Handout #37
CS103
May 2, 2011
Robert Plummer
Problem Set #4 Solutions
1.
Prove by induction that for any natural number n,
3
)
2
)(
1
(
)
)
1
(
(
1
n
n
n
i
i
n
i
Proof: Let P
n
be the proposition that the equality above holds.
We will prove the general
case by induction on n.
BASE CASE:
P
1
states that 1(1+1) = 1(1+1)(1+2)/2.
Since LHS = RHS = 2, the base
case is true.
INDUCTIVE STEP:
Assume that for some k
1, P
k
is true; i.e., that
I.H.
We will show that P
k+1
must be true; i.e., that
Adding (k + 1)(k + 1 + 1) to both sides
of the I.H. and doing simple algebra gives
Applying the definition of
on the left and factoring on the right gives
as desired.
Thus P
k
P
k+1
, and by the Principle of Mathematical Induction, P
n
is true for
all n
0.
■
3
)
2
)(
1
(
)
)
1
(
(
1
k
k
k
i
i
k
i
3
)
3
)(
2
)(
1
(
3
)
2
1
)(
1
1
)(
1
(
)
)
1
(
(
1
1
k
k
k
k
k
k
i
i
k
i
3
)
2
)(
1
(
3
)
2
)(
1
(
)
1
1
)(
1
(
))
)
1
(
(
(
1
1
k
k
k
k
k
k
k
i
i
k
i
3
)
3
)(
2
)(
1
(
)
)
1
(
(
1
1
k
k
k
i
i
k
i
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2
2.
Let x be an integer.
Show that for every integer n
2, x
n
is even if and only if
x is even.
Proving the biconditional will naturally involve two subproofs.
You are required to use
induction in at least one of them, but otherwise, do not use induction where a very simple
noninductive proof will suffice.
This is a biconditional proof.
(
direction)
Show the following is true for all n
2:
P
n
: If x
n
is even, then x is even.
BASE CASE: P
2
states that x
2
even
x is even.
The contrapositive of that is:
x odd
x
2
is odd.
If x is odd, x = 2k + 1 for some integer k, so
x
2
= (2k + 1)(2k + 1) = 4k
2
+ 4k + 1, which is odd.
So P
2
is true.
INDUCTIVE STEP:
Assume P
k
:
x
k
even
x is even for some k
2.
Show
P
k+1:
x
k+1
even
x is even.
Proof of the inductive step:
Suppose x
k+1
is even.
Thus x
k
x is even.
This means that either (case 1) x
k
is even or
(case 2) x is even, since the product of two odd numbers is odd.
But x
k
even
x is even
by the IH, so in both cases x is even, and P
k+1
is true.
By the Principle of Mathematical Induction, P
n
is true for n
2.
(
direction)
Show that for n
2, x even
x
n
even.
Direct proof:
If x is even, x = 2m for some integer m.
So x
n
= (2m)
n
= 2
n
m
n
, which is even since it has
2 as a factor.
■
3.
Prove by induction that for every integer n
2, the complement of the union of
any n sets equals the intersection of the complements of these sets.
Using set notation, we could state this as:
If A
1
, A
2
, ..., A
n
are sets and n
2, then (A
1
A
2
...
A
n
)' = A
1
'
A
2
'
... A
n
' .
Proof by induction.
P
n
: (A
1
A
2
...
A
n
)' = A
1
'
A
2
'
... A
n
'
BASE CASE:
P
2
states that (A
1
A
2
)' = A
1
'
A
2
'.
This is true by De Morgans's Law.
INDUCTIVE STEP:
Assume P
k
:
(A
1
A
2
...
A
k
)' = A
1
'
A
2
'
... A
k
' for some k
2.
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 Spring '11
 PLUMMER
 Mathematical Induction, Natural number, base case, Structural induction, inductive step

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