MODULE
TWO
PROBLEM
SET
1
(p
∧
q)
→
r
∴
(p
∨
q)
→
a
r
p
→
q
q
∴
p
p
→
q
¬p
∴
¬
q
Directions:
Type
your
solutions
into
this
document
and
be
sure
to
show
all
stepsfor
arriving
at
your
solution.
Just
giving
a
final
number
may
not
receive
full
credit.
P
ROBLEM
1
Part
1.
Indicate
whether
the
argument
is
valid
or
invalid.
For
valid
arguments,
prove
that
the
argument
is
valid
using
a
truth
table.
For
invalid
arguments,
give
truth
values
for
the
variables
showing
that
the
argument
is
not
valid.
(1)
Not
valid.
For
p
=
T,
q
=
F,
r
=
F
and
p
=
F,
q
=
T,
r
=
F:
these
two
combinations
prove
the
argument
invalid.
Part
2.
Converse
and
inverse
errors
are
typical
forms
of
invalid
argu-
ments.
Prove
that
each
argument
is
invalid
by
giving
truth
values
for
the
variables
showing
that
the
argument
is
invalid.
You
may
find
it
eas-
ier
to
find
the
truth
values
by
constructing
a
truth
table.
(a)
Converse
error
If
the
truth
value
of
p
is
F
and
q
is
T,
then
p
→
q
and
q
both
are
T
but
p
is
F.
(b)
Inverse
error
If the truth value of p is F and a q is T, then a p → q and ¬p is T but
¬q is F.
The
patient
has
high
blood
pressure
or
diabetes
or
both.
The
patient
has
diabetes
or
high
cholesterol
or
both.
ia ia i∴ia Theia a a patientia a a a hasia a a a highia a a a bloodia a a a pressureia a a a oria a a a highia a a
a cholesterol.
Part
3.
Which
of
the
following
arguments
are
invalid
and
which
are
valid?
Prove
your
answer
by
replacing
each
proposition
with
a
variable
to
obtain
a
the
form
of
the
a
argument.
a
Then
a
prove
a
that
a
the
form
is
validor
invalid.
(a)
p
=
Patients
with
high
blood
pressure
q
=
Patients
with
diabetes
r
=
Patients
with
high
cholesterol
p ∨
q q a a a
∨ r
—
−
−
−
−
−
−
−
−
−
−
−
−
∴ p a a ∨ r
If
p
and
r
are
F
and
q
is
T,
then
both
p
∨
q
and
q
∨
r
are
T.
However,
p
∨
r
is
F.
This
implies
an
invalid
argument.
∃
x
(P(x)
∧
Q(x))
∴ ∃x Q(x) ∧ ∃x
P(x)
A
x
(P
(x)
∨
Q(x))
∴
A
x
Q(x)
∨
A
x
P(x)
P
ROBLEM
2
Part
1.
Which
of
the
following
arguments
are
valid?
Explain
your
reasoning.
(a)
I
have
a
student
in
my
class
who
is
getting
an
A.
Therefore,
John,
a
student
in
my
class,
is
getting
an
A.
Invalid.
We
don’t
know
that
John
is
the
student
in
the
class
who
is
gettingan
A.
(b)
Every
a
Girl
a
Scout
a
who
a
sells
a
at
a
least
a a
30
a a
boxes
a a
of
a
cookies
a
will
get
a
prize.
Suzy,
a
Girl
Scout,
got
a
prize.
Therefore,
Suzy
sold
at
least
30
boxes
of
cookies.
Valid.
Girl
Scouts
who
sell
at
least
30
boxes
of
cookies
get
a
prize
andSuzy
got
a
prize.
Part
2.
Determine
whether
each
argument
is
valid.
If
the
argument
is
valid,
givea
proof
using
the
laws
of
logic.
If
the
argument
is
invalid,
give
values
for
the
pred-
icates
P
and
Q
over
the
domain
a,
b
that
demonstrate
the
argument
is
invalid.
(a)
(
∃
x)
(P
a
(x)
∧
Q(x))
Premise
P(y)
∧
Q(y)
Existential
Inst.1
P(y)
∧
Simplification
2
Q(y)
∧
Simplification
2
(
∃
x)P
(x)
Existential
Gen.
3
(
∃
x)Q(x)
Existential
Gen.
4
((
∃
x)Q(x))
∧
a
((
∃
x)(P
(x))
Intro
5,
6
Valid.
(b)
A
x(P(x)
∨
Q(x))
a
Premise P
(c)
∨
Q(c)
Universal
a
Inst.
1
Invalid
because
Simplification
and
Universal
Gen.
are
not
possible.
P
ROBLEM
3
Prove
the
following
using
a
direct
proof.
Your
proof
should
be
expressed
in
com-plete
English
sentences.
