Answers, Hints, Solutions - Fundamentals of Mathematical Logic

Fundamentals of Mathematical Logic

1.2 Answers, Hints, Solutions

1.122. Prove thecriterion of divisibility by11:For the number N to be divisible by 11, it is necessary and sufficient that the difference between the sums of the digits standing in its decimal notation on even and odd places is divisible by 11.

∗1.123. Fundamental theorem of arithmetic states that each natural number greater than unity is either a prime number or can be presented in the form of the product of prime numbers. Such a presentation is unique to an accuracy of the arrangement of the cofactors[29,77].

Using the mathematical induction principle in the second form, prove the fundamental theorem of arithmetic.

1.124. Prove that for an arbitrary numbern in the natural series 1,2,3, . . ., there existnof consecutive composite numbers.

1.125. Find ten consecutive composite numbers.

1.5.Answer:

(1)not(A1 and A2and. . .andAn)⇔(not A1)or(not A2)or. . .or(notAn);

(2)not(A1orA2or. . .orAn)⇔ (notA1)and(notA2)and. . .and(notAn).

1.6.Solution.

PropositionPmay be presented as disjunction of three propositions:P=PAorPB

orPC, where byPAis denoted proposition “A—true,B—false, andC—false,” and PBandPCfor other two variants of the truth values of propositionsA,B, andCare written similarly. Finally, we obtain:

P =(Aand notBand notC)or (notAandBand notC)or (notAand notBand C).

1.7.Answer: P=AorBorC.

1.8.Answer:

P =not

(AandBand notCand notD)or(Aand notBandCand notD)or (Aand notBand notCandD)or(notAandBandCand notD)or (notAandBand notCandD)or(notAand notBandCandD)

. 1.9.Solution.

Compile a truth table for the statements provided:

A B C AandB B⇒C A⇒(B⇒C) (AandB)⇒C

T T T T T T T

T T F T F F F

T F T F T T T

T F F F T T T

F T T F T T T

F T F F F T T

F F T F T T T

F F F F T T T

The two last columns coincide, this is why the respective expressions are logically equivalent.

1.10.Solution.

Denote: P1= A⇒(B⇒C), P2=(notC)⇒((notA)or (notB)). Truth table for propositions P1andP2is presented below.

From comparison of the two last columns follows the logical equivalence P1

andP2.

A B C B⇒C (notA)or(notB) P₁ P₂

T T T T F T T

T T F F F F F

T F T T T T T

T F F T T T T

F T T T T T T

F T F F T T T

F F T T T T T

F F F T T T T

1.11.Solution.

The problem is solved by compiling a truth table for propositions P1=A or (B and C),P2=(AorB) and (A or C).

A B C AorB AorC BandC P₁ P₂

T T T T T T T T

T T F T T F T T

T F T T T F T T

T F F T T F T T

F T T T T T T T

F T F T F F F F

F F T F T F F F

F F F F F F F F

1.12.Solution.

The problem is solved by compiling a truth table for propositions P1=Aand (BorC),P2=(AandB)or(AandC).

A B C AandB AandC BorC P₁ P₂

T T T T T T T T

T T F T F T T T

T F T F T T T T

T F F F F F F F

F T T F F T F F

F T F F F T F F

F F T F F T F F

F F F F F F F F

1.13.Proof.

Let us introduce the following notations:

P1=(A or B)and (not (AandB)), P2=(Aand(notB))or((notA)andB).

For all possible sets of truth values of propositionsAandBpropositionP1⇔P2

is true, this is why the provided proposition is a tautology.

A B Aand notB notAandB P₁ P₂ P₁⇔P₂

T T F F F F T

T F T F T T T

F T F T T T T

F F F F F F T

1.14.Solution.

Compile a truth table for the compound proposition P =(A⇒(B⇒C))⇒ ((AandB)⇒C):

A B C B⇒C A⇒(B⇒C) AandB (AandB)⇒C P

T T T T T T T T

T T F F F T F T

T F T T T F T T

T F F T T F T T

F T T T T F T T

F T F F T F T T

F F T T T F T T

F F F T T F T T

Analysis of the table shows that for all possible sets of truth values of atomic propositions A,B,andCpropositionP is true. Hence, the provided proposition is a tautology.

1.15.Solution.

