Selected Exercises Goal Explain amp illustrate proof construction of a variety of theorems using strong induction Copyright Peter Cappello 2 Strong Induction Domain of discussion is the ID: 545420
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Slide1
Strong Induction: Selected Exercises
Goal
Explain & illustrate proof construction of
a variety of theorems using strong inductionSlide2
Copyright © Peter Cappello
2
Strong Induction
Domain of discussion is the
positive integers,
Z
+
.
Strong Induction
:
If
p( 1 )
k
(
[
p( 1 )
p( 2 )
…
p(
k
)
]
p(
k
+ 1 ) )
then
n
p(
n
).Slide3
Copyright © Peter Cappello
3
Example: Fundamental Theorem of Arithmetic
Let the universe of discourse be
N
.
Let P(
n
) denote “
n
is a
product of primes.”
Prove that
n
2 P(
n
).
Basis
P(
2
): Since
2
is prime,
2
is a product of primes.
Assume
P
( 2 ), . . . , P(
n
).
(Strong induction hypothesis)Slide4
Copyright © Peter Cappello
4
Show
P(
n
+ 1
):
Case
n
+ 1
is
a prime
:
It is a product of 1 prime: itself.
Case
n
+ 1
is not
a prime
:
n + 1
=
ab
, such that 1
<
a
n
, 1
<
b
n
P(
a
):
a = p
1
p
2
. . .
p
k
,
where the
p
i
s
are primes. (S.I.H.)
P(
b
):
b = q
1
q
2
. . .
q
l
,
where the
q
i
s
are primes
.
(S.I.H.)
n + 1
=
(
p
1
p
2
. . .
p
k
)(
q
1
q
2
. . .
q
l
).Slide5
Copyright © Peter Cappello
5
Exercise 10
A chocolate bar consists of
n
squares arranged in a rectangle.
The bar can be broken into smaller
rectangles
of squares with a vertical or horizontal break.
How many breaks suffice to break the bar into
n
squares?
Consider an example bar consisting of 4 x 3 squares.Slide6
Copyright © Peter Cappello
6
Exercise 10 continued
Use strong induction to prove that
n
- 1
breaks suffice.
Basis
n
= 1
: 1 – 1 = 0 breaks suffice.
Assume
for bars of
up to
n
squares
that
n
– 1 breaks suffice.Show
for bars of n + 1 squares that n breaks suffice.
Break the bar horizontally, if there is > 1 row, or vertically, if there is only 1 row of squares.Slide7
Copyright © Peter Cappello
7
Exercise 10 continued
2. Let
piece 1
have
s
1
squares &
piece 2
have
s
2
squares.
3.
s
1
+ s
2
= n + 1.
4. s1 - 1 breaks suffice to break piece 1 into squares.
(SIH)5. s2 - 1 breaks suffice to break piece 2 into squares.
(SIH)6. 1 + ( s1
– 1 ) + ( s2 – 1 ) = n breaks suffice.Slide8
Copyright © Peter Cappello
8
Exercise 30
Find the flaw with the following “proof” that
a
n
= 1
,
n
≥ 0
, when
a
0
is real.
Proof:
Basis n = 0:
a0 = 1.Inductive step
: Assume that aj = 1, 0
j k
.Show ak+1
= 1:ak+1
= ( ak .a
k ) / ak-1 = 1
.1 / 1 = 1.Slide9
Copyright © Peter Cappello
9
Exercise 30 continued
The proof fails for
n
= 1. Why?Slide10
Copyright © Peter Cappello 2011
10
Exercise 18
Show
: If a convex polygon
P
with consecutive vertices
v
1
,
v
2
, …,
v
n
is
triangulated
into
n
– 2 triangles (Δ), the triangles can be numbered
1, 2, …, n – 2 so that vi
is a vertex of Δi, for i = 1, 2, …,
n – 2.
1
2
3
4
5
6Slide11
Copyright © Peter Cappello 2011
11
Exercise 18
Show
: If a convex polygon
P
with consecutive vertices
v
1
,
v
2
, …,
v
n
is
triangulated
into
n – 2 triangles (
Δ), the triangles can be numbered 1, 2, …, n
– 2 so that vi is a vertex of Δi, for
i = 1, 2, …, n – 2.
1
2
3
4
5
6
1
2
3
4Slide12
ExerciseWhat is the P(
n
) that is claimed?
