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

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

Mathematical Induction 1 Mathematical Induction COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 2 / 17

Mathematical Induction ∀𝑙 ∈ ℤ ∶ ((𝑙 ≥ 𝑐)∧ 𝑄(𝑙)) → 𝑄(𝑙 + 1)) March 26, 2015 Mathematical Induction COMPSCI 230 — Discrete Math ∀𝑜 ∈ ℤ ∶ 𝑜 ≥ 𝑏 → 𝑄(𝑜) : Conclusion : Mathematical Induction Inductive step 𝑄(𝑏) ∧ … ∧ 𝑄(𝑐) : Base case(s) ∀𝑜 ∈ ℤ ∶ 𝑜 ≥ 𝑏 → 𝑄(𝑜) 3 / 17 • Used to prove predicates of the form • Inference rule: Let 𝑐 be an integer with 𝑐 ≥ 𝑏 .

Mathematical Induction : March 26, 2015 Mathematical Induction COMPSCI 230 — Discrete Math [Click here to watch video] Stephen Morris, U. of Toronto ∀𝑜 ∈ ℕ ∶ 𝑄(𝑜) Conclusion The Domino Efgect ∀𝑙 ∈ ℕ ∶ 𝑄(𝑙) → 𝑄(𝑙 + 1) : Inductive step 𝑄(0) : Base case(s) 4 / 17 • Simplest case ( 𝑏 = 𝑐 = 0 ):

Mathematical Induction The Domino Efgect we want to be sure that they all fall, we can show: enough energy to the next to make it fall as well COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 5 / 17 • If we have an infinitely long sequence of dominos and • Base case: We are able to make the first domino fall • Inductive step: The fall of every domino transfers • Let 𝐺(𝑙) mean “domino 𝑙 falls” • Base case: 𝐺(0) • Inductive step: 𝐺(𝑙) → 𝐺(𝑙 + 1)

Mathematical Induction 𝐺(𝑙) to be true and show that under that assumption March 26, 2015 Mathematical Induction COMPSCI 230 — Discrete Math nothing to prove for that case regardless of the truth value of 𝐺(𝑙 + 1) , so there is 𝐺(𝑙 + 1) is true as well 6 / 17 Proof by Induction is true as well? If 𝐺(𝑙) were true, can we conclude that then 𝐺(𝑙 + 1) true: • 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 • So to prove the inductive step we tentatively assume • If 𝐺(𝑙) is false, the conditional is true anyway,

Mathematical Induction Proof by Induction means to topple the first domino when plugging in some small numbers that is, assume domino 𝑙 somehow falls COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 7 / 17 • Proving the base case is typically easy • For dominos, it means showing that we have some • For math, it means showing that the result holds • Most of the work goes into proving the induction step • Make the inductive hypothesis that 𝐺(𝑙) is true • Can we then show that domino 𝑙 + 1 falls as well?

Mathematical Induction Proving the Induction Step some physics argument www.technologyreview.com conditional holds 𝐺(𝑙) → 𝐺(𝑙 + 1) COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 8 / 17 • Proving the induction step for dominos will involve • For math, this proof will involve arguing that some • We don’t know that 𝐺(𝑙) holds, we just assume it to

Mathematical Induction 𝑜 March 26, 2015 Mathematical Induction COMPSCI 230 — Discrete Math so 𝑄(0) is true 1−𝑠 1 − 𝑠 Proving the Geometric-Sum Formula ∑ 𝑘=0 to 1 − 𝑠 [The case 𝑠 = 1 is proven similarly] ∀𝑠 ∈ ℝ ∶ (𝑠 ≠ 1) → 𝑜 9 / 17 𝑘=0 ∑ • Assume 𝑠 ≠ 1 for simplicity • Abbreviate the definition of 𝑄(𝑜) 𝑠 𝑘 = 1 − 𝑠 𝑜+1 𝑠 𝑘 = 1 − 𝑠 𝑜+1 • To be proven: ∀𝑜 ∈ ℕ ∶ 𝑄(𝑜) 𝑘=0 𝑠 𝑘 = 𝑠 0 = 1 and 1−𝑠 𝑜+1 = 1−𝑠 1 • Base case 𝑄(0) : ∑ 0 1−𝑠 = 1

