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Math127121 assignment 3solutions

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Discrete Mathematics (Mathematics 1271)

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Lakehead University

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Math 1271 Assignment 3 Due Friday Oct 1 at 10pm

Construct a truth table for each of the following propositions: a) (P∧Q)⇔((¬Q)⇒P) b) ((Q⇒P)∨((¬Q)⇒(¬P)))⇒(P⇒Q) c) ((P∧Q)∨(¬P))⇒Q d) ((P⇒Q)∨Q)⇒P

Solution:We have the following tables: a)

P Q (P∧Q) ⇔ ((¬Q) ⇒P) T T T T F T T F F F T T F T F F F T F F F T T F

P Q ((Q⇒P) ∨ ((¬Q) ⇒ (¬P))) ⇒ (P⇒Q) T T T T F T F T T T F T T T F F F F F T F T F T T T T F F T T T T T T T

P Q ((P∧Q) ∨ (¬P)) ⇒Q T T T T F T T F F F F T F T F T T T F F F T T F

P Q ((P⇒Q) ∧Q) ⇒P T T T T T T F F F T F T T T F F F T F T

Construct a truth table for each of the following propositions: a) ((P∧Q)⇒R)∨((¬P)⇔(Q∨R)) b) (P∧(R∨Q))⇔((P∨R)∧(P∨Q))

Solution: 1

a) P Q R ((P∧Q) ⇒R) ∨ ((¬P) ⇔ (Q∧R)) T T T T T T F F T T T F T F T F T F T F T F T T F T F T F F F T T F T F F T T F T T T T T F T F F T T T F F F F T F T T T F F F F F F T T T F F

b) P Q R (P∧ (R∨Q)) ⇔ ((P∨R) ∧ (P∨Q)) T T T T T T T T T T T F T T T T T T T F T T T T T T T T F F F F F T T T F T T F T F T T T F T F F T T F F T F F T F T T T F F F F F F F T F F F

For each of the following propositions, construct a logically equivalent proposition that only uses the connectives¬,∨,and∧. a) (R⇒(¬P))⇒((¬P)⇒(¬R)) b) (P⇒(¬Q))⇒(¬(Q⇒P))

Solution:We just specify which combinations of variables giveT,and then join these with∨s. (This is the disjunctive normal form.)

a) We have the truth table:

P R (R ⇒ (¬P)) ⇒( (¬P) ⇒ (¬R)) T T F F T F T F T F T F T F T T F T T T F T F F F F T T T T T T

) A logically equivalent proposition is (P∧R)∨(P∧(¬R))∨((¬P)∧(¬R)).

b) We have the truth table:

P Q (P⇒ (¬Q)) ⇒( ¬(Q ⇒P)) T T F F T F T T F T T F F T F T T F T T F F F T T F F T

A logically equivalent proposition is (P∧Q)∨((¬P)∧Q).

Given setsA={a, c, d},B={a, b, c, e, f}andC={a, d, e, f},findA∩B, A∪C, B∩C,andA\C.

A B C ((A ∧ (¬B)) ∧( ¬C))

T T T F F F F

T T F F F F T

T F T T T F F

T F F T T T T

F T T F F F F

F T F F F F T

F F T F T F F

F F F F T F T

and

A B C A ∧ (¬(B ∨C)) T T T F F T T T F F F T T F T F F T T F F T T F F T T F F T F T F F F T F F T F F T F F F F T F

Since these truth tables are the same, ((A ∧ ¬(B))∧(¬C))⇔(A ∧(¬(B ∨ C))) is a tautology.

We also have

A B C A ∧ ¬((B) ∧( ¬C)) T T T T T F F T T F F F T T T F T T T F F T F F T T F T F T T F T F F F T F F F T T F F T F T F F F F F F T F T

and

A B C (A ∧ ¬(B)) ∨(A ∧C) T T T F F T T T T F F F F F T F T T T T T T F F T T T F F T T F F F F F T F F F F F F F T F T F F F F F F T F F

Since these truth tables are the same, (A ∧ ¬(B))∧(C)⇔(A ∧ ¬(B))∨(A ∧ C) is a tautology.

The set equalities follow from these propositions being tautologies.

Use induction to show that for all strictly positive integersn,

1 ·3 + 3·5 +· · ·+n(n+ 2) =

n(n+ 1)(2n+ 7) 6

.

Solution:As required, we use induction.

Basis Step: Our basis step is the casen= 1 the left hand side we have 1 ·3 = 3,and on the right hand side we have (1)(1 + 1)(2(1) + 7)/6 = 18/6 = 3,so the basis case holds.

Induction Step: Fixk≥1 and assume 1·3 + 2·4 + 3·5 +· · ·+k(k+ 2) = k(k+1)(2k+7) 6 (Induction Hypothesis). We then have 1·3+2·4+3·5+· · ·+k(k+ 2)+ (k+1)((k+1)+2) =k(k+1)(2 6 k+7)+(k+1)((k+1)+2) =k(k+1)(2k+7)+6( 6 k+1)(k+3)= (k+1)(2k 2 +7k+6k+18) 6 =

(k+1)(2k 2 +13k+18) 6 =

(k+1)(k+2)(2k+9) 6 =

(k+1)((k+1)+1)(2(k+1)+7) 6 , which is the proposition forn=k+ 1 completes the induction.

Use induction to show that for all strictly positive integersn,

2 ·2! + 3·3! +· · ·+ (n+ 1)·(n+ 1)! = (n+ 2)!− 2

Solution:As required, we use induction.

Basis Step: Our basis step is the casen= 1 the left hand side we have 2 ·2! = 2·2 = 4,and on the right hand side we have (1 + 2)!−2 = 6−2 = 4,so this case holds.

Induction Step:Fixk≥1 and assume 2·2!+3·3!+· · ·+(k+1)·(k+1)! = (k+2)!− 2 (Induction Hypothesis). We then have 2·2!+3·3!+· · ·+(k+1)·(k+1)!+((k+1)+1)· ((k+1)+1)! = (k+2)!−2+(k+2)·(k+2)! = (1+k+2)·(k+2)!−2 = ((k+1)+2)!− 2 , which is the proposition forn=k+ 1 completes the induction.

Prove that 2m> m 2 for every integerm≥ 5.

Solution:We use induction to show this.

Basis Step: Our basis case ism= 5 this case we have 2m= 2 5 = 32 and m 2 = 5 2 = 25 32> 25 ,the basis case holds.

Induction Step: Now fixk≥5 and assume 2k > k 2 (Induction Hypothesis). We then have 2k+1= 2· 2 k= 2k+ 2k≥ 5 ,we have, using the induction hypothesis, 2k> k 2 ≥ 5 k≥ 2 k+ 1 inequalities, we get 2k+1= 2k+ 2k> k 2 + 2k+ 1 = (k+ 1) 2 ,which is the proposition forn=k+ 1 completes the induction.

Prove that for all even non-negative integersm, m 4 −m 2 is divisible by 12.

Solution:In the casem= 0, m 4 −m 2 = 0,which is divisible by 12 is even andm≥ 2 .Factoring, we havem 4 −m 2 =m 2 (m 2 −1) =m 2 (m−1)(m+ 1).Consider the prime factorization of this expression given by the Fundamental Theorem of Arithmetic. Sincemis even, 2 dividesm,so 2 2 dividesm 2 .Of the three consecutive integersm− 1 , m,andm+ 1,one of the three must be divisible

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