(1) Compile a truth table for the provided propositionP =(A and(B or C))⇒ (A or(B and C)). For all sets of values of propositions,A,B, andC, being parts of P,Ptake the true value.

A B C BorC Aand(BorC)BandC Aor(BandC) P

T T T T T T T T

T T F T T F T T

T F T T T F T T

T F F F F F T T

F T T T F T T T

F T F T F F F T

F F T T F F F T

F F F F F F F T

Thus, propositionPis a tautology.

(2) LetP=(Aand(B⇒C)) and (Bor(A⇒C)). From the analysis of the truth table for proposition Pfollows that for some sets of values of propositions A, B, andC, being a part ofP(e.g., forA=B =T,C=F),Ptakes a false value.

Hence, propositionPis not a tautology.

A B C B⇒C Aand(B⇒C) A⇒C Bor(A⇒C) P

T T T T T T T T

T T F F F F T F

T F T T T T T T

T F F T T F F F

F T T T F T T F

F T F F F T T F

F F T T F T T F

F F F T F T T F

1.16.Answer:

(1) No, it is not.

(2) No, it is not.

1.17.Solution.

Note that the sequence of propositions written in the form of separate sentences with the help of logical operations should be written in the form of conjunction of atomic propositions. Thereby, and using the basic logical operations, we obtain the following symbol form of the compound propositions:

(1)(BorA)⇒C;

(2)(notC)⇒(notA);

(3)((A⇒C) and (not A))⇒((not B) or (not C)).

1.18.Answer:

(1)C⇒(A⇒(notB));

(2) A⇒((Band(notC))or((notB)andC)). 1.19.Solution.

We consider the provided proposition as a compound one. Let us denote the propositions that are parts of it by

A=“the Cardinal’s guards are nearby”;

B=“d’Artagnan is ready to fight”.

Contrapositive to proposition A⇒B by definition is notB⇒notA. Hence, the answer to this exercise isnotB⇒notA=“if d’Artagnan is not ready to fight, then there are no Cardinal’s guards nearby”.

1.20.Solution.

Let us denote the propositions that are parts of the provided compound proposition as

A=“Aramis becomes a bishop”;

B=“Portos obtains a barony”;

C =“d’Artagnan will be presented with a marshal’s baton.”

Contrapositive to proposition (AandB)⇒C is (notC)⇒(not(AandB)), which is logically equivalent to the following:(notC)⇒((notA)or(notB)). Thus, the answer to this exercise is(notC)⇒((notA)or(notB))=“if d’Artagnan is not

presented with a marshal’s baton, then either Aramis will not become a bishop or Portos will not obtain a barony.”

1.21.Answer:

“If Aramis does not disclose the Cardinal’s plot, then either Atos will not intercept the letter, or Portos will not shake off the pursuers.”

1.22.Answer:

(1) P1andP2are not logically equivalent.

(2) Equivalent propositions.

1.23.Answer:

(1) Equivalent propositions;

(2) XandY are not logically equivalent;

(3) Equivalent propositions;

(4) Equivalent propositions.

1.24.Answer:

Identically false are propositions of items (3) and (4).

1.25.Solution.

Let us introduce the notations:

C =“incognito—the Cardinal”;

M =“incognito—Milady”.

Let us write each musketeer’s proposition with the help of logical operations:

(1) d’Artagnan: Cis true, C =T;

(2) Portos: P=(C or M)and not(C and M).

By compilation of a truth table, it is easy to show that Portos’s proposition is logically equivalent to

P =(Cand notM)or(Mand notC).

(3) Aramis: P ⇒C.

(4) Atos: Pand(P⇒C).

According to the condition, Atos’s proposition is true,Pand(P⇒C)=T. Con- junction of two propositions is true if and only if each of the propositions takes the true value, this is why

P=T;

P⇒C=T.

From the second equation follows T⇒C =T, hence, propositionCis true, and the Cardinal was hiding behind the mask.

1.26.Answer: Portos and Aramis had broken their swords.

1.27.Solution.

Let us introduce the notations:

A—“Alexandrov is guilty”;

B—“Bykov is guilty”;

C—“Semenov is guilty.”

Let P =(C⇒B)and not(A⇒B). The problem statement implies that P= T. Let us write the truth table for the propositions in the left side of the obtained equality.

