1) For any two subsets and of a universal set , prove that: a) b) For

1) For any two subsets
and
of a universal set , prove that:
a)
b)
For part a), we must show that
and that
First, let
.
This means that is in the complement of
, so therefore
. This means that
and
(draw a Venn diagram to verify this for yourself if you don’t see it – if is not in the union of
and then cannot be in and also cannot be in ). Since
and
,
and
, so
. Therefore,
Next, let
means
. This means
.
and
, so therefore
. But this means that
cannot be in either
. Thus
or , which
, and so
.
Part b) is similar.
3) Show that
a)
b)
For a), you can work with the left hand side as follows:
Part b) is very similar.
7) Let
, and let
Determine if the relation
is a partial order on set . If it is a partial order, then construct a Hasse diagram.
is reflexive since every element of is related to itself (that is,
is antisymmetric since the only way that both
is transitive since if
and
then
since
and
, we need to also see
The Hasse diagram looks like this:
and
).
is if
.
. Note that here we only need to check this once –
(and we do since
).
4
3
5
8) Let
and
means divides . Determine if the relation
partial order on set . If it is a partial order, then construct a Hasse diagram.
is reflexive since
Also, is antisymmetric since if
and divides is if
.
, i.e.
divides
and
for all
is a
.
, this implies
, i.e. the only way that
divides
Further, is transitive since if
and
, then
. This means if
and
, then
. To
show this, consider what it means for to divide (or
). It means that
such that
.
Similarly,
means
such that
. Using these two facts, we can show that
, i.e. that
such that
. Since
and
, we can replace with
in the first equation. This
gives
. Since
, this means that
(i.e. we have found our such that
, namely
).
Since
is reflexive, antisymmetric and transitive, it is a partial order on the set .
The Hasse diagram is:
72
9) Is the relation
Explain why or why not.
The relation
paired with
is not a function with domain
.
3
2
6
4
24
20
120
a function with domain
?
since there is no element from the codomain that is
10) Prove the results in the following exercises by mathematical induction.
a) For all positive integers
is the statement “For all positive integers ,
show that
is true:
”. First we
is the statement
, which is true since both sides of
the equation equal .
Next we assume
is true for some
We would now like to show that
would like to show that
implies
.
is the statement
is true, based on the assumption that
.
is the statement
is true; that is, we
or
Notice that, because of our assumption, we can replace the first
terms on the left-hand side with
, which gives the equation
All that is left now is to show that this equality is true, so we first work with the left-hand side and show
that it is equal to the right-hand side. Find a common denominator on the left-hand side and add the
two terms together, which gives us
Expand this to get
Expanding the right-hand side gives the same thing, so the equality is true. Thus
by the principle of mathematical induction,
is true for all
.
is true, and so
b) For all positive integers
is the statement “For all positive integers
that
is true:
is the statement
equation are equal to
Next we assume
”. First we show
, which is true since both sides of this
.
is true for some
We would now like to show that
would like to show that
implies
.
is the statement
is true, based on the assumption that
.
is the statement
Notice that, because of our assumption, we can replace the first
terms with
is true; that is, we
, which gives us the
equation
All that is left now is to show that this equality is true, so we work with the left-hand side and show that
it is equal to the right-hand side. Find a common denominator on the left-hand side and add the two
terms together, which gives us
Simplifying this gives us
Thus
is indeed true, and so
is true for all
by mathematical induction.
11) How many subsets of
Here we use the formula
where
contain exactly 5 elements?
(since there are elements in the given set), and
Thus there are 56 subsets of
. This gives us
which contain exactly 5 elements.
12) A basketball coach must choose a 5-person starting team from a roster of 12 players. In how
many ways is this possible?
Again we use the formula for
, where in this case
and
, so we get
Thus there are 792 possible starting teams of 5 from a roster of 12 players.
Note that on the actual exam, you will be required to show all work for problems like #11 and #12 (so
just writing a calculator answer will not be enough).