Mathematical Induction Proving the Geometric-Sum Formula March 26, 2015 Mathematical Induction COMPSCI 230 — Discrete Math (if 𝑠 ≠ 1 ) 1−𝑠 1−𝑠 1−𝑠 1−𝑠 10 / 17 1−𝑠 ? → 𝑄(𝑙 + 1) 1−𝑠 • Inductive step: 𝑄(𝑙) 𝑘=0 𝑠 𝑘 = 1−𝑠 𝑙+1 • 𝑄(𝑙) ∶ ∑ 𝑙 𝑘=0 𝑠 𝑘 = 1−𝑠 𝑙+2 • 𝑄(𝑙 + 1) ∶ ∑ 𝑙+1 • Inductive hypothesis : 𝑄(𝑙) holds • Under the inductive hypothesis, is 𝑄(𝑙 + 1) true? 𝑘=0 𝑠 𝑘 = 𝑠 𝑙+1 + ∑ 𝑙 𝑘=0 𝑠 𝑘 = … [By the inductive hypothesis] • ∑ 𝑙+1 … = 𝑠 𝑙+1 + 1−𝑠 𝑙+1 = 1−𝑠 𝑙+1 +(1−𝑠)𝑠 𝑙+1 = 1−𝑠 𝑙+2 • So if 𝑄(𝑙) is true, then 𝑄(𝑙 + 1) is true as well 𝑘=0 𝑠 𝑘 = 1−𝑠 𝑜+1 • Therefore ∀𝑜 ∈ ℕ ∶ ∑ 𝑜 • Exercise: Do a similar proof for 𝑠 = 1

Mathematical Induction Validity of Mathematical Induction March 26, 2015 Mathematical Induction COMPSCI 230 — Discrete Math ∃𝑜 ∈ ℤ ∶ 𝑜 ≥ 𝑏 ∧ ∼𝑄(𝑜) inductive step holds, and (iii) from assuming that (i) the base cases hold, (ii) the 𝜚 ∧ ∼𝜔 ⇔ De Morgan ∃𝑜 ∈ ℤ ∶ ∼(𝑜 ≥ 𝑏 → 𝑄(𝑜)) 11 / 17 [2 clickers] ∀𝑜 ∈ ℤ ∶ 𝑜 ≥ 𝑏 → 𝑄(𝑜) : Conclusion ∀𝑙 ∈ ℤ ∶ ((𝑙 ≥ 𝑐)∧ 𝑄(𝑙)) → 𝑄(𝑙 + 1)) : Inductive step 𝑄(𝑏) ∧ … ∧ 𝑄(𝑐) : Base case(s) • Cannot use truth tables because of ∀ • Prove this by contradiction • The negation of the conclusion is • ∼(𝜚 → 𝜔) ⇔ ∼(∼𝜚 ∨ 𝜔) • So we need to show that some contradiction derives

Mathematical Induction Proof of Validity March 26, 2015 Mathematical Induction COMPSCI 230 — Discrete Math 𝑄(𝑡 − 1) must be true 12 / 17 • 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, • 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

Mathematical Induction Chessboards and Dominos all 𝑜 ≥ 1 COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 13 / 17 A 2 𝑜 × 2 𝑜 chessboard with 𝑜 = 3 and a 2 × 1 domino • The chessboard can always be tiled with dominos for • 𝑈(𝑜, 𝛽) : arrangement 𝛽 tiles a 2 𝑜 × 2 𝑜 chessboard • ∀𝑜 ≥ 1 ∃ 𝛽 ∶ 𝑈(𝑜, 𝛽) • Direct proof: Each row has an even number of cells, so place 2 𝑜−1 dominos horizontally in each row • Easy because of ‘ ∃ ’: just show a way to do it • However, proof has to hold for all 𝑜

Mathematical Induction ⏟ March 26, 2015 Mathematical Induction COMPSCI 230 — Discrete Math then the whole board is tiled: 𝑈(𝑜 + 1, 𝛾) 𝑄(𝑙+1) ⏟⏟⏟⏟⏟⏟⏟ → ∃ 𝛾 ∶ 𝑈(𝑙 + 1, 𝛾) Also by Induction 𝑄(𝑙) ⏟ ⏟ ⏟ ⏟ ⏟ ⏟ 14 / 17 • Need to state on what: induction on 𝑜 • Base case: 𝑜 = 1 • Inductive step: ∃ 𝛽 ∶ 𝑈(𝑙, 𝛽) • Split a 2 𝑙+1 × 2 𝑙+1 board in four • Each quarter is 2 𝑙 × 2 𝑙 • So if each quarter can be tiled ( 𝑈(𝑙, 𝛽) ) • Done!

Mathematical Induction What if we Remove two Cells? COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 15 / 17 • 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!

Mathematical Induction Solomon Golomb’s Tromino Problem on location of missing cell COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 16 / 17 • Remove a single cell (black or white) from a regular 2 𝑜 × 2 𝑜 chessboard • We can always tile it with a tromino • Proof by induction • Base case 𝑜 = 1 : Check four possibilities, depending

Mathematical Induction Solomon Golomb’s Tromino Problem missing can be tiled so it can be tiled (by the inductive assumption) COMPSCI 230 — Discrete Math Mathematical Induction March 26, 2015 17 / 17 • Inductive step: Consider a 2 𝑙+1 × 2 𝑙+1 board • Assume w.l.o.g. that missing tile is in upper quarter • Inductive assumption: A 2 𝑙 × 2 𝑙 board with one cell • Big idea: Place one tromino as shown • Each quarter is 2 𝑙−1 × 2 𝑙−1 and misses one cell • Done!