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# Strong Induction EECS 203: Discrete PowerPoint Presentation, PPT - DocSlides

briana-ranney | 2019-03-15 | General
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Mathematics. 1. Mathematical . vs. Strong Induction . To prove that . P. (. n. ) is true for all positive . n. .. Mathematical. induction:. Strong. induction:. 2. Climbing the Ladder (Strongly). We want to show that ∀. ID: 756520

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Slide1

Strong Induction

EECS 203: Discrete Mathematics

1

Slide2Mathematical vs Strong Induction

To prove that P(n

) is true for all positive

n

.Mathematical induction:Strong induction:

2

Slide3Climbing the Ladder (Strongly)

We want to show that ∀n≥1 P(n) is true.

Think of the positive integers as a ladder.

1, 2, 3, 4, 5, 6, . . .

You can reach the bottom of the ladder: P(1)Given

all lower steps, you can reach the next.

P(1) → P(2), P(1) ∧

P

(2) →

P(3), . . . ∀k≥1 P(1) ∧ … ∧ P(k) → P(k+1)Then, by strong induction: ∀n≥1 P(n)

3

Slide4Is Strong Induction Really Stronger?

No. Anything you can prove with strong induction can be proved with regular mathematical induction. And vice versa.

B

oth are equivalent to the

well-ordering property.But strong induction can simplify a proof.How?

Sometimes P(k) is not enough to prove P(k+1).But P(1) ∧ . . . ∧ P(k) is strong enough.

4

Slide55

Slide6Coin problem

What is the largest cent-value that cannot be formed using only 3-cent and 5-cent stamps?(A) 2(B) 4

(

C

) 7(D) 8(E) 116

<= Correct answer

Slide7Proof for our Coin problem

Let P(k) = “k cents can be formed using 3-cent and 5-cent stamps.”

Claim

:

∀n≥8 P(n).

Proof by strong induction:Base cases:P

(8): 8 = 3 + 5P(9): 9 = 3 + 3 + 3P(10): 10 = 5 + 5.

7

Slide8Proof for

our Coin problem

Inductive step:

Let

k

be an integer ≥ 11.

Inductive hypothesis: P(j)

is true when 8 ≤

j < k

.

P

(

k

-3) is true.

Therefore, P(

k

) is true. (Add a 3-cent stamp.)

This completes the inductive step.

8

Inductive hypothesis:

P

(

j

) is true whenever

8

≤ j

<

k.

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