Introduction to Mathematical Thinking Coursera Quiz Answers 2023 [💯% Correct Answer]

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About Introduction to Mathematical Thinking Course

Mathematical thinking is not the same as doing mathematics – at least not as mathematics is typically presented in our school system. School math typically focuses on learning procedures to solve highly stereotyped problems.

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Introduction to Mathematical Thinking Quiz Answers

Week 1: Introduction to Mathematical Thinking Coursera Quiz Answers

Quiz 1:

Q1. Is it possible for one of (\phi \wedge \psi) \wedge \theta(ϕ∧ψ)∧θ and \phi \wedge (\psi \wedge \theta)ϕ∧(ψ∧θ) to be true and the other false? (If not, then the associative property holds for conjunction.) [Score: 5 points]

  • Yes
  • No

Q2. Is it possible for one of (\phi \vee \psi) \vee \theta(ϕ∨ψ)∨θ and \phi \vee (\psi \vee \theta)ϕ∨(ψ∨θ) to be true and the other false? (If not, then the associative property holds for disjunction.) [Score: 5 points]

  • Yes
  • No

Q3. Is it possible for one of \phi \wedge (\psi \vee \theta)ϕ∧(ψ∨θ) and (\phi \wedge \psi ) \vee (\phi \wedge \theta)(ϕ∧ψ)∨(ϕ∧θ) to be true and the other false? (If not, then the distributive property holds for conjunction across disjunction.) [Score: 5 points]

  • Yes
  • No

Q4. Is it possible for one of \phi \vee (\psi \wedge \theta)ϕ∨(ψ∧θ) and (\phi \vee \psi ) \wedge (\phi \vee \theta)(ϕ∨ψ)∧(ϕ∨θ) to be true and the other false? (If not, then the distributive property holds for disjunction across conjunction.) [Score: 5 points]

  • Yes
  • No

Q5. Is showing that the negation \neg \phi¬ϕ is true equivalent to showing that \phiϕ is false? [Score: 5 points]

  • Yes
  • No

Q6. Assuming you know nothing more about Alice, which of (a) – (e) is most likely? (Or does (f) hold?) [Score: 5 points]

  • (a) Alice is a rock star and works in a bank.
  • (b) Alice is quiet and works in a bank.
  • (c) Alice is quiet and reserved and works in a bank.
  • (d) Alice is honest and works in a bank.
  • (e) Alice works in a bank.
  • (f) None of the above is more or less likely.

Q7. Assuming you know nothing more about Alice, which of (a) – (e) is most likely? (Or does (f) hold?) [Score: 5 points]

  • (a) Alice is a rock star or she works in a bank.
  • (b) Alice is quiet and works in a bank.
  • (c) Alice is a rock star.
  • (d) Alice is honest and works in a bank.
  • (e) Alice works in a bank.
  • (f) None of the above is more or less likely.

Q8. Identify which of the following are true (where xx denotes an arbitrary real number). If you do not select a particular statement, the system will assume you think it is false. [Score: 5 points]

  • (x\gt 0) \wedge (x \leq 10)(x>0)∧(x≤10) means 0 \leq x \leq 100≤x≤10
  • (x \geq 0) \wedge (x^2 \lt 9)(x≥0)∧(x2<9) means 0 \leq x \lt 30≤x<3
  • (x \geq 0) \wedge (x \leq 0)(x≥0)∧(x≤0) means x=0x=0
  • There is no xx for which (x \lt 4) \wedge (x \gt 4)(x<4)∧(x>4)
  • – 5 \leq x \leq 5−5≤x≤5 means xx is at most 5 units from 0.
  • -5 \lt x \lt 5−5<x<5 implies that xx cannot be exactly 5 units from 0.
  • (x \geq 0) \vee (x \lt 0)(x≥0)∨(x<0)
  • (0 = 1) \vee (x^2 \geq 0)(0=1)∨(x2≥0)
  • If (x \gt 0 \vee x \lt 0)(x>0∨x<0) then x \neq 0x​=0.
  • If x^2 = 9x2=9 then (x = 3 \vee x = -3)(x=3∨x=−3).

