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Math127121 assignment 5solutions
Course: Discrete Mathematics (Mathematics 1271)
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University: Lakehead University
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Math 1271 Assignment 5
Solutions
1. Let Sbe the relation {(a, b)(b, b),(d, c),(a, d)}.Find the converse (inverse) of S.
Solution: The converse of Sis {(x, y)|(y, x)∈S}={(b, a),(b, b),(c, d),(d, a)}.
2. For each of the following relations on Z,determine which of the properties
reflexive, symmetric, transitive, and anti-symmetric it has.
R={(x, y)∈Z|x≤y2}
S={(x, y)∈Z|x−yis divisible by 3}
T={(x, y)∈Z|y≤x−1}
Solution: For any x∈Z,we have x≤x2,so xRx. Thus Ris reflexive. For any
x∈Z, x −x= 0,which is divisible by 3,so xSx. Thus Sis reflexive. For no x∈Z
is it the case that x≤x−1,so xT x never holds, and Tis not reflexive. We have
1≤(2)2but 2 6≤ (1)2,so Ris not symmetric. If x−yis divisible by 3, then so is
−(x−y) = y−x, so xSy implies ySx. Thus Sis symmetric. We have 1 ≤2−1,but
26≤ 1−1,so Tis not symmetric. We have −1≤1 = (1)2= (−1)2,so (−1)R(1)
and (1)R(−1),but −16= 1,so Ris not antisymmetric. We have that 6 −3 = 3 and
3−6 = −3 are both divisible by 3,but 6 6= 3,so Sis not antisymmetric. It is never
the case that y≤x−1 and x≤y−1 both hold, so Tis vacuously antisymmetric.
Let x= 2, y =−3,and z= 1.Then 2 ≤9,so xRy;−3≤1,so yRz; but 2 6≤ 1,so
it is not the case that xRz. Thus Ris not transitive. If x−y= 3mand y−z= 3n,
then x−z=x−y+y−z= 3(n+m),so xSy and ySz together imply xSz. Thus
Sis transitive. if y≤(x−1) and z≤(y−1),then z≤x−2≤x−1,so xT y and
yT z together imply xT z. Thus Tis transitive.
3. Define a relation Son Rby (x, y)∈Sif, and only if, (x+ 1)2= (y+ 1)2. Is S
an equivalence relation? Justify your answer.
Solution: Since (x+ 1)2= (x+ 1)2for any real number, we see that xSx always
holds, so Sis reflexive. If (x+ 1)2= (y+ 1)2,then (y+ 1)2= (x+ 1)2,so
Sis symmetric. Finally, if (x+ 1)2= (y+ 1)2and (y+ 1)2= (z+ 1)2,then
(x+1)2= (z+1)2,so Sis transitive. Since Sis reflexive, symmetric, and transitive,
it is an equivalence relation. (cf. question 6 below)
4. Let Xbe a set with more than two elements. Define a relation Ron P(X),the
power set of X, by (A, B)∈Rif and only if B⊆A. Show that Ris a partial order
on P(X).Is it a well ordering? Is it a total ordering?
Solution: For any set A, A ⊆A, so ARA. Thus Ris reflexive. If ARB and BRA,
then A⊆Band B⊆A, so A=B. Thus Ris antisymmetric. If A, B, and Care
three sets, and ARB and BRC, then B⊆Aand C⊆B, so C⊆Aand ARC. Thus
Ris transitive. Since Ris reflexive, antisymmetric, and transitive, Ris a partial
order. If a, b ∈Xand a6=b, then {a}and {b}are not comparable in the partial
order R. Thus Ris not a total order. Since all well orders are total orders, Ris not
a well order relation either. (Notice that {{a},{b}} has no least element under R.)
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