COT 310001
April 18, 2001
Practice test#3 (Total 105pts)
Name:_________________
SSN:__________________
1)(10pts) Prove by induction that for any integers
a
and
b
(
a

b
)  (
a
n

b
n
), where
n
is a
positive integer,
n
≥
1.
(Hint: use induction on
n
and for the inductive step relate
a
k
+1

b
k
+1
=
a
(
a
k

b
k
)+…?)
Proof.
We are going to prove by induction on
n
≥
1, that for any integers
a
and
b
,
(
a
n

b
n
)
is divisible by (
a

b
). By the definition of divisibility it is sufficient to prove that for any
n
≥
1, for any
a
and
b
there exists an integer
m
such that (
a
n

b
n
)=
m
⋅
(
a

b
).
Basis.
n
=1,
⋅
(
a

b
) = 1
⋅
(
a

b
).
IH
. Assume that proposition is true for
n
=
k
, where
k
is some integer,
k
≥
1. In other words,
we assume that for some
k
≥
1 and any
a
and
b
there exists an integer
m
such that
(
a
k

b
k
)=
m
⋅
(
a

b
).
IS. We need to prove that the proposition holds for
n
=
k
+1, i.e. for any
a
,
b
there exists
some integer
s
, such that (
a
k
+1

b
k
+1
)=
s
⋅
(
a

b
)
To reduce
k
+1 case to the case
n
=
k
, we consider the following identity:
a
k
+1

b
k
+1
=
a
(
a
k

b
k
)+
a b
k

b
k
+1
=
a
(
a
k

b
k
)+
b
k
(
a

b
)
By IH we have for these
a
and
b
some integer
m
such that (
a
k

b
k
)=
m
⋅
(
a

b
), so:
a
(
a
k

b
k
)+
b
k
(
a

b
) =
a m
⋅
(
a

b
) +
b
k
(
a

b
)
So,
a
k
+1

b
k
+1
=
⋅
(
a
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 Spring '09
 Natural number, Prime number, IH. Assume

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