CmSc 365 Theory of Computation
Homework 01 SOLUTION
1. Problem 1.1.3 (a), p.9: Prove the equality A (B C) = (A B) ( A C)
(a) Prove A (B C) = (A B) (A C)
1. We shall show that A (B C) (A B) (A C)
Let x A (B C)
Two cases are to be considered:
a. x A
By def of union and (1) we have: x A B
By def of union and (1) we have: x A C
From (2) and (3) => x (A B) (A C)
b. x (B C)
(1)
(2)
(3)
(4)
(5)
By def of intersection and (5) => x B
By def of union and (6) => x A B
By def of intersection and (5) => x C
By def of union and (8) => x A C
From (7) and (9) => x (A B) (A C)
(6)
(7)
(8)
(9)
(10)
From (4) and (10):
A (B C) (A B) (A C)
(11)
2. We shall show that (A B) (A C) A (B C)
Let x (A B) (A C)
(12)
By def of intersection we have
x (A B)
x (A C)
(13)
(14)
Let x A
By def of union and (15) =>
x A (B C)
Consider now the case when x A
Then from (13) x B
From (14) x C
(15)
(16)
(17)
(18)
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From (18) and (17) x B C
From (19) and def of union x A (B C)
(19)
(20)
From (20)
(A B) (A C) A (B C)
(21)
From (21) and (11) we have
(A B) (A C) = A (B C)
2. Problem 1.2.2, p.13: Let R = {(a,b),(a,c),(c,d),(a,a),(b,a)}. What is R R, the
composition of R with itself? What is R-1, the inverse of R? Is R, R R, or R-1 a
function?
R is not a function
R R = {(a,a),(a,d),(a,b),(a,c),(b,b),(b,c),(b,a)}
R-1 = {(b,a),(c,a),(d,c),(a,a),(a,b)}
not a function
not a function
In all relations there is an element that has two images in the relation, and therefore they
are not functions.
3. Problem 1.3.6, p.19. Let R A x A be a binary relation as defined below. In which
cases R is a partial order? A total order? Prove your answers by examining the
properties of the relations
a. A = the positive integers; (a,b) R iff b is divisible by a
Week partial order: reflexive, anti-symmetrical, transitive
a. reflexivity:
Let a is any positive integer. a is divisible by a (property of positive
integers) therefore (a,a) R
b. anti-symmetrical
Let (a,b) R and a b
a is divisible by b, therefore b is not divisible by a,
therefore (b,a) R
c. transitive
Let (a,b) R and (b,c) R. a is divisible by b, and b is divisible by c.
Therefore a is divisible by c (property of positive integers)
Therefore (a,c) R
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Therefore R is partial order.
It is not a total order, because there are elements of A that are not in the
relation R, e.g. (3,7) and (7,3) are not in R
b. A = N; (a,b) R iff b = a or b = a+1
Not a partial order because it is not transitive.
Let (a,b) R and (b,c) R, and let b = a+1, and c = b+1
Consider (a,c): c = b + 1 = a + 2, therefore (a,c) R
c. A is the set of all English words; (a,b) R iff a is the same as b or occurs more
frequently than b in our textbook.
Partial order
Reflexive: (a,a) R by the definition of R
Anti-symmetric:
Let (a,b) R, and a b. Therefore a occurs more frequently than b, therefore b
does not occur more frequently than a, therefore (b,a) is not in the relation.
Transitivity:
Let (a,b) R and (b,c) R
A occurs more frequently than b, and b occurs more frequently than c, therefore a
occurs more frequently than c, therefore (a,c) R
Not a total order, because different words that occur with same frequency cannot
be ordered.
4. Problem 1.3.8 (a), p.20
Prove that if S is any collection of sets, then RS = {(A,B): A, B S and A B} is a
partial order
Reflexivity:
Let A S
By definition of subsets, A A,
Therefore (A,A) RS
Anti-symmetric:
Let (A,B) RS and A B
By def of Rs, A B.
Since A B, B A, therefore (B,A) RS
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Transitivity:
Let (A,B) RS and (B,C) RS
A B , and B C
Therefore A C (by definition of subsets)
Therefore (A,C) RS
5. Prove that the set of all finite strings of the letter a is countable. Hint: show how the
elements can be ordered, i.e. find a bijection that maps elements to N
Let w be a finite string consisting of the letter “a”. Consider the function
F(w) = k, where k is the number of the letters in the string.
This function is a bijection to the set N, therefore the set of all finite strings of the letter
“a” is countable.
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