Mathematical Induction - University of British Columbia ...elyse/220/2016/10Induction.pdf · Induction 10.1 Strong Induction 10.2 Smallest Counterex-ample Mathematical Induction S

10. Mathe-maticalInduction

10.1StrongInduction

10.2SmallestCounterex-ample

Mathematical Induction

S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

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Like dominoes!

https://www.youtube.com/watch?v=ZzRcGDJaxsU


10.1StrongInduction



S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

S0 S1 S2 S3S4 S5 S6 S7 S8 S9 S10

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Like dominoes!



10.1StrongInduction



S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10S0 S1 S2 S3S4 S5 S6 S7 S8 S9 S10

S0 S1 S2 S3

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S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

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Like dominoes!



10.1StrongInduction



S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10S0 S1 S2 S3S4 S5 S6 S7 S8 S9 S10S0 S1 S2 S3

S4

S5 S6 S7 S8 S9 S10

S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

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Like dominoes!



10.1StrongInduction




S4

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S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

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Like dominoes!



10.1StrongInduction




S4

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S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

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10.1StrongInduction




S4

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S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

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10.1StrongInduction




S4

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S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

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10.1StrongInduction




S4

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S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

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Like dominoes!



10.1StrongInduction




S4

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S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

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10.1StrongInduction



S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 · · ·

S0 S1 S2 S3 S4 S5 S7 S8 S9S6 · · ·S0 S1 S2 S3 S4 S5 S8 S9S6 S7 · · ·S1 S2 S3 S4 S5 S6 S7 S8 S9S0 · · ·S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 · · ·

For any k ∈ N, if Sk is true, then Sk+1 is true.(This is called the “inductive step.” The assumption Sk is true is calledthe “inductive hypothesis.”)

Sa is true for some a ∈ N.(This is called the “base case.”)

Therefore, Sk is true for all k ≥ a.


10.1StrongInduction



S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 · · ·

S0 S1 S2 S3 S4 S5 S7 S8 S9S6 · · ·

S0 S1 S2 S3 S4 S5 S8 S9S6 S7 · · ·S1 S2 S3 S4 S5 S6 S7 S8 S9S0 · · ·S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 · · ·





10.1StrongInduction



S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 · · ·S0 S1 S2 S3 S4 S5 S7 S8 S9S6 · · ·

S0 S1 S2 S3 S4 S5 S8 S9S6 S7 · · ·

S1 S2 S3 S4 S5 S6 S7 S8 S9S0 · · ·S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 · · ·





10.1StrongInduction



S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 · · ·S0 S1 S2 S3 S4 S5 S7 S8 S9S6 · · ·S0 S1 S2 S3 S4 S5 S8 S9S6 S7 · · ·

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S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 · · ·





10.1StrongInduction



S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 · · ·S0 S1 S2 S3 S4 S5 S7 S8 S9S6 · · ·S0 S1 S2 S3 S4 S5 S8 S9S6 S7 · · ·S1 S2 S3 S4 S5 S6 S7 S8 S9S0 · · ·

S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 · · ·





10.1StrongInduction



Proposition: Suppose n ∈ N. Then 5|(n5 − n)

Proof: (by induction)

Let k ∈ N, and suppose 5|(k5 − k). Then there exists some x ∈ Z suchthat k5 − k = 5x . Then:

(k + 1)5 − (k + 1) = [k5 + 5k4 + 10k3 + 10k2 + 5k + 1]− (k + 1)

= [k5 − k] + [5k4 + 10k3 + 10k2 + 5k]

= 5x + [5k4 + 10k3 + 10k2 + 5k]

= 5[x + k4 + 2k3 + 2k2 + k]

So, 5∣∣∣((k + 1)5 − (k + 1)

)When n = 1, n5 − n = 0, so 5|(n5 − n).

It follow by induction that, for every n ∈ N, 5|(n5 − n).


10.1StrongInduction



Proposition: Suppose n ∈ N. Then 5|(n5 − n)


Let k ∈ N, and suppose 5|(k5 − k). Then there exists some x ∈ Z suchthat k5 − k = 5x . Then:

(k + 1)5 − (k + 1) = [k5 + 5k4 + 10k3 + 10k2 + 5k + 1]− (k + 1)

= [k5 − k] + [5k4 + 10k3 + 10k2 + 5k]

= 5x + [5k4 + 10k3 + 10k2 + 5k]

= 5[x + k4 + 2k3 + 2k2 + k]

So, 5∣∣∣((k + 1)5 − (k + 1)

)When n = 1, n5 − n = 0, so 5|(n5 − n).