A B C C⇒B A⇒Bnot(A⇒B) P

T T T T T F F

T T F T T F F

T F T F F T F

T F F T F T T

F T T T T F F

F T F T T F F

F F T F T F F

F F F T T F F

The compound proposition (C⇒B)and not(A⇒B)takes the true value only in one case, namely when propositionsBandCare false and proposition Ais true.

The conclusion follows from here that Alexandrov is guilty of the robbery.

1.28.Solution.

First method

Let Si =T, ifith machine is working, wherei =1,2,3,4, and Si =F otherwise.

Let us compile a truth table for the logical expressions corresponding to machine’s operating conditions (Table1.3). From the problem statement, we conclude that the following compound propositions have the true value:

P1=S1⇒(S2andS3), P2=S3⇔S4,

P3=S2⇒notS4, P4=(S1and notS2)or(notS1andS2).

From the truth table, it follows that

S1=S3=S4=F, S2=T.

We obtain that only the second machine is working at the moment.

Second method

Since the number of rows in the truth table is 2ⁿ, wherenis the number of variables in the logical expression [4], then asngrows, the size of the truth table grows rapidly, and it becomes difficult to use the truth table. Thereby, the preferred method is the method of simplification of logical expressions using equivalence transformations.

Consider logical expression

P =P1andP2andP3andP4.

Table 1.3 To Exercise1.28

S₁ S₂ S₃ S₄ P₁ P₂ P₃ P₄ P

T T T T T T F F F

T T T F T F T F F

T T F T F F F F F

T T F F F T T F F

T F T T F T T T F

T F T F F F T T F

T F F T F F T T F

T F F F F T T T F

F T T T T T F T F

F T T F T F T T F

F T F T T F F T F

F T F F T T T T T

F F T T T T T F F

F F T F T F T F F

F F F T T F T F F

F F F F T T T F F

According to the problem conditions, the given expression takes the true value.

Simplify P, using the laws of the algebra of logic (Table1.2).

Since the change of the order of the expressions bound by the conjunction operation takes place in a logically equivalent formula (commutative law), then let us present expressionPin the form

P =P1andP2andP3andP4=((P1andP4)andP2)andP3. Now, we sequentially simplify the obtained expression:

P1andP4=(S1⇒(S2andS3))and[(S1and notS2)or(notS1andS2)]⇔

⇔(notS1or(S2andS3))and[(S1and notS2)or(notS1andS2)]⇔

⇔ [(notS1or(S2andS3)and(S1and notS2)]or

or[(notS1or(S2andS3))and(notS1andS2)]⇔

⇔ [(notS1andS1and notS2)or(S2andS3andS1and notS2)]or or[(notS1and notS1andS2)or(S2andS3and notS1andS2)]⇔

⇔(ForF)or(notS1andS2)or(notS1andS2andS3)=notS1andS2.

Then, we calculate(P1andP4)andP3:

(P1andP4)andP3=(notS1andS2)and(S2⇒notS4)⇔

⇔(notS1andS2)and(notS2or notS4)⇔

⇔(notS1andS2and notS2)or(notS1andS2and notS4)⇔

⇔ notS1andS2and notS4. Let us perform the final logical multiplication:

P=((P1andP4)andP3)andP2=

=(notS1andS2and notS4)and [(S3andS4)or(notS3and notS4)]⇔

⇔(notS1andS2and notS4andS3andS4)or

or(notS1andS2and notS4and notS3and notS4)⇔

⇔For(notS1andS2and notS4and notS3and notS4)⇔

⇔notS1andS2and notS3and notS4. The obtained equivalent formula

P=P1andP2andP3andP4=notS1andS2and notS3and notS4

allows easily finding the values of logical expressionsS1–S4: S1=S3=S4=F, S2=T.

Hence, only the second machine is working at the moment.

1.29.Solution.

First method

LetSi =T, ifith machine is working, wherei =1,2,3,4, andSi =F otherwise. The answer to the problem’s question can be obtained by compiling a truth table for the logical expressions corresponding to the machine’s operating conditions (Table1.4).

From the problem statement, we conclude that the following compound propositions have the true value:

P1=S1⇒S2, P2=S2⇔S3,

P3=S4⇒ notS3, P4=(S1andS2)or(notS1and notS2).

