MATH 135
Fall 2006
Proofs, Part IV
We’ve spent a couple of days looking at one particular technique of proof: induction.
Let’s look at a few more.
Direct Proof
Here we start with what we’re given and proceed in a direct line to the conclusion.
Example
If
n
is an odd integer, then
n
2

1 is divisible by 4.
Proof
Recall that every even integer can be written as 2
m
for some integer
m
and every odd integer can
be written as 2
k
+ 1 for some integer
k
.
Since
n
is odd,
n
= 2
k
+ 1 for some
k
∈
Z
.
Thus
n
2

1 = (2
k
+ 1)
2

1 = 4
k
2
+ 4
k
+ 1

1 = 4
k
2
+ 4
k
= 4(
k
2
+
k
)
so is divisible by 4.
Tips
1) Most steps should be justiﬁed, with the exception of straightforward algebraic steps.
2) Always introduce/deﬁne new variables and give their domain.
3) Start with what you’re given, not with what you want to prove.
4) Keep one eye forward and one behind; always be aware of where you’re going.
Example
If
a
,
b
,
c
form an arithmetic sequence, then (
b

c
)
x
2
+ (
c

a
)
x
+ (
a

b
) = 0 has equal roots.
Proof
Since
a
,
b
,
c
form an arithmetic sequence, then
b
=
a
+
d
,
c
=
a
+ 2
d
for some
d
∈
R
.
Thus our equation is

dx
2
+ 2
dx

d
= 0

d
(
x
2

2
x
+ 1) = 0

d
(
x

1)
2
= 0
and so has two equal roots.
Example
If
AB
is a diameter of a circle and
C
is on the circle, then
∠
ACB
=
π
2
.
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Let
O
be the centre of the circle and join
CA
,
CO
, and
CB
.
Suppose
∠
ACO
=
x
.
Since
AO
,
BO
and
CO
are radii, then
AO
=
BO
=
CO
.
Since
AO
=
CO
, then
4
ACO
is isosceles, so
∠
CAO
=
∠
ACO
=
x
.
Therefore,
∠
COA
=
π

∠
ACO

∠
CAO
=
π

2
x
.
Also,
∠
COB
=
π

∠
COA
=
π

(
π

2
x
) = 2
x
.
Since
CO
=
BO
, then
4
BCO
is isosceles, so
∠
BCO
=
∠
CBO
.
Thus, looking at the angles in
4
COB
, 2
x
+ 2
∠
BCO
=
π
, so
∠
BCO
=
π
2

x
.
Therefore,
∠
ACB
=
∠
ACO
+
∠
BCO
=
x
+
π
2

x
=
π
2
, as required.
A
C
B
O
x
Compound Statements
If
A
and
B
are mathematical statements, we often see compound statements such as
“
A
and
B
”
“
A
or
B
”
For “
A
and
B
” to be TRUE, both
A
and
B
must be TRUE.
Otherwise (when one is FALSE or both are FALSE), “
A
and
B
” is FALSE.
For “
A
or
B
” to be TRUE, either or both of
A
and
B
must be TRUE.
Otherwise (when both are FALSE), “
A
or
B
” is FALSE.
Example
A
=“2 is a prime number”,
B
=“5 is a perfect square”
Is “
A
and
B
” TRUE or FALSE?
Is “
A
or
B
” TRUE or FALSE?
Aside Regarding Sets
Recall that if
A
and
B
are sets, then
A
∪
B
is the set of elements that are in either
A
or
B
, and
A
∩
B
is the set of elements that are in both
A
and
B
.
So
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 Spring '08
 ANDREWCHILDS
 Logic, Quantification, Irrational number, Proof by contradiction

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