CS 70
Vivian Fang
August 31, 2016
Lecture 4
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Strengthening Induction Hypothesis
Theorem: The sum of the first n odd numbers is a2 (a perfect square, a ∈ N).
Just plugging things in:
n = 1 :1 = 12
n = 2 :1 + 3 = 22
n = 3 :1 + 3 + 5 = 32
Now we can try prove:
Theorem: The sum of the first n odd numbers is n2 . (If this is proved, the first theorem
holds)
Proof : By Induction.
• kth odd number is 2k − 1 for k ≥ 1.
• Base Case: 1 (first odd number) is 12 .
• Induction Hypothesis: Sum of the first n odds is perfect square a2 = n2 .
• Induction Step: To prove that sum of first n + 1 odds is (n + 1)2 .
1. The (n + 1)st odd number is 2(n + 1) − 1 = 2n + 1
2. Sum of the first n + 1 odds is a2 + 2n + 1 = n2 + 2n + 1
3. n2 + 2n + 1 = (n + 1)2 which proves P (n + 1).
Can we tile any 2n × 2n with L-tiles (with a hole) for every n?
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Theorem: Any tiling of 2n × 2n square has to have one hole.
Assume we have M tiles, we will be able to over M × 3 squares (each l-tile is 3 tiles).
Proof :
• Need to establish 2n × 2n = 22n mod 3 = 1.
• Base Case: True for k = 0. 20 = 1.
• Inductive Hypothesis: n = k. 22k = 3a + 1 for integer a.
22(k+1) = 22k × 22
= 4 × 22k
= 4 × (3a + 1)
= 12a + 3 + 1
= 3(4a + 1) + 1
a integer =⇒ (4a + 1) is an integer. We have the remainder 1, so we have proved
that 2n × 2n = 22n mod 3 = 1.
Theorem: You can tile the 2n × 2n to leave a hole adjacent to the center.
Proof :
• Base Case: A single tile works fine for a 2 square.
• Induction Hypothesis:Any 2n × 2n square can be tiled with a hole at the center.
• Inductive Step: This is hard to prove. Hard to orient center of 2n × 2n squares to be
the middle of 2n+1 × 2n+1 square.
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So let’s prove a stronger theorem:
Theorem: You can tile the 2n × 2n to leave a hole adjacent anywhere.
Proof :
• Base Case: A tile is file. You can rotate the L-tile to cover 3 out of the 4 squares.
• Induction Hypothesis: Any 2n × 2n square can be tiled with a hole anywhere.
Consider a 2n+1 × 2n+1 square. You can tile the three inner squares to have their hole
in the center (of the big square), and fill that in with an L-tile.
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Strong Induction
Theorem: Every natural number n > 1 can be written as a product of primes.
Fact: A prime n has exactly 2 factors, 1 and n.
Proof :
• Base Case: n = 2.
• P (n) = “n can be written as a product of primes.”
Either n + 1 is a prime or n + 1 = a · · · b where 1 < a, b < n + 1.
P (n) says nothing about a, b.
Strong Induction Principle: If P (0) and ((∀k ∈ N)(P (0) ∧ . . . ∧ P (k)) =⇒ P (k + 1))),
then ∀k ∈ N(P (k)).
Strong Induction Hypothesis: a and b are products of primes.
Strong Induction is just a rewriting of Regular Induction.
Let Q(k) = P (0) ∧ P (1) . . . P (k).
By the induction principle: If Q(0), and (∀k ∈ N)(Q(k) =⇒ Q(k+1)) then (∀k ∈ N)(Q(k)).
Also, Q(0) ≡ P (0), and (∀k ∈ N)(Q(k)) ≡ (∀k ∈ N)(P (k)).
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(∀k ∈ N)(Q(k) =⇒ Q(k + 1))
≡ (∀k ∈ N)((P (0) ∧ . . . ∧ P (k)) =⇒ ((P (0) ∧ . . . ∧ P (k) ∧ P (k + 1))
≡ (∀k ∈ N)((P (0) ∧ . . . ∧ P (k)) =⇒ P (k + 1)
because A =⇒ A ∧ B ≡ A =⇒ B
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Well Ordering Principle
Well Ordering Principle states tht for any subset of N, there is a smallest element.
We know: (∀n)P (n) is not true, then (∃n)¬P (n).
∃n, s.t. ¬P (n).
Cases:
• n = 0: ¬P (0)
• n > 0:
– Let m be the smallest n s.t. ¬P (m).
– P (m − 1) =⇒ P (m) is false.
This proves the induction principle:
• Induction Principle: [P (0) ∧ (∀k ∈ N)(P (k) =⇒ P (k + 1))] =⇒ ∀nP (n)
• Contrapositive: ¬∀nP (n) =⇒ ¬[P (0) ∧ (∀k ∈ N)(P (k) =⇒ P (k + 1))]
• can be written as: ∃n¬P (n) =⇒ [¬P (0) ∨ ¬(∀k ∈ N)(P (k) =⇒ P (k + 1))]
• ¬(∀k ∈ N)(P (k) =⇒ P (k + 1)) holds because P (m − 1) =⇒ P (m) is false. k = m.
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