8.1e

Report
Relations & Their Properties:
Selected Exercises
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) a is taller than b.
b) a and b were born on the same day.
c) a has the same first name as b.
d) a and b have a common grandparent.
Copyright © Peter Cappello 2011
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Exercise 20
Must an asymmetric relation be antisymmetric?
A relation is asymmetric  a b ( aRb  (b, a)  R ).
Copyright © Peter Cappello 2011
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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:
1. Assume R is asymmetric.
2.
a  b ( ( a, b )  R  ( b, a )  R ). (step 1. & defn of  )
3.
a  b ( ( aRb  bRa )  a = b )
(implication premise is false.)
4. Therefore, asymmetry implies antisymmetry.
Copyright © Peter Cappello 2011
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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.
Copyright © Peter Cappello 2011
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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:
1. Let R = { (a, a) }.
2. R is antisymmetric.
3. R is not asymmetric: aRa  (a, a)  R is false.
Antisymmetry thus does not imply asymmetry.
Copyright © Peter Cappello 2011
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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?
R
SR
S
1
1
2
2
3
4
3
4
Copyright © Peter Cappello 2011
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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:
1.
R is antisymmetric  R  R-1  { ( a, a ) | a  A }
We prove this by contradiction.
2.
R  R-1  { ( a, a ) | a  A }  R is antisymmetric.
We prove this by contradiction.
Copyright © Peter Cappello 2011
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Exercise 50
Prove R is antisymmetric  R  R-1  { ( a, a ) | a  A }.
1.
Proceeding by contradiction, we assume that:
1.
R is antisymmetric: a b ( ( aRb  bRa )  a = b ).
2.
It is not the case that R  R-1  { ( a, a ) | a  A }.
2.
a b (a, b)  R  R-1, where a  b.
3.
Let (a, b)  R  R-1, where a  b.
(Step 2)
4.
aRb , where a  b.
(Step 3)
5.
aR-1b, where a  b.
(Step 3)
6.
bRa, where a  b.
7.
R is not antisymmetric, contradicting step 1. (Steps 4 & 6)
8.
Thus, R is antisymmetric  R  R-1  { ( a, a ) | a  A }.
Copyright © Peter Cappello 2011
(Step 1.2)
(Step 5 & defn of R-1)
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Exercise 50 continued
Prove R  R-1  { ( a, a ) | a  A }  R is antisymmetric.
1.
Proceeding by contradiction, we assume that:
1.
R  R-1  { ( a, a ) | a  A }.
2.
R is not antisymmetric: ¬a b ( ( aRb  bRa )  a = b )
2.
Assume a b ( aRb  bRa  a  b )
3.
bR-1a, where a  b.
(Step 2s & defn. of R-1)
4.
( b, a )  R  R-1 where a  b, contradicting step 1.
(Step 2 & 3)
5.
Therefore, R is antisymmetric.
Copyright © Peter Cappello 2011
(Step 1.2)
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