COT 3100H Spring Homework #3 Solutions
1)
THIS IS WHAT I MEANT TO BE QUESTION A...
a) (B  C)  (B – A)  (C – A).
We must prove the following: (B – A)  (C – A).
This means, for an arbitrarily chosen element x, we must show the following:
if x(B – A), then x(C – A).
We will prove this using the direct proof method.
1) Assume x(B – A).
2) By definition of set difference, this means xB and x  A.
3) Since B  C and xB, by definition of subset, we know that xC.
4) By definition of set difference since, xC and x  A, x(A – C), as desired.
REAL QUESTION A
a) ((A  B)  (B  C))  (A  C).
This statement is true. This means, we must prove that A  C.
We must show that for an arbitrary xA that xC and that there exists some
element y such that yA but yC.
First, let's consider an arbitrary xA. Since A  B, by the definition of proper
subset we can conclude that xB. Similarly, using the given information again,
since B  C, we can conclude that xC. This proves the first part of what we
need to show.
Now, since we know that A  B, in follows that there exists an element y such
that yA but yB. But, if yB, since B  C, it follows that yC. Thus, we have
shown that there exists an element y such that yA but yC as desired.
REAL QUESTION B
b) ((A  B)  (A  C))  ((B  C)  (C  B)).
This statement is false as well. Consider the following counter example:
A = , B = {1}, C={2}
Here clearly, (A  B)  (A  C) since the empty set is a proper subset of all
non-empty sets. But both of (B  C) and (C  B) are false here.
I MEANT FOR THIS TO BE THE QUESTION C...
c) ((A  C)  (B  C))  A  B  C.
This statement is false. Consider the following counter example:
A = {1}, B = {2}, C={1,2}. Clearly we have that (A  C)  (B  C). But, in
this example we have that A  B = C, thus it is false that A  B  C.
REAL QUESTION C AND D - ANSWERS ARE IN QUESTION #3 because I
goofed up!!!
2) Let A, B and C be arbitrary sets taken from the set of positive integers.
(a) Show that the sets A  B  C and A  B  C are equal, using any method.
(b) Prove or disprove: If C  B, then ( A  B)  ( B  C )  C  ( A  B) .
(a) In this solutions, set laws will be used to show the equivalence:
A  B  C  A  B  C , using DeMorgan's Law
 A  ( B  C ) , using Law of Double Complement
 A  B  C , using Associative Law
Another solution uses a Membership Table.
(b) We must prove the following: ( A  B)  ( B  C )  C  ( A  B)
Thus, We must prove that if an arbitrary element x  ( A  B)  ( B  C ) , then
x  C  ( A  B) .
We split our work into two cases: (1) x  ( A  B) , and (2) x  ( B  C ) .
Case (1): x  ( A  B) , by definition of set difference, we have that x  A , and x  B .
Since the latter is true, and since C  B, it follows that x  C . Hence, x  C , based on
the definition of set complement. Also, since x  A , it naturally follows that
x  ( A  B) , by definition of set union. Since x  C and x  ( A  B) , by definition of
set intersection, we conclude that x  C  ( A  B) .
Case (2): x  ( B  C ) , by definition of set difference, we have that x  B , and x  C .
Hence, x  C , based on the definition of set complement. Also, since x  B , it naturally
follows that x  ( A  B) , by definition of set union. Since x  C and x  ( A  B) , by
definition of set intersection, we conclude that x  C  ( A  B) .
3) Let A, B and C be any three sets. Prove or disprove the following propositions:
a)
b)
c)
d)
e)
If A B C, then either AB or AC.
(A  C) (C B) = 
Power(A)Power(B)  Power (A B)
If AB, then AxCBxC.
If AxCBxC, then AB.
a) If A B C, then either AB or AC.
It can be disproved by the following counter example. Take A = {1, 2}, B
= {1, 3} and C = {2, 4}. Then A  B  C = {1, 2, 3, 4}, but A is neither a subset
of B , nor a subset of C.
b) (A  C) (C B) = 
Proof by contradiction. Assume that (A  C) (C B)   to show that it results
to contradiction. (A  C) (C B)   means that there exists some x  (A  C)
(C B). By the definition of intersection we can imply, that there exists x for
which the following proposition is true: p = (x A  C)  (x C  B). Using the
definition of set difference we can rewrite p as
: p = (x A) (x C)  (x C) (x  B). But (x C)  (x C) =
False, so
p = (x A) [(x C)  (x C)] (x  B) = (x A) False (x  B) =
False.
So, the assumption that intersection (A  C) (C B)   is not empty
results to contradiction which proves that this assumption is false, i.e.
intersection is empty.
Other solution is to use the membership table.
c) Power(A)Power(B)  Power (A B)
This is false. To disprove take a counterexample: A = {1, 2}, B = {2, 3}, Power(A) = {,
{1}, {2}, {1, 2}}, Power(B) = {, {2}, {3}, {2, 3}}, Power(A)  Power(B) = {{1}, {1,
2}}, A B = {1}, Power (A B )={, {1}}. Thus, {1, 2} Power(A)  Power(B), but {1,
2} Power (A B ), so the proposition Power(A)Power(B)  Power (A B) is
disproved.
d) If AB, then AxCBxC.
This is true. We must show the following: AxCBxC. This means, for an arbitrarily
chosen ordered pair (x,y) we must show the following:
if (x,y) AxC, then (x,y)BxC.
We use direct proof.
1) Assume (x,y) AxC.
2) If this is the case, then we have that xA and yC.
3) We are given that AB and xA. By definition of subset, it follows that xB.
4) By definition of Cartesian Product, we can conclude that (x,y)BxC, as desired.
e) If AxCBxC, then AB.
This is false, let A = {1}, B = {2}, and C = {}. In this case AxC = {}, BxC = {}, but A is
NOT a subset of B.