MODULE
TWO
PROBLEM
SET
1
(p
∧
q)
→
r
∴
(p
∨
q)
→
d
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 d q is T, then d 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.
id id i∴id Theid d d patientid d d d hasid d d d highid d d d bloodid d d d pressureid d d d orid d d d highid
d d d 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
the
form
of
the
argument.
Then
prove
that
the
form
is
validor
invalid.
(a)
p
=
Patients
with
high
blood
pressure
q
=
Patients
with
diabetes
r
=
Patients
with
high
cholesterol
p ∨
q q d d d
∨ r
—
−
−
−
−
−
−
−
−
−
−
−
−
∴ p d d ∨ 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
Girl
Scout
who
sells
at
least
30
boxes
of
cookies
will
get
a
prize.
Suzy,
a
Girl
Scout,
got
a
prize.
Therefore,
Suzy
sold
at
d
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.
d
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
d
(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))
∧
d
((
∃
x)(P
(x))
Intro
5,
6
Valid.
(b)
A
x(P(x)
∨
Q(x))
d
Premise P
(c)
∨
Q(c)
Universal
d
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
d
(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
d
+
(n
2
+
2n
+
1)
d
+
(n
2
+
4n
+
4)
d
+
(n
2
+
6n
+
9)
=
d
4n
2
+
12n
+
14
=
2(2n
2
+
6n
+
7)
=
2
∗
m,
for
m
=
d
2(2n
2
+
6n
+
7)
d
∈
Z
∴ the sum of d the squares d of 4 consecutive 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 = d b d
, d d d
b
/
=
0,
a,
b
∈ Z
c
x
= ,
d
d
0,
c,
d
d
∈
Z
c a ac
therefore
x y
d
=
d d
(
)(
d
)
=
d
d
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
d d d
d
=
b
∗
(d
−
1
+
1)
b
∗
d d
(d
−
1
+
1)
d d
=
d d
b
∗
(d
−
1)
+
b
since,
d
is
odd,
d
—
1
is
d
even,
so
b
∗
d
(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
=
1
=
d
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
d d
+
d
2
2
d
+
3
2
.... +
d
(k
−
1)
2
d
+
d
k
2
d
+
d
(k
+
d d
1)
2
=
((k +
1)
∗
(k
+
1
+
d
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
+
d
9
+
...
+
n
2
=
n
(n
+
1
)(2n
+
1)/6
for
all
positive
integers
n.
