1. Relations & Their Properties: Selected Exercises

2. Copyright © Peter Cappello
Exercise 10
Which relations in Exercise 4 are irreflexive?
A relation is irreflexive  a  A (a, a)  R.
Ex. 4 relations on the set of all people:
a is taller than b.
a and b were born on the same day.
a has the same first name as b.
a and b have a common grandparent.
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Exercise 20
Must an asymmetric relation be antisymmetric?
A relation is asymmetric  a b ( aRb  (b, a)  R ).
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Exercise 20
Must an asymmetric relation be antisymmetric?
A relation is asymmetric  a b ( aRb  (b, a)  R ).
To Prove:
(a  b ( aRb  (b, a)  R ) )  (a  b ( (aRb  bRa )  a = b ) )
Proof:
Assume R is asymmetric.
a  b ( ( a, b )  R  ( b, a )  R ). (step 1. & defn of  )
a  b ( ( aRb  bRa )  a = b ) (implication premise is false.)
Therefore, asymmetry implies antisymmetry.
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Exercise 20 continued
Must an antisymmetric relation be asymmetric?
(a b ( ( aRb  bRa )  a = b ) )  a  b ( aRb  ( b, a )  R )?
Work on this question in pairs.
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Exercise 20 continued
Must an antisymmetric relation be asymmetric ?
(a b ( (aRb  bRa )  a = b ) )  a  b ( aRb  (b, a)  R ) ?
Proof that the implication is false:
Let R = { (a, a) }.
R is antisymmetric.
R is not asymmetric: aRa  (a, a)  R is false.
Antisymmetry thus does not imply asymmetry.
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Exercise 30
Let R = { (1, 2), (1, 3), (2, 3), (2, 4), (3, 1) }.
Let S = { (2, 1), (3, 1), (3, 2), (4, 2) }.
What is S  R? 1 2 3 4 R S
S  R
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Exercise 50
Let R be a relation on set A.
Show:
R is antisymmetric  R  R-1  { ( a, a ) | a  A }.
To prove:
R is antisymmetric  R  R-1  { ( a, a ) | a  A }
We prove this by contradiction.
R  R-1  { ( a, a ) | a  A }  R is antisymmetric.
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Exercise 50
Prove R is antisymmetric  R  R-1  { ( a, a ) | a  A }.
Proceeding by contradiction, we assume that:
R is antisymmetric: a b ( ( aRb  bRa )  a = b ).
It is not the case that R  R-1  { ( a, a ) | a  A }.
a b (a, b)  R  R-1, where a  b (Step 1.2)
Let (a, b)  R  R-1, where a  b (Step 2)
aRb , where a  b (Step 3)
aR-1b, where a  b (Step 3)
bRa, where a  b (Step 5 & defn of R-1)
R is not antisymmetric, contradicting step 1. (Steps 4 & 6)
Thus, R is antisymmetric  R  R-1  { ( a, a ) | a  A }.
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Exercise 50 continued
Prove R  R-1  { ( a, a ) | a  A }  R is antisymmetric.
Proceeding by contradiction, we assume that:
R  R-1  { ( a, a ) | a  A }.
R is not antisymmetric: ¬a b ( ( aRb  bRa )  a = b )
Assume a b ( aRb  bRa  a  b ) (Step 1.2)
bR-1a, where a  b (Step 2s & defn. of R-1)
( b, a )  R  R-1 where a  b, contradicting step (Step 2 & 3)
Therefore, R  R-1  { ( a, a ) | a  A }  R is antisymmetric.
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