It follow by induction that, for every n ∈ N, 5|(n5 − n).


10.1StrongInduction



Proposition: Suppose n ∈ N. Thenn−1∑k=0

2k = 2n − 1.


Let m ∈ N, and supposem−1∑k=0

2k = 2m − 1. Then:

m∑k=0

2k = 2m +m−1∑k=0

2k

= 2m + (2m − 1) = 2(2m)− 1 = 2m+1 − 1

So, if the statement holds for n = m, then it also holds for n = m + 1.

When n = 1,n−1∑k=0

2k = 20 = 1, and 1 = 2n − 1, so the statement holds

when n = 1.

It follow by induction that, for every n ∈ N,n−1∑k=0

2k = 2n − 1.


10.1StrongInduction



Proposition: Suppose n ∈ N. Thenn−1∑k=0

2k = 2n − 1.


Let m ∈ N, and supposem−1∑k=0

2k = 2m − 1. Then:

m∑k=0

2k = 2m +m−1∑k=0

2k

= 2m + (2m − 1) = 2(2m)− 1 = 2m+1 − 1

So, if the statement holds for n = m, then it also holds for n = m + 1.

When n = 1,n−1∑k=0

2k = 20 = 1, and 1 = 2n − 1, so the statement holds

when n = 1.

It follow by induction that, for every n ∈ N,n−1∑k=0

2k = 2n − 1.


10.1StrongInduction



Proposition: Let f (x) = x ln x . Then for all n ∈ N such that n ≥ 2,

f (n)(x) = (−1)n(n − 2)!x1−n.

(Recall 0! = 1.)


Suppose k ∈ N and f (k)(x) = (−1)k(k − 2)!x1−k . Then

f (k+1)(x) =d

dx

{f (k)(x)

}=

d

dx

{(−1)k(k − 2)!x1−k

}= (1− k)(−1)k(k − 2)!x1−k−1

= −(k − 1)(−1)k(k − 2)!x1−(k+1)

= (−1)k+1(k − 1)!x1−k+1

So, if the statement holds for n = k, then is also holds for n = k + 1.

Note f ′(x) = ln x + 1 (by the product rule), sof (2)(x) = 1

x= (−1)20!x1−2, so the statement holds for n = 2.

By induction, we conclude the statement holds for all natural n ≥ 2.


10.1StrongInduction



Proposition: Let f (x) = x ln x . Then for all n ∈ N such that n ≥ 2,

f (n)(x) = (−1)n(n − 2)!x1−n.

(Recall 0! = 1.)


Suppose k ∈ N and f (k)(x) = (−1)k(k − 2)!x1−k . Then

f (k+1)(x) =d

dx

{f (k)(x)

}=

d

dx

{(−1)k(k − 2)!x1−k

}= (1− k)(−1)k(k − 2)!x1−k−1

= −(k − 1)(−1)k(k − 2)!x1−(k+1)

= (−1)k+1(k − 1)!x1−k+1

So, if the statement holds for n = k, then is also holds for n = k + 1.

Note f ′(x) = ln x + 1 (by the product rule), sof (2)(x) = 1

x= (−1)20!x1−2, so the statement holds for n = 2.

By induction, we conclude the statement holds for all natural n ≥ 2.


10.1StrongInduction


Cautionary Tale

Proposition: Let f (x) = ex Then f (n)(x) = 0 for all n ∈ N.

Proof:

Suppose f (n)(x) = 0. Then f (n+1)(x) = ddx{0} = 0. So, if the statement

holds for n, then it also holds for n + 1.

We conclude by induction that f (n)(x) = 0 for all n ∈ N.

We need a base case! This is clearly false.


10.1StrongInduction


Cautionary Tale

Proposition: Let f (x) = ex Then f (n)(x) = 0 for all n ∈ N.

Proof:

Suppose f (n)(x) = 0. Then f (n+1)(x) = ddx{0} = 0. So, if the statement

holds for n, then it also holds for n + 1.

We conclude by induction that f (n)(x) = 0 for all n ∈ N.

We need a base case! This is clearly false.