From the truth table, it follows that

S1=S4=F, S2=S3=T.

We finally obtain that the second and the third machines are working at the moment.

Table 1.4 To Exercise1.29

S₁ S₂ S₃ S₄ P₁ P₂ P₃ P₄ P

T T T T T T F F F

T T T F T T T F F

T T F T T F T F F

T T F F T F T F F

T F T T F F F T F

T F T F F F T T F

T F F T F T T T F

T F F F F T T T F

F T T T T T F T F

F T T F T T T T T

F T F T T F T T F

F T F F T F T T F

F F T T T F F F F

F F T F T F T F F

F F F T T T T F F

F F F F T T T F F

Second method

Consider logical expression

P =P1andP2andP3andP4.

According to the problem statement, the following expression takes a true value.

Let us try to simplify P.

Applying the commutative law, we present expression Pin the form P =P1andP2andP3andP4=((P1andP4)andP2)andP3. Now, we sequentially simplify the obtained expression:

P1andP4=(S1⇒S2)and[(S1and notS2)or(notS1andS2)]⇔

⇔(notS1orS2)and[(S1and notS2)or(notS1andS2)]⇔

⇔ [(notS1orS2)and(S1and notS2)]or[(notS1orS2)and(notS1andS2)]⇔

⇔ [(notS1andS1and notS2)or(S2andS1and notS2)]or or[(notS1and notS1andS2)or(S2and notS1andS2)]⇔

⇔(ForF)or(notS1andS2)or(S2and notS1)⇔notS1andS2.

Then, we calculate(P1andP4)andP2:

(P1andP4)andP2=(notS1andS2)and[S2andS3or(notS2and notS3)]⇔

⇔(notS1andS2andS2andS3)or

or(notS1andS2and notS2and notS3)⇔

⇔(notS1andS2andS3)orF⇔notS1andS2andS3. Now, we only need to perform the last logical multiplication:

P=((P1andP4)andP2)andP3=(notS1andS2andS3)and(S4⇒notS3)⇔

⇔(notS1andS2andS3)and(notS3or notS4)⇔

⇔(notS1andS2andS3and notS3)or(notS1andS2andS3and notS4)⇔

⇔For(notS1andS2andS3and notS4)⇔notS1andS2andS3and notS4. Of course, we may choose any other order of logical multiplication of expressions P1–P4, if this will provide for convenience of calculations.

The obtained equivalent formula

P =P1andP2andP3andP4=notS1andS2andS3and notS4

allows easily finding the values of logical expressionsS1–S4: S1=S4=F, S2=S3=T.

Hence, the second and the third machines are working at the moment.

1.30.Answer:

Bad were the compute server and the first cooling unit.

1.31.Answer:

The administrator had to repair the main server and the first and the second cooling units.

1.32.Solution.

The problem can be solved by compiling a truth table or by the method of equivalence transformations of logical expressions. However, it is easier to note that since the standby airport operates when and only when the respective main airport is closed, it is easy to conclude: operatesB1,B1=T. Out of the remaining airportsAandA1, one and only one can receive planes. Further, operation of A1is necessary for B1, this is whyA1=T. Thus, the operating airports areA1andB1.

1.33.Solution.

Let us introduce logical variables:

A— “Atos is going”;

B— “Portos is going”;

C— “Aramis is going.”

From the problem statement, we conclude that

(A or B)and(BorC)and(B⇒C)and(A⇒not C)=T.

Let us simplify the expression in the left side of the obtained equality:

(AorB)and(BorC)and (B⇒C)and(A⇒ notC)⇔

⇔(Aor B)and(BorC)and(notBorC)and (notA or notC)⇔

⇔(A or B) and [(Band notB) or C] and (notAor notC)⇔

⇔(Aor B)and C and (notA or notC)⇔

⇔(A or B) and [(C and notA) or (C and notC)] ⇔

⇔(A or B) and (notA and C)⇔

⇔(A and (notA and C))or (notA and B and C)⇔

⇔notA and B and C.

Hence, Portos and Aramis are going.

1.34.Answer: Aramis and Portos.

1.35.Solution.

Since each vacationer is either taking a trip or bathing at the beach, we may introduce logical variablesA,B,V, andG, whose value determines the occupation of each student. DenoteA=T, if Alice is taking a trip, andA=F, if Alice is at the beach.