Week 2: Introduction to Mathematical Thinking Coursera Quiz Answers

Quiz 1: Problem Set 2

Q1. Which of the following conditions are necessary for the natural number nn to be divisible by 6? Select all those you believe are necessary. [6 points]

  • nn is divisible by 3.
  • nn is divisible by 9.
  • nn is divisible by 12.
  • n=24n=24.
  • n^2n
    2
    is divisible by 3.
  • nn is even and divisible by 3.

Q2. Which of the following conditions are sufficient for the natural number nn to be divisible by 6? Select all those you believe are sufficient. [6 points]

  • nn is divisible by 3.
  • nn is divisible by 9.
  • nn is divisible by 12.
  • n=24n=24.
  • n^2n
    2
    is divisible by 3.
  • nn is even and divisible by 3.

Q3. Which of the following conditions are necessary and sufficient for the natural number nn to be divisible by 6? Select all those you believe are necessary and sufficient. [6 points]

  • nn is divisible by 3.
  • nn is divisible by 9.
  • nn is divisible by 12.
  • n=24n=24.
  • n^2n
    2
    is divisible by 3.
  • nn is even and divisible by 3.

Q4. Identify the antecedent in the conditional ”If the apples are red, they are ready to eat.” [1 point]

  • THE APPLES ARE RED
  • THE APPLES ARE READY TO EAT

Q5. Identify the antecedent in the conditional ”The differentiability of a function ff is sufficient for ff to be continuous.” [1 point]

  • ff IS DIFFERENTIABLE
  • ff IS CONTINUOUS

Q6. Identify the antecedent in the conditional ”A function ff is bounded if ff is integrable.” [1 point]

  • ff IS BOUNDED
  • ff IS INTEGRABLE

Q7. Identify the antecedent in the conditional ”A sequence S is bounded whenever S is convergent.” [1 point]

  • S IS BOUNDED
  • S IS CONVERGENT

Q8. Identify the antecedent in the conditional ”It is necessary that nn is prime in order for 2^n – 12
n−1 to be prime.”

  • nn IS PRIME
  • 2^n – 12
    n
    −1 IS PRIME

Q9. Identify the antecedent in the conditional ”The team wins only when Karl is playing.” [1 point]

  • THE TEAM WINS
  • KARL IS PLAYING

Q10. QIdentify the antecedent in the conditional ”When Karl plays the team wins.” [1 point]

  • THE TEAM WINS
  • KARL PLAYS

Q11. Identify the antecedent in the conditional ”The team wins when Karl plays.” [1 point]

  • THE TEAM WINS
  • KARL PLAYS

Q12. For natural numbers m, nm,n, is it true that mnmn is even iff mm and nn are even? [2 points]

  • Yes
  • No

Q13. Is it true that mnmn is odd iff mm and nn are odd? [2 points]

  • Yes
  • No

Q14. Which of the following pairs of propositions are equivalent? Select all you think are equivalent. [6 points]

  • \neg P \vee Q \ , \ P \Rightarrow Q¬P∨Q , P⇒Q
  • \neg (P \vee Q) \ , \ \neg P \wedge \neg Q¬(P∨Q) , ¬P∧¬Q
  • \neg P \vee \neg Q \ , \ \neg (P \vee \neg Q)¬P∨¬Q , ¬(P∨¬Q)
  • \neg (P \wedge Q) \ , \ \neg P \vee \neg Q¬(P∧Q) , ¬P∨¬Q
  • \neg (P \Rightarrow (Q \wedge R)) \ , \ \neg (P \Rightarrow Q) \vee \neg (P \Rightarrow R)¬(P⇒(Q∧R)) , ¬(P⇒Q)∨¬(P⇒R)
  • P \Rightarrow (Q \Rightarrow R) \ , \ (P \wedge Q) \Rightarrow RP⇒(Q⇒R) , (P∧Q)⇒R

Q15. A major focus of this course is learning how to assess mathematical reasoning. How good you are at doing that lies on a sliding scale. Your task is to evaluate this purported proof according to the course rubric.

Enter your evaluation (which should be a whole number between 0 and 24, inclusive) in the box. An answer within 4 points of the instructor’s evaluation counts as correct. [5 points]

You should read the website section “Using the evaluation rubric” (and watch the associated short explanatory video) before attempting this question. There will be many more proof evaluation questions as the course progresses.

