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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
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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
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