By a similar rule, determine the values of variablesB,V,G.

According to the problem statement, expression

(BorV)and((notV and notG)⇒notA)and (not(notA⇒(VorG))) takes the true value:

(BorV)and((notVand notG)⇒notA)and(not(notA⇒(VorG)))⇔

⇔(BorV)and(V orGor notA)and(notAand notVand notG)⇔

⇔((Band notAand notVand notG)or or(Vand notAand notVand notG))and and(VorGor notA)⇔

⇔((notAandBand notVand notG)orF)and(VorGor notA)⇔

⇔(notAandBand notVand notG)and(notAorVorG)⇔

⇔(notAandBand notVand notGand notA)or or(notAandBand notVand notGandV)or or(notAandBand notVand notGandG)⇔

⇔(notAandBand notVand notG)orForF⇔

⇔ notAandBand notVandnotG.

SincenotAandBand notVand notG=T, then we obtain the following result:

Bogdan is taking a trip, and Alice, Valeria, and George are at the beach.

1.36.Answer: Alice.

1.37.Answer:

A B A|B A↓B

T T F F

T F T F

F T T F

F F T T

1.38.Solution.

(1) Using transformations for logically equivalent expressions for operations of negation, disjunction, conjunction, and implication, we obtain:

not A⇔(not A)or(not A)=A| A;

A or B⇔not (not A)or not (not B)=(A| A)|(B|B);

A and B⇔not (not A or not B)=(A|B)|(A| B);

A⇒B ⇔(not A)or B⇔(not A) or not (not B)=A|(B| B).

(2) The task is fulfilled similarly to item 1):

not A⇔(not A) and (not A)= A↓ A;

A or B⇔not (not A and not B)=(A↓ B)↓(A↓B);

A and B⇔ not (not A) and not (not B)⇔ not A↓not B=(A↓ A)↓(B↓ B);

A⇒B ⇔(not A) or B ⇔not (A and not B)⇔not (not A↓B)⇔

⇔not ((A↓A)↓B)=((A↓ A)↓B)↓((A↓ A)↓B).

1.39.Answer:

(1) A↓B⇔((A| A)|(B|B))|((A| A)|(B| B));

(2) A|B⇔((A↓ A)↓(B↓B))↓((A↓ A)↓(B ↓B)).

1.40.Answer: Predicates are sentences (2)–(4).

1.41.Answer:

(1)not P(n)= {natural numbernis odd}; (2)not P(x,y)= {xy}.

1.42.Solution.

Let us write the negation of the expression T and use logical equivalences not∀x(P(x)) ⇔ ∃xnot(P(x)) andnot∃x(P(x)) ⇔ ⇔ ∀xnot(P(x)):

(1) notT = not∀x(∃y(P(x,y))⇔ ∃x(not∃y(P(x,y)))⇔

⇔ ∃x(∀y(notP(x,y)));

(2) notT = not∃x(∃y(P(x,y))⇒Q(x,y))⇔

⇔ ∀x(not(∃y(P(x,y)))⇒Q(x,y))⇔

⇔ ∀x(∃y(P(x,y))and notQ(x,y)).

1.43.Solution.

(1) The proposition∀x(∃y(x>y))means “for each real numberxthere exists a certain numbery, that is less thanx”. This is a true proposition.

(2) The proposition∀x(x =0⇒ ∃y(x y=1))means “for each nonzero valuex there exists a reciprocal numbery, such thatx y=1”. This is a true proposition.

1.44.Answer:

(1) True;

(2) False.

1.45.Solution.

Using the definition of a limit of numerical sequence, we obtain

nlim→∞an =a ⇔ not ∀ε >0∃n0∈N∀n ∈N(nn0⇒ |an−a|< ε).

We sequentially introduce the negation operation under the signs of the quantifiers:

nlim→∞an =a ⇔ not ∀ε >0∃n0∈N∀n∈N(nn0⇒ |an−a|< ε)⇔

⇔ ∃ε >0 not ∃n0∈N∀n ∈N(n n0⇒ |an−a|< ε)⇔

⇔ ∃ε >0∀n0∈N not ∀n∈N(nn0⇒ |an−a|< ε)⇔

⇔ ∃ε >0∀n0∈N∃n ∈Nnot (nn0⇒ |an−a|< ε)⇔

⇔ ∃ε >0∀n0∈N∃n ∈N(nn0 and |an−a|ε).

