[PPT] - Mathematical Induction COMPSCI 230 Discrete Math March 26, 2015 PowerPoint Presentation

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Mathematical Induction

COMPSCI 230 — Discrete Math March 26, 2015

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 1 / 17

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Mathematical Induction

1 Mathematical Induction

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 2 / 17

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Mathematical Induction

Used to prove predicates of the form

∀𝑜 ∈ ℤ ∶ 𝑜 ≥ 𝑏 → 𝑄(𝑜)

Inference rule: Let 𝑐 be an integer with 𝑐 ≥ 𝑏.

Base case(s) : 𝑄(𝑏) ∧ … ∧ 𝑄(𝑐) Inductive step : ∀𝑙 ∈ ℤ ∶ ((𝑙 ≥ 𝑐)∧ 𝑄(𝑙)) → 𝑄(𝑙 + 1)) Conclusion : ∀𝑜 ∈ ℤ ∶ 𝑜 ≥ 𝑏 → 𝑄(𝑜)

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 3 / 17

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Mathematical Induction

The Domino Efgect

Simplest case (𝑏 = 𝑐 = 0):

Base case(s) : 𝑄(0) Inductive step : ∀𝑙 ∈ ℕ ∶ 𝑄(𝑙) → 𝑄(𝑙 + 1) Conclusion : ∀𝑜 ∈ ℕ ∶ 𝑄(𝑜)

Stephen Morris, U. of Toronto [Click here to watch video] COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 4 / 17

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Mathematical Induction

The Domino Efgect

If we have an infinitely long sequence of dominos and

we want to be sure that they all fall, we can show:

Base case: We are able to make the first domino fall
Inductive step: The fall of every domino transfers

enough energy to the next to make it fall as well

Let 𝐺(𝑙) mean “domino 𝑙 falls”
Base case: 𝐺(0)
Inductive step: 𝐺(𝑙) → 𝐺(𝑙 + 1)

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 5 / 17

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Mathematical Induction

Proof by Induction

Let 𝐺(𝑙) mean “domino 𝑙 falls”
Base case: 𝐺(0)
Inductive step: 𝐺(𝑙) → 𝐺(𝑙 + 1)
Proving the conditional does not require 𝐺(𝑙) to be

true: If 𝐺(𝑙) were true, can we conclude that then 𝐺(𝑙 + 1) is true as well?

So to prove the inductive step we tentatively assume

𝐺(𝑙) to be true and show that under that assumption 𝐺(𝑙 + 1) is true as well

If 𝐺(𝑙) is false, the conditional is true anyway,

regardless of the truth value of 𝐺(𝑙 + 1), so there is nothing to prove for that case

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 6 / 17

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Mathematical Induction

Proof by Induction

Proving the base case is typically easy
For dominos, it means showing that we have some

means to topple the first domino

For math, it means showing that the result holds

when plugging in some small numbers

Most of the work goes into proving the induction step
Make the inductive hypothesis that 𝐺(𝑙) is true

that is, assume domino 𝑙 somehow falls

Can we then show that domino 𝑙 + 1 falls as well?

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 7 / 17

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Mathematical Induction

Proving the Induction Step

Proving the induction step for dominos will involve

some physics argument

www.technologyreview.com

For math, this proof will involve arguing that some

conditional holds 𝐺(𝑙) → 𝐺(𝑙 + 1)

We don’t know that 𝐺(𝑙) holds, we just assume it to

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 8 / 17

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Mathematical Induction

Proving the Geometric-Sum Formula

Assume 𝑠 ≠ 1 for simplicity

[The case 𝑠 = 1 is proven similarly]

Abbreviate the definition of 𝑄(𝑜)

∀𝑠 ∈ ℝ ∶ (𝑠 ≠ 1) →

𝑜

∑

𝑘=0

𝑠𝑘 = 1 − 𝑠𝑜+1 1 − 𝑠 to

𝑜

∑

𝑘=0

𝑠𝑘 = 1 − 𝑠𝑜+1 1 − 𝑠

To be proven: ∀𝑜 ∈ ℕ ∶ 𝑄(𝑜)
Base case 𝑄(0): ∑0

𝑘=0 𝑠𝑘 = 𝑠0 = 1 and 1−𝑠𝑜+1 1−𝑠

= 1−𝑠1

1−𝑠 = 1

so 𝑄(0) is true

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 9 / 17

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Mathematical Induction

Proving the Geometric-Sum Formula

Inductive step: 𝑄(𝑙)

?

→ 𝑄(𝑙 + 1)

𝑄(𝑙) ∶ ∑𝑙

𝑘=0 𝑠𝑘 = 1−𝑠𝑙+1 1−𝑠

𝑄(𝑙 + 1) ∶ ∑𝑙+1

𝑘=0 𝑠𝑘 = 1−𝑠𝑙+2 1−𝑠

Inductive hypothesis: 𝑄(𝑙) holds
Under the inductive hypothesis, is 𝑄(𝑙 + 1) true?
∑𝑙+1

𝑘=0 𝑠𝑘 = 𝑠𝑙+1 + ∑𝑙 𝑘=0 𝑠𝑘 = … [By the inductive hypothesis]

… = 𝑠𝑙+1 + 1−𝑠𝑙+1

1−𝑠

= 1−𝑠𝑙+1+(1−𝑠)𝑠𝑙+1

1−𝑠

= 1−𝑠𝑙+2

1−𝑠

So if 𝑄(𝑙) is true, then 𝑄(𝑙 + 1) is true as well
Therefore ∀𝑜 ∈ ℕ ∶ ∑𝑜

𝑘=0 𝑠𝑘 = 1−𝑠𝑜+1 1−𝑠

(if 𝑠 ≠ 1)

