LECTURE 7: CASES AND COMMON ERRORS (
§
3.43.5)
1.
Proof by Cases
Occasionally it is useful to break a proof into smaller pieces that can each be dealt with.
Example 1.
If the domain is
N
we might consider
x
= 1 as one case, and
x >
1 as another case.
If the domain is
Z
we might consider the two cases:
x
is odd,
x
is even.
If the domain is
R
we might consider
x
= 0,
x >
0, and
x <
0 as three cases.
Result.
If
n
∈
Z
then
n
2
+ 3
n
+ 5
is odd.
Proof.
We proceed by cases, according to the parity of
n
.
Case 1:
n
is even. Then
n
= 2
x
for some
x
∈
Z
. Thus
n
2
+3
n
+5 = 4
x
2
+6
x
+5 = 2(2
x
2
+3
x
+2)+1
is odd.
Case 2:
n
is odd. Then
n
= 2
y
+ 1 for some
y
∈
Z
. Thus
n
2
+ 3
n
+ 5 = (2
y
+ 1)
2
+ 3(2
y
+ 1) + 5 =
4
y
2
+ 4
y
+ 1 + 6
y
+ 3 + 5 = 2(2
y
2
+ 5
y
+ 4) + 1 is odd.
We say that two integers have the
same parity
if they are both even, or both odd. Otherwise, we
say they have
opposite parity
.
Result.
Let
x, y
∈
Z
. Then
x
and
y
are of the same parity if and only if
x
+
y
is even.
Proof.
First, assume that
x
and
y
are of the same parity. We have two cases.
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 Spring '11
 Smith
 Even and odd functions, Parity, Evenness of zero, Z.

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