When calculating the negation of implication, we used equivalence P⇒Q ⇔ not P or Q.

1.49.Proof.

LetP(n)be the predicate “1+3+5+. . .+(2n−1)=n²”.

B a s i s s t e p

Forn=1 we obtain 1=1², i.e.,P(1)is true.

I n d u c t i v e s t e p

Let forn =k, the proposition 1+3+5+. . .+(2k−1)=k²be true. Let us prove the truth ofP(k+1):

1+3+5+. . .+(2k−1)+(2(k+1)−1)=

=k²+(2(k+1)−1)=k²+2k+1=(k+1)². Thus, for any natural k the following implication is valid P(k)⇒P(k+1).

Hence, by the mathematical induction principle the predicate P(n)has the true value for all naturaln.

1.50.Proof.

Denote predicate “1+4+7+. . .+(3n−2)= 1

2n(3n−1)” byP(n). B a s i s s t e p

Forn=1 we obtain 1= 1

2·1·(3−1); thereforeP(1)is true.

I n d u c t i o n s t e p

Let forn=k, the proposition 1+4+7+. . .+(3k−2)= 1

2k(3k−1)be true.

Then

1+4+7+. . .+(3k−2)+(3(k+1)−2)=

= 1

2k(3k−1)+(3(k+1)−2)= 1

2(3k²+5k+2)=

= 1

2(k+1)(3(k+1)−1).

Thus, for any k=1,2, . . . the implication P(k)⇒ P(k+1)is valid. Hence, according to the mathematical induction principle the predicate P(n)as the true value for all naturaln.

1.52.Solution.

It is required to prove the identity

1³+2³+ · · · +n³=(1+2+ · · · +n)²

for all naturaln. Denote byP(n)the predicate 1³+2³+ · · · +n³=(1+2+ · · · + n)².

B a s i s s t e p

Forn=1 we obtain 1=1; therefore,P(1)is true.

I n d u c t i v e s t e p

Let forn =k, the propositionP(k)be true. Then

1³+2³+ · · · +k³+(k+1)³=(1+2+ · · · +k)²+(k+1)³. As is known, 1+2+ · · · +k= 1

2k(k+1); hence 1³+2³+ · · · +k³+(k+1)³= 1

2k(k+1)2

+(k+1)³=

=(k+1)² 1

4k²+(k+1)

(k+1)(k+2) 2

. Taking into account that 1+2+ · · · +k+(k+1)= 1

2(k+1)(k+2), we finally obtain:

1³+2³+ · · · +k³+(k+1)³=(1+2+ · · · +k+(k+1))².

Hence, according to the mathematical induction principle the predicate P(n)has the true value for all naturalnand the sum of cubes of the firstnnatural numbers is equal to the square of their sum.

1.54.Proof.

Let us use the mathematical induction method. Denote byP(n)the predicateSn= a1+an

2 n.

B a s i s s t e p

If we assume thatn=1, then the predicate P(n)takes the formS1=a1+a1

2 ·

1=a1, which is the true proposition.

I n d u c t i v e s t e p

LetP(k),k=1,2,3, . . .be true. Let us consider the sum of the firstk+1 terms of arithmetic progression:

Sk+1=Sk+ak+1. According to inductive suppositionSk =a1+ak

2 k. Then Sk+1=a1+ak

2 k+ak+1=1 2

(a1+ak)k+2ak+1

2[(a1+a1+(k−1)d)k+2(a1+kd)]= 1

2[2a1(k+1)+k(k+1)d]=

2(a1+a1+kd)(k+1)= a1+ak+1

2 (k+1).

By the mathematical induction method, it is proved thatP(n)takes the true value for all naturaln.

1.55.Hint. Sk+1=Sk+b1q^k. 1.56.Proof.

Denote f(n)=n²−nandP(n)—the predicate “f(n)is divisible by 2”.

B a s i s s t e p

Forn=1, we obtain f(1)=1²−1=0—even number; thereforeP(1)is true.