10.1StrongInduction


Induction: Triominoes

Triomino


10.1StrongInduction



Triomino

4× 4 chessboard


10.1StrongInduction



Triomino

4× 4 chessboard


10.1StrongInduction



Triomino

4× 4 chessboard


10.1StrongInduction



Triomino

4× 4 chessboard


10.1StrongInduction



Triomino

4× 4 chessboard


10.1StrongInduction



Triomino


10.1StrongInduction



Proposition: For any n ∈ N, a chessboard of dimensions 2n × 2n that ismissing any single square can be tiled by triominoes.

Proof : (by induction)

When n = 1, the board is exactly the shape of a single triomino, so it can betiled.

Suppose any board of dimensions 2n × 2n missing one square can be tiled.Consider a board of dimension 2n+1 × 2n+1, missing any one square. Then wecan cut the board into four equal-sized boards, each of dimension 2n × 2n. Oneof these contains the missing square; by the inductive hypothesis, this can betiled. The other three smaller boards do not contain any missing pieces. Place atriomino on the three squares belonging to the three smaller boards that meetin the middle of the board (as in the picture below). Now the smaller boards areall of size 2n × 2n, and we need to tile all but one of their squares. By theinductive hypothesis, we can finish the tiling.

We conclude that the proposition holds.


10.1StrongInduction









10.1StrongInduction






Suppose any board of dimensions 2n × 2n missing one square can be tiled.Consider a board of dimension 2n+1 × 2n+1, missing any one square.

Then wecan cut the board into four equal-sized boards, each of dimension 2n × 2n. Oneof these contains the missing square; by the inductive hypothesis, this can betiled. The other three smaller boards do not contain any missing pieces. Place atriomino on the three squares belonging to the three smaller boards that meetin the middle of the board (as in the picture below). Now the smaller boards areall of size 2n × 2n, and we need to tile all but one of their squares. By theinductive hypothesis, we can finish the tiling.



10.1StrongInduction






Suppose any board of dimensions 2n × 2n missing one square can be tiled.Consider a board of dimension 2n+1 × 2n+1, missing any one square. Then wecan cut the board into four equal-sized boards, each of dimension 2n × 2n. Oneof these contains the missing square; by the inductive hypothesis, this can betiled. The other three smaller boards do not contain any missing pieces.

Place atriomino on the three squares belonging to the three smaller boards that meetin the middle of the board (as in the picture below). Now the smaller boards areall of size 2n × 2n, and we need to tile all but one of their squares. By theinductive hypothesis, we can finish the tiling.



10.1StrongInduction









10.1StrongInduction


Triominoes tiling


10.1StrongInduction


Triominoes tiling


10.1StrongInduction


Triominoes tiling


10.1StrongInduction


Triominoes tiling


10.1StrongInduction


Triominoes tiling


10.1StrongInduction


Triominoes tiling


10.1StrongInduction


Triominoes tiling


10.1StrongInduction


Triominoes tiling


10.1StrongInduction


Triominoes tiling


10.1StrongInduction


Triominoes tiling


10.1StrongInduction


Triomino Corollary

Corollary: For any n ∈ N, 3|22n − 1.

Proof: 22n − 1 is the number of squares on a chessboard of dimension 2n × 2n

that’s missing one piece. Since these squares can be covered completely bytriomino pieces that have three squares each, 3|22n − 1.


10.1StrongInduction


Triomino Corollary

Corollary: For any n ∈ N, 3|22n − 1.

Proof: 22n − 1 is the number of squares on a chessboard of dimension 2n × 2n

that’s missing one piece. Since these squares can be covered completely bytriomino pieces that have three squares each, 3|22n − 1.


10.1StrongInduction


Prime Divisors

Proposition: Let p be prime. For any integers a1, . . . , an, if p divides theproduct a1 · · · an, then p divides some ai , 1 ≤ i ≤ n.

Lemma: If a and b are integers with gcd(a, b) = 1, then there exist k, ` ∈ Zsuch that ak + b` = 1.

Proof: We proceed by induction on the number of integers. For the base case,suppose n = 2, and p|a1a2. If p|a1, we’re done. Suppose p 6 | a1. Since p isprime, gcd(p, a1) = 1, so there exist k, ` ∈ Z such that kp + a1` = 1.Multiplying by a2, we see

a2 = kpa2 + a1a2`.

