Set Theory -Famous Identitiestodd/cse240/cse240_lecture8_2-11-19.pdf · 5Extensible Networking Platform-CSE 240 –Logic and Discrete Mathematics 5 Set Theory -Famous Identities •DeMorgan

Extensible Networking Platform 11 - CSE 240 – Logic and Discrete Mathematics

Announcements

• Homework 3 was due this morning

• Exam 1 is in one week! – Proofs– Logical equivalence– Sets– Basic induction

• On Wednesday we will review for the exam

• Read Section 3.2 (Growth of Functions), 5.4-5.5 (Recursive Algorithms and Program correctness) by Wednesday


Set Theory - Famous Identities

Identity

Domination

Idempotent

A Ç U = AA U Æ = A

A U U = UA ÇÆ = Æ

A U A = AA Ç A = A



• Excluded Middle

• Uniqueness

• Double complement

A U A = U

A Ç A = Æ

A = A



• Commutativity

• Associativity

• Distributivity

A U B =

(A U B) U C =

A Ç B =B U A B Ç A

(A Ç B) Ç C =

A U (B U C)

A Ç (B Ç C)

A U (B Ç C) =

A Ç (B U C) =

(A U B) Ç (A U C)

(A Ç B) U (A Ç C)



• DeMorgan�s I

• DeMorgan�s II

A B

Hand waving is good for

intuition, but we aim for a more formal proof.

(A U B) = A Ç B

(A Ç B) = A U B


Set Theory – 4 Ways to prove identities

• Show that A Í B and that B Ê A.

• Use a membership table.

• Use previously proven identities.

• Use logical equivalences to prove equivalent set definitions.

Like truth tables

Like º

Not hard, a little tedious



Prove that

1. (Í) (x Î A U B) ® (x Ï A U B) ®

(x Ï A and x Ï B) ® (x Î A Ç B)

2.( ) (x Î A Ç B) ® (x Ï A and x Ï B)

® (x Ï A U B) ® (x Î A U B)

(A U B) = A Ç B

⊇



Prove that using a membership table.

0 : x is not in the specified set1 : otherwise

(A U B) = A Ç B

A B A B A Ç B A U B A U B

1 1 0 0 0 1 01 0 0 1 0 1 00 1 1 0 0 1 00 0 1 1 1 0 1

Haven’t we seen this before?



Prove that using logically equivalent set definitions.

(A U B) = A Ç B

(A U B) = {x : ¬(x Î A v x Î B)}

= {x : ¬(x Î A) Ù ¬(x Î B)}

= A Ç B

= {x : (x Î A) Ù (x Î B)}


Set Theory - Generalized Union

€

Ai

i=1

n

= A1∪ A2∪…∪ An

Ex. Let U = N, and define:

€

Ai = {x : ∃k > 1, x = ki,k ∈ Ν}

A1 = {2,3,4,…}A2 = {4,6,8,…}A3 = {6,9,12,…}


Set Theory - Generalized Union

€

Ai

i=1

n

= A1∪ A2∪…∪ An


€

Ai = {x : ∃k > 1, x = ki,k ∈ Ν}

Then

€

Ai

i= 2

∞

= ?a) Primesb) Compositesc) Æd) Ne) I have no clue.

primes


Set Theory - Generalized Intersection

€

Ai

i=1

n

= A1∩ A2∩…∩ An


€

Ai = {x : ∃k, x = ki,k ∈ Ν}

A1 = {1,2,3,4,…}A2 = {2,4,6,…}A3 = {3,6,9,…}


Set Theory - Generalized Intersection

€

Ai

i=1

n

= A1∩ A2∩…∩ An


€

Ai = {x : ∃k, x = ki,k ∈ Ν}

Then

€

Ai

i=1

n

= ? Multiples of LCM(1,…,n)


Set Theory - Inclusion/Exclusion

Example:How many people are wearing a watch?How many people are wearing sneakers?

How many people are wearing a watch OR sneakers?

AB|A È B| = |A| + |B| - |A Ç B|


Set Theory - Inclusion/Exclusion

Example:There are 83 cs majors.40 are taking cs240.31 are taking cs101.22 are taking both.

How many are taking neither?

83 - (40 + 31 - 22) = 34

4031


Set Theory - Generalized Inclusion/Exclusion

Suppose we have:

And I want to know |A U B U C|

A B

C

|A U B U C| = |A| + |B| + |C|

+ |A Ç B Ç C| - |A Ç B| - |A Ç C| - |B Ç C|


Review: Mathematical Induction

Use induction to prove that the sum of the first n odd integers is n2.

Prove a base case (n=1)

Base case (n=1): the sum of the first 1 odd integer is 12. Yes, 1 = 12. Prove P(k)®P(k+1)

Assume P(k): the sum of the first k odd ints is k2. 1 + 3 + … + (2k - 1) = k2

Prove that 1 + 3 + … + (2k - 1) + (2k + 1) = (k+1)2

Inductive hypothesis

1 + 3 + … + (2k-1) + (2k+1) = k2 + (2k + 1) By inductive hypothesis

= (k+1)2By arithmetic


Mathematical Induction - a cool example

Deficient TilingA 2n x 2n sized grid is deficient if all but

one cell is tiled.