I n d u c t i v e s t e p

Suppose that f(k)is even for naturalk1. Let us prove that this implies evenness of f(k+1):

f(k+1)=(k+1)²−(k+1)=k²+2k+1−k−1=(k²−k)+2k= f(k)+2k. Since in the right side of the obtained relation stands a sum of two even numbers, then f(k+1)is divisible by 2.

Note. The statement of the problem becomes apparent if we presentk²−kin the formk²−k=k(k−1). Out of two consecutive natural numbers, one is necessarily even, and their product is divisible by 2.

1.57.Proof.

(1) Denote the predicate “4n³+14nis divisible by 3” byP(n).

B a s i s s t e p

Forn=1 we obtain: 18 is divisible by 3, which is the true proposition.

I n d u c t i v e s t e p

Let n=kand 4k³+14k be divisible by 3. Show that from the last statement follows the truth ofP(k+1):

4(k+1)³+14(k+1)=4(k³+3k²+3k+1)+14(k+1)=

=(4k³+14k)+3(4k²+4k+6).

The first summand in the sum is divisible by 3 by the inductive supposition, and the second includes number 3 as a cofactor.

Thus, according to the mathematical induction principle the predicate P(n)has the true value for all naturaln.

(2) Denote the predicate “n⁵−n−10 is divisible by 5” asP(n). Since 10=2·5, then it is sufficient to prove that f(n)=n⁵−n is divisible by 5 for all natural valuesn.

B a s i s s t e p

Forn=1 we obtain: f(1)=0 is divisible by 5.

I n d u c t i v e s t e p

Suppose that f(k)=k⁵−k=5l for some integerl. From this supposition, it follows that f(k+1)is divisible by 5. Let us prove this statement.

Actually,

f(k+1)=(k+1)⁵−(k+1)=(k+1)

(k+1)⁴−1

=(k+1)

(k+1)²(k+1)²−1

=(k+1)

(k²+2k+1)(k²+2k+1)−1

=(k+1)

(k⁴+4k³+6k²+4k+1)−1

=(k+1)(k⁴+4k³+6k²+4k)=

=k⁵+5k⁴+10k³+10k²+4k=(k⁵−k)+5(k⁴+2k³+2k²+k).

Using the inductive supposition, we obtain that f(k+1)is divisible by 5. There- fore,n⁵−n+10 is divisible by 5 for all naturaln.

(3) B a s i s s t e p

Forn=1 we obtain: 6²ⁿ⁻¹−6=0 is divisible by 7.

I n d u c t i v e s t e p

Suppose that 6^2k⁻¹−6 is divisible by 7, k=1,2, . . .

Therefore, 6^2k⁻¹−6=7l, and 6^2k⁻¹=7l+6 for some integerl. Let us prove that 6²⁽^k⁺¹⁾⁻¹−6=6^2k⁺¹−6 is divisible by 7:

6^2k⁺¹−6=6^2k⁻¹⁺²−6=36·6^2k⁻¹−6=

=36(7l+6)−6=36·7l+35·6.

Note that both summands in the obtained sum are divisible by 7; therefore, the expression 6^2k⁺¹−6 is divisible by 7. The inductive transition is proved. Therefore, 6²ⁿ⁻¹−6 is divisible by 7 for all natural numbersn.

(4) Denote f(n)=n(n−1)(2n−1).

B a s i s s t e p

Forn=1 we obtain: f(1)=0 is divisible by 6.

I n d u c t i v e s t e p

Let f(k)=k(k−1)(2k−1)be divisible by 6. Let us prove that from thus supposition follows divisibility by 6 of the value f(k+1):

f(k+1)=(k+1)k(2(k+1)−1)=k(k+1)(2k+1)=

=(k²+k)(2k+1)=2k³+3k²+k.

Calculate the difference f(k+1)− f(k):

f(k+1)− f(k)=2k³+3k²+k−k(k−1)(2k−1)=

=2k³+3k²+k−(2k³−3k²+k)=6k². In the equality f(k+1)= f(k)+6k², the right side is divisible by 6 according to the inductive supposition. Thereby, f(k+1)is divisible by 6, and according to the mathematical induction method, f(n)=n(n−1)(2n−1)is divisible by 6 for all naturaln.

1.58.Proof.

B a s i s s t e p

Forn=1 we obtain:a−1 is divisible bya−1—the true proposition.