Since p|a1a2, we see p divides the right hand side of the equation, so p|a2.This finishes the base case.

Suppose now that the corollary holds whenever n ≤ m for some m ∈ Z.Suppose p|a1 · · · amam+1. Let A = a1 · · · am. If p|am+1, we’re done. If not,then since p|A · am+1, by the base step, p|A. Then by the inductivehypothesis, p divides some ai , 1 ≤ i ≤ m. This finishes the proof.


10.1StrongInduction


Prime Divisors




a2 = kpa2 + a1a2`.




10.1StrongInduction


Prime Divisors




a2 = kpa2 + a1a2`.




10.1StrongInduction


Strong Induction

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Outline for proof by strong induction:

Suppose Sn is true for every n ≤ k. Then Sk+1 is true as well.This part is the inductive step;the inductive hypothesis is that Sn is true for every n ≤ k.

Sk is true for all k ≤ a.Base case(s)

We conclude by (strong) induction that Sk is true for all k.


10.1StrongInduction


Strong Induction

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10.1StrongInduction


Strong Induction

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10.1StrongInduction


Strong Induction

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10.1StrongInduction


Strong Induction

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S3 S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5

S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5

S6

S7 S8 S9 S10






10.1StrongInduction


Strong Induction

S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3 S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5

S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5

S6

S7 S8 S9 S10






10.1StrongInduction


Strong Induction

S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3 S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5

S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5

S6

S7 S8 S9 S10






10.1StrongInduction


Strong Induction

S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3 S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5

S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5

S6

S7 S8 S9 S10






10.1StrongInduction


Strong Induction

S0 S1 S2 S3 S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3 S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4 S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5 S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5

S6 S7 S8 S9 S10

S0

S1

S2

S3

S4

S5

S6

S7 S8 S9 S10






10.1StrongInduction


Strong Induction

Proposition: Let {an} be the sequence of numbers defined as follows:

a1 = 2

a2 = 4

a3 = 8

an = an−1 + an−2 + 2an−3 for every natural n ≥ 4.

Then an = 2n for every n ∈ N.

Proof:

Suppose an = 2n for every natural n ≤ k, and k ≥ 3. Then k + 1 ≥ 4, so:

ak+1 = ak + ak−1 + 2ak−2

= 2k + 2k−1 + 2 · 2k−2 = 2k + 2 · 2k−1

= 2 · 2k = 2k+1

Then ak+1 = 2k+1 for every natural n ≤ k + 1.

Note a1 = 21, a2 = 22, and a3 = 23.

By strong induction, we conclude an = 2n for every n ∈ N.


10.1StrongInduction


Strong Induction

Proposition: Let {an} be the sequence of numbers defined as follows:

a1 = 2

a2 = 4

a3 = 8

an = an−1 + an−2 + 2an−3 for every natural n ≥ 4.

Then an = 2n for every n ∈ N.

Proof:

Suppose an = 2n for every natural n ≤ k, and k ≥ 3. Then k + 1 ≥ 4, so:

ak+1 = ak + ak−1 + 2ak−2

= 2k + 2k−1 + 2 · 2k−2 = 2k + 2 · 2k−1

= 2 · 2k = 2k+1

Then ak+1 = 2k+1 for every natural n ≤ k + 1.

Note a1 = 21, a2 = 22, and a3 = 23.

By strong induction, we conclude an = 2n for every n ∈ N.


10.1StrongInduction


Strong Induction: Nim

Game of Nim

There are two piles of matches, and two players. A turn consists of a playermay take as many matches as they like (but at least one) from one of thepiles. The player who takes the last match wins.

Proposition: In a game of Nim where both piles of matches are initially thesame size, the second player can always win if they take an equal number ofmatches as their opponent did in their last term, but from the otherpile.

Proof : (by induction on the initial size of the piles)

Suppose we know Player 2 can win when the sizes of the piles are less than mfor some m ∈ N. (Inductive Hypothesis) Player 1 takes k matches from onepile, for some k ≤ m. If k = m, then Player 2 finishes off the remaining pile andwins. If k 6= m, then after Player 2’s move, each pile has size m − k, and1 ≤ m − k < m, so by the IH, Player 2 can now win.

When each pile has size 1, Player 2 wins on their first move.

We conclude by induction that the proposition is true.