2n

2n



• We want to show that all 2n x 2n sized deficient grids can be tiled with tiles, called triominoes, shaped like:



• Is it true for all 21 x 21 grids?

Yes!



Inductive Hypothesis:We can tile any 2k x 2k deficient board

using our fancy designer tiles.

Use this to prove:We can tile any 2k+1 x 2k+1 deficient

board using our fancy designer tiles.



2k

2k 2k

2k

2k+1

OK!! (by IH)

?

?

?



2k

2k 2k

2k

2k+1

OK!! (by IH)

OK!! (by IH)

OK!! (by IH)

OK!! (by IH)






Mathematical Induction - why does it work?

Definition:A set S is �well-ordered� if every non-empty

subset of S has a least element.

Given (we take as an axiom): the set of natural numbers (N) is well-ordered.

Is the set of integers (Z) well ordered?No.

{ x Î Z : x < 0 } has no least element.



Is the set of non-negative reals (R) well ordered?

No. { x Î R : x > 1 } has no

least element.



Proof of Mathematical Induction:

We prove that (P(0) Ù ("k P(k) ® P(k+1))) ® ("n P(n))

Proof by contradiction.Assume1. P(0)2. "k P(k) ® P(k+1)3. ¬"n P(n) $n ¬P(n)



Assume1. P(0)2. "n P(n) ® P(n+1)3. ¬"n P(n) $n ¬P(n)

Let S = { n : ¬P(n) } Since N is well ordered, S has a least element. Call it k.

What do we know?-P(k) is false because it’s in S.-k ¹ 0 because P(0) is true.-P(k-1) is true because P(k) is the least element in S.

But by (2), P(k-1) ® P(k). ContradictsP(k-1) true, P(k) false.

Done.


Strong Mathematical Induction

IfP(0) and"n³0 (P(0) Ù P(1) Ù … Ù P(n)) ® P(n+1)

Then"n³0 P(n) In our proofs, to show P(k+1), our inductive

hypothesis assumes that ALL of P(0), P(1), … P(k) are true, so we can use ANY

of them to make the inference.


Game with Matches

• Two players take turns removing any number of matches from one of two piles of matches. The player who removes the last match wins

• Show that if two piles contain the same number of matches initially, then the second player is guaranteed a win


Strategy for Second Player• Let P(n) denote the statement �the second player wins

when they are initially n matches in each pile�

• Basis step: P(1) is true, because only 1 match in each pile, first player must remove one match from one pile. Second player removes other match and wins

• Inductive step: suppose P(j) is True for all j 1<=j <= k.

• Prove that P(k+1) is true, that is the second player wins when each piles contains k+1 matches


Strategy for Second Player

• Suppose that the first player removes r matches from one pile, leaving k+1 –r matches there

• By removing the same number of matches from the other pile the second player creates the situation of two piles with k+1-r matches in each. Apply the inductive hypothesis and the second player wins each time.

How is this different than

regular induction?


Postage Stamp Example

• Prove that every amount of postage of 12 cents or more can be formed using just 4-cent and 5-cent stamps

• P(n) : Postage of n cents can be formed using 4-cent and 5-cent stamps

• All n >= 12, P(n) is true


Postage Stamp Proof• Base Case: n = 12, n = 13, n = 14, n = 15

– We can form postage of 12 cents using 3, 4-cent stamps– We can form postage of 13 cents using 2, 4- cent stamps and 1

5-cent stamp– We can form postage of 14 cents using 1, 4-cent stamp and 2 5-

cent stamps– We can form postage of 15 cents using 3, 5-cent stamps

• Induction Step– Let n >= 15– Assume P(k) is true for 12 <= k <= n, that is postage of k cents

can be formed with 4-cent and 5-cent stamps (Inductive Hypothesis)

– Prove P(n+1)– To form postage of n +1 cents, use the stamps that form

postage of n-3 cents (from I.H) with a 4-cent stamp

Why doesthis work?


Recursive Definitions

We completely understand the function f(n) = n!, right?

As a reminder, here�s the definition:n! = 1 · 2 · 3 · … · (n-1) · n, n ³ 1

Inductive (Recursive) Definition

But equivalently, we could define it like this:

Recursive Case

Base Case îíì

=³-•

= 0 n if 1

1n if )!1(! nnn


Recursive Definitions

Another VERY common example:

Fibonacci Numbers

Recursive Case

Base Cases

ïî

ïí

ì

>-+-==

=1 if )2()1(1 if 10 if 0

)(nnfnfnn

nf

Is there a non-recursive definition for the Fibonacci

Numbers?

€

f (n) =15

1+ 52

"

# $

%

& '

n

−1− 52

"

# $

%

& '

n)

* + +

,

- . .

Documents

Set Theory -Famous Identitiestodd/cse240/cse240_lecture8_2-11-19.pdf · 5Extensible Networking Platform-CSE 240 –Logic and Discrete Mathematics 5 Set Theory -Famous Identities •DeMorgan