Given a relation R on set A= {1, 2, 3, 5, 6, 10, 15, 30} such that R = {(a, b): a is divisor of b and a ϵ A, b ϵ A} Show that it is a POSET. Draw its digraph and Hasse Diagram. Compute Maximal and Minimal element of Hasse Diagram. Assume f(x) = x+3, g(x) = x-3 and h(x) = 4x, x ϵ R. Find the composition (i) g o f (ii) h o f and (iii) f o h o g.
Answers
SOLUTION
EVALUATION
ANSWER TO QUESTION : 1
Here the given set is
A= {1, 2, 3, 5, 6, 10, 15, 30}
The relation R is such that
R = {(a, b) | a is divisor of B }
CHECKING FOR REFLEXIVE
Let a ∈ A
Since a is divisor of a
So (a, a) ∈ R
So R is Reflexive
CHECKING FOR ANTISYMMETRIC
Let a, b ∈ A and (a, b) ∈ R and (b, a) ∈ R
(a, b) ∈ R implies a is divisor of b
(b, a) ∈ R implies b is divisor of a
Above gives a = b
Thus (a, b) ∈ R and (b, a) ∈ R implies a = b
So R is antisymmetric
CHECKING FOR TRANSITIVE
Let a, b, c ∈ A
Also let (a, b) ∈ R and (b, c) ∈ R
⇒ a is divisor of b & b is divisor of c
⇒a is divisor of c
⇒(a, c) ∈ R
Thus (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R
R is transitive
Hence R is a poset
ANSWER TO QUESTION : 2
Diagram : Diagram is referred to the attachment
ANSWER TO QUESTION : 3
The least element a is the element such that a ≤ x for all x ∈ A
The greatest element b is the element such that x ≤ b for all x ∈ A
Thus 1 is the minimal element and 30 is the maximal element
ANSWER TO QUESTION : 4
f(x) = x + 3 , g(x) = x - 3 and h(x) = 4x, x ϵ R.
(i) g o f(x)
= g(f(x))
= g(x + 3)
= x + 3 - 3
= x
g o f(x) = x
(ii) h o f(x)
= h(f(x))
= h(x +3)
= 4(x + 3)
= 4x + 12
h o f(x) = 4x + 12
(iii) f o h o g(x)
= f(h(g(x)))
= f(h(x - 3))
= f(4x - 12)
= 4x - 12 + 3
= 4x - 9
f o h o g(x) = 4x - 9
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