10.1StrongInduction



Game of Nim








10.1StrongInduction



Game of Nim








10.1StrongInduction


Strong Induction

Proposition: For any convex polygon with n ≥ 3 edges, the sum of theinterior angles is (n − 2)180 degrees.

Lemma: If you draw a line between any two (non-consecutive) points on aconvex polygon, you divide that convex polygon into two (smaller) convexpolygons.

Lemma: The interior angles of any triangle sum to 180 degrees.

First, note that a triangle has angles summing to 180 degrees.

Suppose the proposition holds whenever n ≤ k − 1. Then consider anypolygon with k edges. Draw a point between two ponts that areseparated by only one point, so the polygon is cut into a triangle and aconvex polygon on k − 1 points. Then the sum of the angles inside thetriangle is 180, while the sum of the angles inside the other polygon is(by the inductive hypothesis) 180(k − 3). Then the sum of the insideangles of the original polygon is 180 + 180(k − 3) = 180(k − 2), so theproposition also holds for n = k.

By strong induction, we conclude the proposition holds for all naturaln ≥ 3.


10.1StrongInduction


Strong Induction

Proposition: For any convex polygon with n ≥ 3 edges, the sum of theinterior angles is (n − 2)180 degrees.

Lemma: If you draw a line between any two (non-consecutive) points on aconvex polygon, you divide that convex polygon into two (smaller) convexpolygons.

Lemma: The interior angles of any triangle sum to 180 degrees.

First, note that a triangle has angles summing to 180 degrees.

Suppose the proposition holds whenever n ≤ k − 1. Then consider anypolygon with k edges. Draw a point between two ponts that areseparated by only one point, so the polygon is cut into a triangle and aconvex polygon on k − 1 points. Then the sum of the angles inside thetriangle is 180, while the sum of the angles inside the other polygon is(by the inductive hypothesis) 180(k − 3). Then the sum of the insideangles of the original polygon is 180 + 180(k − 3) = 180(k − 2), so theproposition also holds for n = k.

By strong induction, we conclude the proposition holds for all naturaln ≥ 3.


10.1StrongInduction


Proof by Smallest Counterexample

Proposition: The statements S1, S2, . . . are all true.

Proof:

S1 is true

Suppose (by way of contradiction) that not every Sn is true

Then there exists some smallest n ∈ N such that Sn is false

Then Sn is false and Sn−1 is true

...

Contradiction.


10.1StrongInduction




Proof:

S1 is true




...

Contradiction.


10.1StrongInduction




Proof:

S1 is true




...

Contradiction.


10.1StrongInduction




Proof:

S1 is true




...

Contradiction.


10.1StrongInduction




Proof:

S1 is true




...

Contradiction.


10.1StrongInduction


Proof by Smallest Counterexample: Fibonnaci Numbers

Fibonnaci Sequence

F0 = 1 • F1 = 1 • F2 = 2 • For n ≥ 3 in N, Fn = Fn−1 + Fn−2

Let ϕ be the golden ratio, ϕ =1 +√

5

2.

Proposition: For every n ∈ N, Fn < ϕn. Proof:

Note that φ > 32, so F1 < φ1 and F2 <

94< φ2. (Base cases.)

Suppose Fn is the first Fibonnaci number such that Fn ≥ ϕn. Then n ≥ 3(by the base cases) and so Fn−1 < ϕn−1 and Fn−2 < ϕn−2. Then:

ϕn ≤ Fn = Fn−1 + Fn−2 < ϕn−1 + ϕn−2

ϕ2 < ϕ+ 1(1 +√

5

2

)2

<1 +√

5

2+ 1

6 + 2√

5

4<

3 +√

5

2

3 +√

5 < 3 +√

5

This is a contradiction.

We conclude that there does not exist any n ∈ N such that Fn ≥ ϕ.Therefore, the proposition is true.


10.1StrongInduction



Fibonnaci Sequence

F0 = 1 • F1 = 1 • F2 = 2 • For n ≥ 3 in N, Fn = Fn−1 + Fn−2


5

2.

Proposition: For every n ∈ N, Fn < ϕn.

Proof:




ϕn ≤ Fn = Fn−1 + Fn−2 < ϕn−1 + ϕn−2

ϕ2 < ϕ+ 1(1 +√

5

2

)2

<1 +√

5

2+ 1

6 + 2√

5

4<

3 +√

5

2

3 +√

5 < 3 +√

5




10.1StrongInduction




5

2.

