McCombs Math 381
Methods of Proof
1
Direct Proof:
To prove
p
→
q
:
Assume
p
is true
, then show this leads to
q
being true.
Proof by Contrapositive:
To prove
p
→
q
:
Assume
q
is false
, then show this leads to
p
being false.
That is, show that
¬
q
→ ¬
p
.
Proof by Contradiction:
To prove
p
→
q
:
Assume
p
is true and
q
is false
, then show this leads to a
contradiction of the original assumption, or of some other
already established result.
That is, show that
p
∧ ¬
q
(
)
→
contradiction
.
This method demonstrates
¬
p
→
q
(
)
≡
F
.
Proof by Cases:
To prove
p
→
q
, where
p
=
p
1
∨
p
2
∨
p
3
∨ ⋅⋅⋅∨
p
n
:
Show
p
1
→
q
,
p
2
→
q
,
p
3
→
q
,
⋅⋅⋅
p
n
→
q
.
That is, show each case comprising
p
leads to
q
.
Note: This method requires a
finite
number of cases.
Constructive Existence
To prove
∃
xP x
( )
:
Proof:
Find (i.e. “construct”) a specific element of the domain
set for which
P
is true.
NonConstructive Existence
To prove
∃
xP x
( )
:
Proof:
Demonstrate that
P
is true without specifically finding
an element of the domain set.
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McCombs Math 381
Methods of Proof
2
Prove each of the following by the
direct proof
.
1.
Every odd integer is the difference of two squares.
Given integer
n
.
:
A
n
is odd
:
B
2
2
n
p
q
=
−
for some pair of integers
,
p q
Prove
:
A
B
→
.
Consider the odd integer
n
. Since
n
is odd,
2
1
n
k
=
−
for some integer
k
.
Note that
(
)
2
2
2
2
1
2
1
2
1
k
k
k
k
k
k
−
−
=
−
+
−
=
−
. Since
k
is an integer,
1
k
−
is an
integer. Thus,
n
is the difference of two squares.
2.
The sum of an even integer and an odd integer is odd.
Given integers
n
and
m
. If
n
is even, and
m
is odd, then
n
m
+
is odd.
:
A n
is even, and
m
is odd
:
B
n
m
+
is odd.
Prove
:
A
B
→
.
Assume integer
n
is even, and integer
m
is odd. This means
2
n
k
=
for some
integer
k
, and
2
1
m
p
=
+
for some integer
p
. Thus
(
)
2
2
1
2
1
n
m
k
p
k
p
+
=
+
+
=
+
+
.
Since
k
and
p
are integers,
k
p
+
is an integer. Therefore,
n
m
+
is odd.
Prove each of the following by the
contrapositive
method.
1. Given integers
x
and
y
. If
x
⋅
y
is even, then at least one of the two must be even.
:
A
x y
⋅
=
even
:
A
¬
x y
⋅
=
odd
:
B
x
=
even or
y
=
even
:
B
¬
x
=
odd and
y
=
odd
To prove
A
B
→
, we need to show that
B
A
¬
→ ¬
.
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 Summer '11
 NA
 Math, Prime number, Divisor, 2Km, McCombs Math

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