I n d u c t i v e s t e p

Let a^k−1 be divisible by a−1 for some natural k. Therefore, there exists such polynomial p(a), the equalitya^k−1=(a−1)p(a)is valid, ora^k=(a− 1)p(a)+1. Let us prove thata^k⁺¹−1 is divisible bya−1:

a^k⁺¹−1=a·a^k−1=a[(a−1)p(a)+1]−1=a(a−1)p(a)+(a−1).

From the last equality follows

a^k⁺¹−1=(a−1)(a p(a)+1)=(a−1)p(a), wherep(a)is some polynomial.

This proves that the polynomialaⁿ−1 is divisible by(a−1)for all naturaln.

Note. The statement of this point is easy to prove by directly dividing the polyno- mialaⁿ−1 by(a−1)by the method known from the algebra course [41]:

aⁿ −1 a−1

−

aⁿ − aⁿ⁻¹ aⁿ⁻¹+aⁿ⁻²+ · · · +a+1 aⁿ⁻¹

−

aⁿ⁻¹ − aⁿ⁻² ...

a²

− a²−a

a − 1

− a − 1

Note that the formula is obtained for the sum of geometric progression (see Exer- cise1.55) in the particular caseb1=1.

1.59.Hint. Use the mathematical induction method.

1.60.Hint. 79·4ⁿ−4(−1)ⁿ =80·4ⁿ−4ⁿ−4(−1)ⁿ. 1.61.Solution.

For the proof, let us use the mathematical induction principle. Denotean= 79 30× 4ⁿ+(−1)ⁿ

5 −1

3. B a s i s s t e p

Forn=1 we obtain:a1=10 is integer.

I n d u c t i v e s t e p

Suppose thatakis integer for some integer valuek. Let us prove thatak+1is also integer. For this, calculate the differenceak+1−ak:

ak+1−ak= 79

30·4^k⁺¹+(−1)^k⁺¹

5 −1

− 79

30·4^k+(−1)^k

5 −1

= 79

10·4^k−2(−1)^k

5 = 1

10(79·4^k−4(−1)^k).

The number 1

10(79·4^k−4(−1)^k)is integer (see the previous exercise). Therefore, an= 1

10(79·4ⁿ−4(−1)ⁿ)is integer for all values ofn=1,2, . . . 1.65.Proof.

First method

Denote byP(n)the predicate 1

3·7+ 1

7·11+ · · · + 1

(4n−1)(4n+3)= n 3(4n+3) for somen.

B a s i s s t e p

The inductive base consists in the statement P(1). For n=1 the sum on the left side consist of the only summand: 1

3·7 = 1

3·(4+3); therefore P(1)is true.

I n d u c t i v e s t e p

Let us prove the truth of the predicate P(k+1):

3·7+ 1

7·11+ · · · + 1 (4k−1)(4k+3)

k/(3(4k+3))

+ 1

(4(k+1)−1)(4(k+1)+3) =

= k

3(4k+3)+ 1

(4k+3)(4k+7)= 1

3(4k+3)(4k+7)(k(4k+7)+3)=

= k+1 3(4(k+1)+3). Therefore, the initial identity is true for alln=1,2, . . .

Second method

This method is only based in algebraic transformations. Note that each summand in the sum is presented in the form of the difference

(4k−1)(4k+3)= 1 4

4k−1 − 1 4k+3

Rewrite the sum in the following form:

3·7 + · · · + 1

(4n−1)(4n+3) =

= 1 4

1 3 −1

+ 1

7 − 1 11

+ · · · + +

4k−1 − 1 4k+3

4k+3 − 1 4k+7

+ · · · + +

4n−1− 1 4n+3 . Regroup the summands in the obtained sum so that to unite the summands opposite in sign:

3·7+ · · · + 1

(4n−1)(4n+3) =

= 1 4

1 3 +

−1 7 +1

+ · · · +

− 1

4k+3 + 1 4k+3

+ +

− 1

4k+7 + 1 4k+7

+ · · · − 1 4n+3

=1 4

3− 1

4n+3

= n 3(4n+3). 1.66.Hint. See the solution of the previous exercise.

1.67.Answer:

n k=1

(ak+b)(ak+a+b) = n

(a+b)(an+a+b).

Dalam dokumen Sergei Kurgalin Sergei Borzunov - Digital Library Univ STEKOM (Halaman 36-59)