HOMEWORK 3: COMMENTS
1.
Nonbook problems
(1)
(a) The contrapositive of the statement is “if 3 does not divide
n
then 3 does not
divide
n
2
.” Now, if 3 does not divide 3, then either
n
= 3
k
+ 1 for some integer
k
, so that
n
2
= (3
k
+ 1)
2
= 9
k
2
+ 6
k
+ 1 = 3(3
k
2
+ 2
k
) + 1
,
and so 3 doesn’t divide
n
2
; or
n
= 3
k
+ 2 for some integer
k
, so that
n
2
= (3
k
+ 2)
2
= 9
k
2
+ 12
k
+ 4 = 3(3
k
2
+ 4
k
+ 1) + 1
,
and so 3 doesn’t divide
n
2
again.
Book Problems.
Problem 12
Solution.
We proceed by induction on
n
. The base case is
n
= 1. We have 3

9 is true; this
establishes the base case. Assume as inductive hypothesis that 3

(4
k
+ 5). Then there exists
a
q
∈
Z
such that 4
k
+ 5 = 3
q
. Our goal is to show that 3

(4
k
+1
+ 5). We compute
4
k
+1
+ 5 = 4
·
4
k
+ 5
= 4
·
(4
k
+ 5

5) + 5
= 4(3
q

5) + 5
by inductive hypothesis
= 12
q

20 + 5 = 3(4
q

5)
.
Hence 3

(4
k
+1
+ 5), and by induction 3

(4
n
+ 5) for all
n
∈
N
.
±
Problem 17
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 Spring '07
 COURTNEY
 Division

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