Proposition: For every n ∈ N, Fn < ϕn. Proof:




ϕn ≤ Fn = Fn−1 + Fn−2 < ϕn−1 + ϕn−2

ϕ2 < ϕ+ 1(1 +√

5

2

)2

<1 +√

5

2+ 1

6 + 2√

5

4<

3 +√

5

2

3 +√

5 < 3 +√

5




10.1StrongInduction


Proof by Smallest Counterexample: 4|5n − 1

Proposition: For every n ∈ N, 4|5n − 1.

Proof:

Note that when n = 1, 5|51 − 1.

Suppose n is the smallest natural number so that 4 6 | 5n − 1. Then n ≥ 2(by the base case), so n − 1 ∈ N (that is–it isn’t zero), so by hypothesis,4|5n−1 − 1. That is, there exists some x ∈ N such that 4x = 5n−1 − 1.Now:

5n − 1 = 5 · 5n−1 − 1 = 5 (4x + 1)− 1

= 5(4x) + 5− 1 = 4(5x) + 4 = 4(5x + 1)

This contradicts the assumption that 4 6 | 5n − 1.

We conclude that there is no natural n such that 4 6 | 5n − 1, so theproposition is true.


10.1StrongInduction


Proof by Smallest Counterexample: 4|5n − 1

Proposition: For every n ∈ N, 4|5n − 1.

Proof:

Note that when n = 1, 5|51 − 1.

Suppose n is the smallest natural number so that 4 6 | 5n − 1. Then n ≥ 2(by the base case), so n − 1 ∈ N (that is–it isn’t zero), so by hypothesis,4|5n−1 − 1. That is, there exists some x ∈ N such that 4x = 5n−1 − 1.Now:

5n − 1 = 5 · 5n−1 − 1 = 5 (4x + 1)− 1

= 5(4x) + 5− 1 = 4(5x) + 4 = 4(5x + 1)

This contradicts the assumption that 4 6 | 5n − 1.

We conclude that there is no natural n such that 4 6 | 5n − 1, so theproposition is true.


10.1StrongInduction


Proof by Smallest Counterexample:Fundamental Theorem of Arithmetic

Fundamental Theorem of Arithmetic

Every n ∈ N greater than 1 has a unique prime factorization. That is, ifn = p1 · p2 · p3 · · · pk and n = a1 · a2 · a3 · · · a` are two prime factorizations of n,then k = `, and the primes pi and ai are the same (but they might be in adifferent order).

This proof involves both existence and uniqueness.

Proof: First, we will use strong induction to show that every integer n > 1 hasa prime factorization. For the base case, suppose n = 2. Then 2 = 2 is theprime factorization of n. For the inductive step, suppose there exists aninteger k ≥ 2 such that that every integer n in [2, k] has a prime factorization.We consider the integer k + 1. If k + 1 is prime, then it is its own primefactorization. If it is not prime, then k + 1 = ab for some positive integers aand b, both of which are greater than 1 and less than k + 1. By the inductivehypothesis, a = p1 · · · pk and b = a1 · · · a` for primes pi , ai . Thenk + 1 = p1 · · · pk · a1 · · · a`. So, k + 1 has a prime factorization. We concludeby induction that every integer greater than 1 has a primefactorization.


10.1StrongInduction








10.1StrongInduction








10.1StrongInduction


Fundamental Theorem of Arithmetic, Cont’d

Now we will prove uniqueness, using proof by smallest counterexample. Forthe base case, note that n = 2 has only one prime factorization, since the onlyprime divisor of 2 is itself. Suppose for the sake of contradiction that there issome integer n > 2 with different prime factorizations. Let n be, in particular,the smallest such integer, and label the factorizations n = p1 · · · pk anda1 · · · a`. Then p1|n, so p1|a1 · · · a`. Then by a previous theorem, p1|ai forsome 1 ≤ i ≤ `. Then, since ai is prime, ai = p1. But now the integer n

p1has

two different prime factorizations, p2 · · · pk and a1 · · · ai−1ai+1 · · · a`. Thiscontradicts that n was the smallest such integer, and finishes the proof.

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Mathematical Induction - University of British Columbia ...elyse/220/2016/10Induction.pdf · Induction 10.1 Strong Induction 10.2 Smallest Counterex-ample Mathematical Induction S