Strong Induction:
Selected Exercises
Goal
Explain & illustrate proof construction of a
variety of theorems using strong induction
Strong Induction
Domain of discussion is the positive integers, Z+.
Strong Induction:
If
1. p( 1 )
2. k ( [ p( 1 )  p( 2 )  …  p( k ) ]  p( k + 1 ) )
then
n p( n ).
Copyright © Peter Cappello
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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)
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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:
1.
n + 1 = ab, such that 1 < a  n, 1 < b  n
2. P( a ): a = p1p2 . . . pk, where the pis are primes. (S.I.H.)
3. P( b ): b = q1q2 . . . ql, where the qis are primes. (S.I.H.)
4.
n + 1 = ( p1p2 . . . pk )( q1q2 . . . ql ).
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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.
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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.
1. Break the bar horizontally, if there is > 1 row, or
vertically, if there is only 1 row of squares.
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Exercise 10 continued
2. Let piece 1 have s1 squares & piece 2 have s2 squares.
3. s1 + s2 = 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.
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Exercise 30
Find the flaw with the following “proof” that an = 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 .ak ) / ak-1 = 1.1 / 1 = 1.
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Exercise 30 continued
The proof fails for n = 1. Why?
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Exercise 18
Show: If a convex polygon P with consecutive vertices v1, v2, …, vn 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.
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Exercise 18
Show: If a convex polygon P with consecutive vertices v1, v2, …, vn 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
6
1
5
2
4
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Exercise
• What is the P( n ) that is claimed?
• For what n is it claimed to be true?
• What is a recursive formulation of P( n )?
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1
Exercice18 Proof
1
3
Basis n = 3: True, as shown.
2
Assume proposition for polygons of k vertices, where 3  k  n - 1.
Show proposition for polygons with n vertices.
1.
For n > 3, every triangulation of P includes a diagonal from vn or vn-1.
(If not, P is not triangulated.)
2.
Case: A diagonal connects vn to vk:
6
1
5
2
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3
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18 Proof continued
Subdivide P into 2 smaller polygons, P1 & P2, defined by this diagonal:
P1 has vertices v1, v2, …, vk, vn.
Renumber vn of P1 as vk+1; (k – 1 triangles when triangulated)
4
1
2
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18 Proof continued
Subdivide P into 2 smaller polygons, P1 & P2, defined by this diagonal:
P1 has vertices v1, v2, …, vk, vn.
Renumber vn of P1 as vk+1; (k – 1 triangles when triangulated)
Triangulate P1 with triangles properly numbered 1, …, k – 1. (SIH)
4
1
1
2
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2
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18 Proof continued
P2 has vertices vk, vk+1, …, vn.
P2 has n – k + 1 vertices;
(n – k – 1 triangles)
For each vi of P2, renumber vertices v’i = vi – k +1.
4= 6 - 2
3= 5 - 2
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2= 4 - 2
1= 3 – 2
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18 Proof continued
P2 has vertices vk, vk+1, …, vn.
For each vi of P2, renumber vertices v’i = vi – k +1.
P2 has n – k + 1 vertices;
(n – k – 1 triangles)
Triangulate it with triangles properly numbered 1, …, n – k – 1. (SIH)
4
3
2
1
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1
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18 Proof continued
P2 has vertices vk, vk+1, …, vn.
For each vi of P2, renumber vertices v’i = vi – 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.
4
3
4
3
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1
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18 Proof continued
P2 has vertices vk, vk+1, …, vn.
For each vi of P2, renumber vertices v’i = vi – 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 vi now participates in Δi, i = k, …, n – 2.
6
1
1
5
4
2
3
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3
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18 Proof continued
3. Case: A diagonal connects vn-1 to vk:
This case is handled similarly. Do this as an exercise.
This constructive proof is essentially a recursive
algorithm for computing these triangles & their
numbers.
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End
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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.
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40 Proof
1. y – x > 0.
2. 1/(y – x) > 0.
3. Let a  Z, a > 1/(y – x).
4. y – x > 1/a.
5. 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.
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Characters
• ≥≡~┌┐└┘
• ≈
• 
•  Ω Θ
• Σ
• 
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