the set of integers. Then the relation R = {(a, b) : a, b ∈ Z and (a + b) is even} defined on Z is(a) Only symmetric (b) Symmetric and transitive only (c) An equivalence relation (d) None of the above

If R be a relation defined as a R b ⇔ | a | &lt; b, then R is(a) Reflexive only (b) Symmetric only (c) Transitive only (d) Reflexive and transitive but not symmetric

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment

If R be a relation defined as a R b ⇔ | a | < b, then R is(a) Reflexive only (b) Symmetric only (c) Transitive only (d) Reflexive and transitive but not symmetric