NOTE: The scoring system for proof evaluation questions is somewhat arbitrary, due to limitations of the platform. But the goal is to provide
opportunities for you to reflect on what makes an argument a good proof, and you are allowed to repeat the
Problem Sets as many times as it takes to be able to progress. Your “score” is simply feedback information.
Moreover, the “passing grade” for Problem Sets is a low 35%.

Enter answer here

Week 3: Introduction to Mathematical Thinking Coursera Quiz Answers

Quiz 1: Problem Set 3

Q1. Let xx be a variable ranging over doubles tennis matches, and tt be a variable ranging over doubles tennis matches when Rosario partners with Antonio. Let W(x)W(x) mean that Rosario and her partner (whoever it is) win the doubles match xx. Select the following English sentences that mean the same as the symbolic formula \exists tW(t)∃tW(t).

  • Rosario and Antonio win every match where they are partners.
  • Rosario and her partner sometimes win the match when she partners with Antonio.
  • Whenever Rosario partners with Antonio, they win the match.
  • Rosario and Antonio win exactly one match when they are partners.
  • Rosario and Antonio win at least one match when they are partners.
  • If Rosario and her partner win the match, she must be partnering with Antonio.

Q2. Let xx be a variable ranging over doubles tennis matches, and tt be a variable ranging over doubles tennis matches when Rosario partners with Antonio. Let W(x)W(x) mean that Rosario and her partner (whoever it is) win the doubles match xx. Select the following English sentences that mean the same as the symbolic formula \forall tW(t)∀tW(t).

  • Rosario and Antonio win every match where they are partners
  • Rosario always partners with Antonio.
  • Whenever Rosario partners with Antonio, they win the match.
  • Sometimes, Rosario and her partner win the match.
  • Rosario and her partner win the match whenever she partners with Antonio.
  • If Rosario and her partner win the match, she must be partnering with Antonio.

Q3. Which of the following formal propositions says that there is no largest prime. (There may be more than one. You have to select all correct propositions.) The variables denote natural numbers. [6 points]

Q4. The symbol \exists ! x∃!x means “There exists a unique xx such that …” Which of the following accurately defines the expression \exists ! x \phi(x)∃!xϕ(x)?

  • \exists x \forall y [\phi(x) \wedge [\phi(y) \Rightarrow (x \neq y)]]∃x∀y[ϕ(x)∧[ϕ(y)⇒(x


    =y)]]
  • \exists x [\phi(x) \wedge (\exists y)[\phi(y) \Rightarrow (x \neq y)]]∃x[ϕ(x)∧(∃y)[ϕ(y)⇒(x


    =y)]]
  • \exists x \exists y [(\phi(x) \wedge \phi(y)) \Rightarrow (x = y)]∃x∃y[(ϕ(x)∧ϕ(y))⇒(x=y)]
  • [\exists x \phi(x)] \wedge (\forall y)[\phi(y) \Rightarrow (x = y)][∃xϕ(x)]∧(∀y)[ϕ(y)⇒(x=y)]
  • \exists x [\phi(x) \wedge (\forall y)[\phi(y) \Rightarrow (x = y)]]∃x[ϕ(x)∧(∀y)[ϕ(y)⇒(x=y)]]

Q5. Which of the following means “The arithmetic operation x !\uparrow! yx↑y is not commutative.” (\uparrow↑ is just some arbitrary binary operation.)

  • \forall x \forall y [x !\uparrow! y \neq y !\uparrow! x]∀x∀y[x↑y


    =y↑x]
  • \forall x \exists y [x !\uparrow! y \neq y !\uparrow! x]∀x∃y[x↑y


    =y↑x]
  • \exists x \exists y [x !\uparrow! y \neq y !\uparrow! x]∃x∃y[x↑y


    =y↑x]
  • \exists x \forall y [x !\uparrow! y \neq y !\uparrow! x]∃x∀y[x↑y


    =y↑x]

Q6. Evaluate this purported proof, and evaluate it according to the course rubric.

Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box. An answer within 4 points of the instructor’s grade counts as correct. [5 points]

You should read the website section “Using the rubric” (and watch the explanatory video) before attempting this question. There will be many more proof evaluation questions as the course progresses.

Enter answer here

Week 4: Introduction to Mathematical Thinking Coursera Quiz Answers

Quiz 1: Problem Set 4

Q1. Which of the following is equivalent to \neg \forall x[P(x) \Rightarrow (Q(x) \vee R(x))]¬∀x[P(x)⇒(Q(x)∨R(x))]? (Only one is.)