If
a,
b,
and
c
are
integers
such
that
b
is
a
multiple
of
a
3
and
c
is
a
multiple
of
b
2
,then
c
is
a
multiple
of
a
6
.
b
is
a
multiple
of
a
3
,
b
=
ma
3
,
m
∈
Z
and
c
is
a
multiple
of
b
2
then
c
is
a
multiple
of
a
(ma
3
)
2
so,
c
is
a
multiple
of
m
2
a
6
∴ c is a multiple of a6
P
ROBLEM
4
Prove
the
following
using
a
direct
proof:
The
sum
of
the
squares
of
4
consecutive
integers
is
an
even
integer.consecutive
integers:
n,
n
+
1,
n
+
2,
n
+
3
claim:
n
2
+
(n
+
1)
2
+
(n
+
2)
2
+
(n
+
3)
2
=
an
even
integer
consider:
n
2
+
(n
+
1)
2
+
(n
+
2)
2
+
(n
+
3)
2
=
n
2
a
+
(n
2
+
2n
+
1)
a
+
(n
2
+
4n
+
4)
a
+
(n
2
+
6n
+
9)
=
a
4n
2
+
12n
+
14
=
2(2n
2
+
6n
+
7)
=
2
∗
m,
for
m
=
a
2(2n
2
+
6n
+
7)
a
∈
Z
∴ the sum of the squares a of 4 consecutive a integers is an even
integer
P
ROBLEM
5
∗
Prove
the
following
using
a
proof
by
contrapositive:
Let
x
be
a
rational
number.
Prove
that
if
xy
is
irrational,
then
y
is
irrational.
To
prove
by
contrapositive:
show
that
if
y
is
rational,
then
xy
is
rational.
a
y = a b a
, a a a
b
/
=
a
0,
a,
b
∈ Z
c
x
= ,
d
d
0,
c,
d
a
∈
Z
c a ac
therefore
x y
a
=
a
(
)(
a
)
=
d
a
b bd
where
a,
c
∈
Z
→
ac
∈
Z
and
b,
d
∈
Z
→
bd
∈
Z
therefore
ac
is
rational,
and
thus
xy
is
rational
bd
∴
if
y
is
rational
then
xy
is
rational
∴
if
xy
is
irrational
then
y
is
irrational
P
ROBLEM
6
Prove
the
following
using
a
proof
by
contradiction:
The
average
of
four
real
numbers
is
greater
than
or
equal
to
at
least
one
of
the
numbers.
a+b+c+d
<
a+b+c+d
which
is
not
possible
and
therefore
a
contradiction.
P
ROBLEM
7
Let
q
=
a
and
r
=
b e
c
be
two
rational
numbers
written
in
lowest
terms.
Let
d
s
=
q
+
r
and
s
=
f
be
written
in
lowest
terms.
Assume
that
s
is
not
0.
Prove
or
disprove
the
following
two
statements.
a.
If
b
and
d
are
odd,
then
f
is
odd.
b
∗
I
a a a
d
=
b
∗
(d
−
1
+
1)
b
∗
a a
(d
−
1
+
1)
a a
=
a a
b
∗
(d
−
1)
+
b
since,
d
is
odd,
d
—
1
is
even,
so
b
∗
a
(d
−
1)
is
even.
∴ since b is odd and the sum of an odd number and even number is always odd
the
above
product
is
odd
b.
If
b
and
d
are
even,
then
f
is
even.
False.
If
b
and
d
are
even,
f
is
not
always
an
even
number.
Let
q
=
Since 2 is even, then b and d are
even.
1 1
and
r
= .
2 2
s
=
1
+
1
=
2
a
=
1
=
a
e
2 2 2 1
f
∴
f
=
1
which
is
an
odd
number.
∴
P
ROBLEM
8
Define
P(n)
to
be
the
assertion
that:
n
n(n
+
1)(2n
+
1)
∑
j
2
=
6
j=1
(a)
Verify
that
P(3)
is
true.
1
2
+2
2
+
3
2
=(
3
∗
(3
+
1)(2
∗
3
+
1))/6
1
+
4
+
9
=3
14
=
14
(b)
Express
P
(k).
∗ 4 ∗ 7/6
1
2
+
2
2
+
3
2
....
+
(k
−
1)
2
+
k
2
=
(k
∗
(k
+
1)(2
∗
k
+
1))/6
(c)
Express
P(k
+
1).
1
2
a a
+
a
2
2
a
+
a
3
2
.... +
a
(k
−
1)
2
a
+
a
k
2
a
+
a
(k
+
a a a
1)
2
=
((k +
1)
∗
(k
+
1
+
a
1)(2
∗
(k
+
1)
+
1))/6
(d)
In
an
inductive
proof
that
for
every
positive
integer
n,
n
∑j
2
=
j=1
n(n
+
1)(2n
+
1)
6
what
must
be
proven
in
the
base
case?
1
2
=
(1
1
=
1
∗
(1
+
1)(2
∗
1
+1))/6
(e)
In
an
inductive
proof
that
for
every
positive
integer
n,
n
∑j
2
=
P
ROBLEM
9
j=1
n(n
+
1)(2n
+
1)
6
what
must
be
proven
in
the
inductive
step?
With
any
induction
proof
we
must
start
with
base
case(base
step)
1.
Show
true
for
n
=
1
base
case
2.
Assume
true
for
n
=
k
assumption
3.
Show
true
for
n
=
k
+
1
its
a
inductive
step
so
its
must
we
prove
for
your
answer
(f)
What
would
be
the
inductive
hypothesis
in
the
inductive
step
from
your
previous
answer?
n
=
k
(g)
Prove
by
induction
that
for
any
positive
integer
n,
n
∑j
2
=
j=1
n(n
+
1)(2n
+
1)
6
Therefore,
by
the
principle
of
mathematical
induction,
1
+
4
+
a
9
+
...
+
n
2
=
n
(n
+
1
)(2n
+
1)/6
for
all
positive
integers
n.
