Advertisements
Advertisements
Question
On the set of natural numbers let R be the relation defined by aRb if 2a + 3b = 30. Write down the relation by listing all the pairs. Check whether it is transitive
Advertisements
Solution
Given N = set of natural numbers
R is the relation defined by a R b if 2a + 3b = 30
3b = 30 – 2a ⇒ b = `(30 - 2a)/3` a, b ∈ N
a = 1, b = `(30 - 2)/3 = 28/3 ∉ "N"`
a = 2, b = `(30 - 4)/3 = 26/3 ∉ "N"`
a = 3, b = `(30 - 6)/3 = 24/3` = 8 ∈ N
∴ (3, 8) ∈ R
a = 4, b = `(30 - 8)/3 = 22/3 ∉ "N"`
a = 5, b = `(30 - 10)/3 = 20/3 ∉ "N"`
a = 6, b = `(30 - 12)/3 = 18/3` = 6 ∈ N
∴ (6, 6) ∈ R
a = 7, b = `(30 - 14)/3 = 16/3 ∉ "N"`
a = 8, b = `(30 - 16)/3 = 14/3 ∉ "N"`
a = 9, b = `(30 - 18)/3 = 12/3` = 4 ∈ N
∴ (9, 4) ∈ R
a = 10, b = `(30 - 20)/3 = 10/3 ∉ "N"`
a = 11, b = `(30 - 22)/3 = 8/3 ∉ "N"`
a = 12, b = `(30 - 24)/3 = 6/3` = 2 ∈ N
∴ (12, 2) ∈ R
a = 13, b = `(30 - 26)/3 = 4/3 ∉ "N"`
a = 14, b = `(30 - 28)/3 = 2/3 ∉ "N"`
a = 15, b = `(30 - 30)/3 = 0/3` = 0 ∈ N
When a > 15, b negative and does not belong to N.
∴ R = {(3, 8), (6, 6), (9, 4), (12, 2)}.
Clearly R is transitive since we cannot find elements (a, b), (b, c) in R such that (a, c) ∉ R
APPEARS IN
RELATED QUESTIONS
Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.
Find the inverse relation R−1 in each of the cases:
(i) R = {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}
Determine the domain and range of the relations:
(i) R = {(a, b) : a ∈ N, a < 5, b = 4}
Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b2}. Is the statement true?
(a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R
Justify your answer in case.
Let A = [1, 2, 3, 5], B = [4, 6, 9] and R be a relation from A to B defined by R = {(x, y) : x − yis odd}. Write R in roster form.
R is a relation from [11, 12, 13] to [8, 10, 12] defined by y = x − 3. Then, R−1 is
If (x − 1, y + 4) = (1, 2) find the values of x and y
Let A = {1, 2, 3, 4), B = {4, 5, 6}, C = {5, 6}. Verify, A × (B ∩ C) = (A × B) ∩ (A × C)
Let A = {1, 2, 3, 4), B = {4, 5, 6}, C = {5, 6}. Verify, A × (B ∪ C) = (A × B) ∪ (A × C)
Write the relation in the Roster Form. State its domain and range
R6 = {(a, b)/a ∈ N, a < 6 and b = 4}
Answer the following:
R = {1, 2, 3} → {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} Check if R is reflexive
Let X = {a, b, c, d} and R = {(a, a), (b, b), (a, c)}. Write down the minimum number of ordered pairs to be included to R to make it symmetric
Choose the correct alternative:
Let R be the universal relation on a set X with more than one element. Then R is
Choose the correct alternative:
Let f : R → R be defined by f(x) = 1 − |x|. Then the range of f is
Given R = {(x, y) : x, y ∈ W, x2 + y2 = 25}. Find the domain and Range of R.
Is the given relation a function? Give reasons for your answer.
f = {(x, x) | x is a real number}
If R = {(x, y): x, y ∈ Z, x2 + 3y2 ≤ 8} is a relation on the set of integers Z, then the domain of R–1 is ______.
Let S = {x ∈ R : x ≥ 0 and `2|sqrt(x) - 3| + sqrt(x)(sqrt(x) - 6) + 6 = 0}`. Then S ______.
Let N denote the set of all natural numbers. Define two binary relations on N as R1 = {(x, y) ∈ N × N : 2x + y = 10} and R2 = {(x, y) ∈ N × N : x + 2y = 10}. Then ______.