  • \exists x[P(x) \vee \neg Q(x) \vee \neg R(x)]∃x[P(x)∨¬Q(x)∨¬R(x)]
  • \exists x[\neg P(x) \wedge Q(x) \wedge R(x)]∃x[¬P(x)∧Q(x)∧R(x)]
  • \exists x[P(x) \wedge \neg Q(x) \wedge \neg R(x)]∃x[P(x)∧¬Q(x)∧¬R(x)]
  • \exists x[P(x) \wedge (\neg Q(x) \vee \neg R(x))]∃x[P(x)∧(¬Q(x)∨¬R(x))]
  • \exists x[P(x) \vee (\neg Q(x) \wedge \neg R(x))]∃x[P(x)∨(¬Q(x)∧¬R(x))]

Q2. Let p,qp,q be variables denoting tennis players, let tt be a variable denoting games of tennis, and let W(p,q,t)W(p,q,t) mean that pp plays against qq in game tt and wins. Which of the following claims about tennis players mean the same as the symbolic formula \forall p \exists q \exists t W(p,q,t)∀p∃q∃tW(p,q,t)? Select all that have that meaning.

  • Everyone wins a game.
  • Everyone loses a game.
  • For every player there is another player they beat all the time.
  • There is a player who loses every game.
  • There is a player who wins every game.

Q3. Let p,qp,q be variables denoting the tennis players in a club, let tt be a variable denoting the club’s games of tennis, and let W(p,q,t)W(p,q,t) mean that pp plays against qq in game tt and wins. Assuming that there are at least two tennis players and games between them do take place, which (if any) of the following symbolic formula cannot possibly be true? Select all you think cannot possibly be true. [3 points]

  • \forall p \exists q \exists t W(p,q,t)∀p∃q∃tW(p,q,t)
  • \forall p \forall q \exists t W(p,q,t)∀p∀q∃tW(p,q,t)
  • \forall q \exists p \exists t W(p,q,t)∀q∃p∃tW(p,q,t)

Q4. Which (one) of the following means “Everybody loves a lover”, where L(x,y)L(x,y) means (person) xx loves (person) yy and a lover is defined to be someone in a mutual loving relationship? [5 points] If English is not your native language, you might want to discuss this sentence with a native English speaker before you answer. It’s an idiomatic expression.]

  • \forall x \forall y [\exists z(L(x,z) \wedge L(z,x)) \Rightarrow L(y,x)]∀x∀y[∃z(L(x,z)∧L(z,x))⇒L(y,x)]
  • \forall x \forall y [\forall z(L(x,z) \vee L(z,x)) \Rightarrow L(y,x)]∀x∀y[∀z(L(x,z)∨L(z,x))⇒L(y,x)]
  • \forall x [\exists z(L(x,z) \wedge L(z,x)) \wedge \forall y L(y,x)]∀x[∃z(L(x,z)∧L(z,x))∧∀yL(y,x)]

Q5. Which of the following statements about the order relation on the real line is/are false?

  • \forall x \forall y \forall z[(x \leq y) \wedge (y \leq z) \Rightarrow (x \leq z)]∀x∀y∀z[(x≤y)∧(y≤z)⇒(x≤z)]
  • \forall x \forall y [(x \leq y) \wedge (y \leq x) \Rightarrow (x = y)]∀x∀y[(x≤y)∧(y≤x)⇒(x=y)]
  • \forall x \exists y [(x \leq y) \wedge (y \leq x)]∀x∃y[(x≤y)∧(y≤x)]
  • \exists x \forall y [(y \lt x) \vee (x \lt y)]∃x∀y[(y<x)∨(x<y)]

Q6. A student produced this purported proof while trying to understand Euclid’s proof of the infinitude of the primes. Evaluate it according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box. An answer within 4 points of the instructor’s grade counts as correct. [5 points]

You should read the website section “Using the rubric” and watch the associated short explanatory video before attempting this question.

Enter answer here

Week 5: Introduction to Mathematical Thinking Coursera Quiz Answers

Quiz 1: Problem Set 5

Q1. Let m, nm,n denote any two natural numbers. Is the following a valid proof that mnmn is odd iff mm and nn are odd?