For what n is it claimed to be true?What is a recursive formulation of P( n )?Copyright © Peter Cappello12Slide13
Copyright © Peter Cappello 2011
13
Exercice18 Proof
Basis
n
= 3
: True, as shown.
Assume
proposition for polygons of
k
vertices, where
3
k
n - 1
.
Show
proposition for polygons with
n vertices.
For n > 3, every triangulation of P includes a diagonal from
vn or vn-1.
(If not, P is not triangulated.)Case: A diagonal connects v
n to vk:
1
2
3
1
1
2
3
4
5
6Slide14
Copyright © Peter Cappello 2011
14
18 Proof continued
Subdivide
P
into 2
smaller
polygons,
P
1
&
P
2
, defined by this diagonal:
P
1
has vertices
v
1
, v
2
, …, vk,
vn. Renumber vn
of P1 as vk+1;
(k – 1 triangles when triangulated)
1
2
3
4Slide15
Copyright © Peter Cappello 2011
15
18 Proof continued
Subdivide
P
into 2
smaller
polygons,
P
1
&
P
2
, defined by this diagonal:
P
1
has vertices
v
1
, v
2
, …, v
k
, vn
. Renumber vn
of P1 as vk+1
; (k – 1 triangles when triangulated)Triangulate
P1 with triangles properly numbered
1, …, k – 1. (SIH)
1
2
3
1
2
4Slide16
Copyright © Peter Cappello 2011
16
18 Proof continued
P
2
has vertices
v
k
, v
k+1
, …, v
n
.
P
2
has
n – k
+ 1
vertices;
(
n – k
– 1 triangles)
For each
v
i of P
2, renumber vertices v’i = v
i – k +1
.
4
= 6 - 2
1
= 3 – 2
2
= 4 - 2
3
= 5 - 2Slide17
Copyright © Peter Cappello 2011
17
18 Proof continued
P
2
has vertices
v
k
, v
k+1
, …, v
n
.
For each
v
i
of
P
2
, renumber vertices
v’
i
=
v
i – k +1
.P2 has n – k + 1
vertices; (n – k – 1 triangles)
Triangulate it with triangles properly numbered 1, …, n – k – 1
. (SIH)
1
2
4
1
2
3Slide18
Copyright © Peter Cappello 2011
18
18 Proof continued
P
2
has vertices
v
k
, v
k+1
, …, v
n
.
For each
v
i
of
P
2
, renumber vertices
v’
i
=
v
i – k +1
.P2 has n – k + 1
vertices; (n – k – 1 triangles)
Triangulate it with triangles properly numbered 1, …, n – k – 1
. (SIH)Add k
– 1 to each triangle number: k, k + 1, …, n – 2.
3
4
4
1
2
3Slide19
Copyright © Peter Cappello 2011
19
18 Proof continued
P
2
has vertices
v
k
, v
k+1
, …, v
n
.
For each
v
i
of
P
2
, renumber vertices
v’
i
=
v
i – k +1
.P2 has n – k + 1
vertices; (n – k – 1 triangles)
Triangulate it with triangles properly numbered 1, …, n – k – 1
. (SIH)Add k
– 1 to each triangle number: k, k + 1, …, n – 2.
Original vertex
v
i
now
participates in
Δ
i
, i = k, …, n
– 2
.
1
2
3
4
5
6
1
2
3
4Slide20
Copyright © Peter Cappello
20
18 Proof continued
3.
Case: A diagonal connects
v
n-1
to
v
k
:
This case is handled similarly. Do this as an exercise.
This
constructive proof
is essentially a
recursive algorithm
for computing these triangles & their numbers.Slide21
Copyright © Peter Cappello
21
EndSlide22
Copyright © Peter Cappello 2011
22
40
Well-ordering principle
: Every nonempty set of nonnegative integers has a least element.
Use the
well-ordering principle
to show that
[
x, y
R
x < y
]
r
Q
,
x < r < y.Slide23
23
40 Proof
y – x
> 0.
1/(y – x) > 0.Let a Z, a > 1/(y – x). y – x > 1/a.
Choose the least positive j
such that
└
x
┘
+
j/a > x
. (
WOP
)
6. Let
r
=
└
x
┘+ j/a.
7. r Q. (
└x┘
, j, a Z)
8. x < r. (
defn of r)
9. └x┘+ (
j – 1)/a < x. (choice of
j)10.
r – 1/a < x.
(
defn
of
r
.)
11.
r – (y – x) < x.
(step 10 & 4)
12. r < y.Slide24
Copyright ©
Peter Cappello
24
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