Exercise: Do a similar proof for 𝑠 = 1

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 10 / 17

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Mathematical Induction

Validity of Mathematical Induction

Base case(s) : 𝑄(𝑏) ∧ … ∧ 𝑄(𝑐) Inductive step : ∀𝑙 ∈ ℤ ∶ ((𝑙 ≥ 𝑐)∧ 𝑄(𝑙)) → 𝑄(𝑙 + 1)) Conclusion : ∀𝑜 ∈ ℤ ∶ 𝑜 ≥ 𝑏 → 𝑄(𝑜)

Cannot use truth tables because of ∀

[2 clickers]

Prove this by contradiction
The negation of the conclusion is

∃𝑜 ∈ ℤ ∶ ∼(𝑜 ≥ 𝑏 → 𝑄(𝑜))

∼(𝜚 → 𝜔) ⇔ ∼(∼𝜚 ∨ 𝜔)

De Morgan

⇔ 𝜚 ∧ ∼𝜔

So we need to show that some contradiction derives

from assuming that (i) the base cases hold, (ii) the inductive step holds, and (iii) ∃𝑜 ∈ ℤ ∶ 𝑜 ≥ 𝑏 ∧ ∼𝑄(𝑜)

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 11 / 17

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Mathematical Induction

Proof of Validity

If we assume base case: 𝑄(𝑏) ∧ … ∧ 𝑄(𝑐)
and assume inductive step: ∀𝑙 ∈ ℤ ∶ ((𝑙 ≥ 𝑐)∧ 𝑄(𝑙)) → 𝑄(𝑙 + 1))
and negate the conclusion: ∃𝑜 ∈ ℤ ∶ 𝑜 ≥ 𝑏 ∧ ∼𝑄(𝑜)
then we obtain a contradiction
So let 𝑡 be the smallest 𝑜 for which 𝑜 ≥ 𝑏 ∧ ∼𝑄(𝑜)
Then, 𝑡 ≥ 𝑏 and 𝑄(𝑡) is false
Since 𝑄(𝑏) ∧ … ∧ 𝑄(𝑐), we have 𝑡 > 𝑐 and 𝑡 − 1 ≥ 𝑐
Since 𝑡 is the smallest 𝑜 for which 𝑄(𝑜) is false,

𝑄(𝑡 − 1) must be true

So (𝑡 − 1 ≥ 𝑐) ∧ 𝑄(𝑡 − 1) is true
But then by the inductive step with 𝑙 = 𝑡 − 1 we have 𝑄(𝑡) is true
Contradiction! Mathematical induction is valid

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 12 / 17

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Mathematical Induction

Chessboards and Dominos

A 2𝑜 × 2𝑜 chessboard with 𝑜 = 3 and a 2 × 1 domino

The chessboard can always be tiled with dominos for

all 𝑜 ≥ 1

𝑈(𝑜, 𝛽): arrangement 𝛽 tiles a 2𝑜 × 2𝑜 chessboard
∀𝑜 ≥ 1 ∃ 𝛽 ∶ 𝑈(𝑜, 𝛽)
Direct proof: Each row has an even number of cells,

so place 2𝑜−1 dominos horizontally in each row

Easy because of ‘∃’: just show a way to do it
However, proof has to hold for all 𝑜

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 13 / 17

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Mathematical Induction

Also by Induction

Need to state on what: induction on 𝑜
Base case: 𝑜 = 1
Inductive step: ∃ 𝛽 ∶ 𝑈(𝑙, 𝛽)

⏟ ⏟ ⏟ ⏟ ⏟ ⏟ ⏟

𝑄(𝑙)

→ ∃ 𝛾 ∶ 𝑈(𝑙 + 1, 𝛾) ⏟⏟⏟⏟⏟⏟⏟

𝑄(𝑙+1)

Split a 2𝑙+1 × 2𝑙+1 board in four
Each quarter is 2𝑙 × 2𝑙
So if each quarter can be tiled (𝑈(𝑙, 𝛽))

then the whole board is tiled: 𝑈(𝑜 + 1, 𝛾)

Done!

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 14 / 17

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Mathematical Induction

What if we Remove two Cells?

Cannot tile! Argument 1: Base case 𝑜 = 1 fails
Suffjcient but weak. Only proves ∃𝑜 ≥ 1 ∀ 𝛽 ∶ ∼𝑈(𝑜, 𝛽)
Can we prove ∀𝑜 ≥ 1 ∀ 𝛽 ∶ ∼𝑈(𝑜, 𝛽)?
By contradiction, assume that a tiling exists for some 𝑜
The new board has more white cells than black cells
Each domino must cover one black cell and one white cell
Therefore, the board has as many white cells as black cells
Contradiction!

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 15 / 17

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Mathematical Induction

Solomon Golomb’s Tromino Problem

Remove a single cell (black or white) from a regular

2𝑜 × 2𝑜 chessboard

We can always tile it with a tromino
Proof by induction
Base case 𝑜 = 1: Check four possibilities, depending
n location of missing cell

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 16 / 17

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Mathematical Induction

Solomon Golomb’s Tromino Problem

Inductive step: Consider a 2𝑙+1 × 2𝑙+1 board
Assume w.l.o.g. that missing tile is in upper quarter
Inductive assumption: A 2𝑙 × 2𝑙 board with one cell

missing can be tiled

Big idea: Place one tromino as shown
Each quarter is 2𝑙−1 × 2𝑙−1 and misses one cell

so it can be tiled (by the inductive assumption)

Done!

COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 17 / 17