If m,nm,n are odd there are integers p,qp,q such that m=2p+1, n=2q+1m=2p+1,n=2q+1. Then mn=(2p+1)(2q+1) = 2(2pq+p+q)+1mn=(2p+1)(2q+1)=2(2pq+p+q)+1, so mnmn is odd. That completes the proof.

  • Valid
  • Invalid

Q2. Take the sentence:

You can fool some of the people some of the time, but you cannot fool all of the people all the time.

Let xx be a variable for a person, tt a variable for a period of time, and let F(x,t)F(x,t) mean you can fool xx at time tt.

Which of the following mathematical formulas is equivalent to the given statement?

  • \exists x \exists t F(x,t) \wedge \exists x \exists t \neg F(x,t)∃x∃tF(x,t)∧∃x∃t¬F(x,t)
  • \exists x \exists t F(x,t) \wedge \neg \forall x \exists t F(x,t)∃x∃tF(x,t)∧¬∀x∃tF(x,t)
  • \exists x \exists t F(x,t) \wedge \neg \exists x \exists t F(x,t)∃x∃tF(x,t)∧¬∃x∃tF(x,t)
  • None of the above.

Q3. True or false? For any two statements \phiϕ and \psiψ, either \phi \Rightarrow \psiϕ⇒ψ or its converse is true (or both).

  • True
  • False

Q4. Are the following two statements equivalent?

\neg(\phi \Rightarrow \psi)¬(ϕ⇒ψ) and \phi \wedge (\neg\psi)ϕ∧(¬ψ)

  • Yes.
  • No.

Q5. Are the following two statements equivalent?

(\phi \vee \psi) \Rightarrow \theta(ϕ∨ψ)⇒θ and (\phi \Rightarrow \theta) \wedge (\psi \Rightarrow \theta)(ϕ⇒θ)∧(ψ⇒θ)

  • Yes.
  • No.

Q6. True or false? There are infinitely many natural numbers nn for which \sqrt{n}
n is rational. (Before entering your answer, you should construct a proof of the statement or its negation, so you are sure.)

  • True
  • False

Q7. This argument claims to prove that 1=2.

Obviously it is incorrect. Identify exactly what the error is, and evaluate the purported proof according to the course rubric.

Remember, this is not a regular mathematics course of the kind you are probably familiar with. We are working on various elements of mathematical thinking, mathematical exposition, and the communication of mathematics. The rubric is designed to focus attention on all of those factors. Your “Overall valuation” figure is the grade you would assign a student if s/he submitted this proof in a first-year college mathematics course.

Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box. An answer within 4 points of the instructor’s grade counts as correct.

You should read the website section “Using the rubric” and view the associated short explanatory video before attempting this question.

Enter answer here

Week 6: Introduction to Mathematical Thinking Coursera Quiz Answers

Quiz 1: Problem Set 6

Q1. Is the following proof valid or not?

Theorem: For any natural number nn, 2^n > 2n2
n

2n.

Proof: By induction. The case n=1n=1 is obviously true, so assume the inequality holds for nn.

That is, assume 2^n>2n2
n

2n. Then?

2^{n+1} = 2 \cdot 2^n > 2 \cdot 2n2
n+1
=2⋅2
n

2⋅2n (by the induction hypothesis) = 4n = 2n + 2n \geq 2n + 2=4n=2n+2n≥2n+2 (since n \geq 1) = 2(n+1)n≥1)=2(n+1)

This establishes the inequality for n+1n+1. Hence, by induction, the inequality holds for all nn.

  • Valid
  • Invalid

Q2. Is the following proof valid or not?

Theorem: If a nonempty finite set XX has nn elements, then XX has exactly 2^n2
n
distinct subsets.

Proof: By induction on nn.

The case n=1n=1 is true, since if XX is a set with exactly one element, say X = {a}X={a}, then XX has the two subsets \emptyset∅ and XX itself.

Assume the theorem is true for nn. Let XX be a set of n+1n+1 elements. Let a \in Xa∈X and let Y = X – {a}Y=X−{a} (i.e., obtain YY by removing aa from XX). Then YY is a set with nn elements. By the induction hypothesis, YY has 2^n2
n
subsets. List them as Y_1,\ldots,Y_{2^n}Y
1

,…,Y
2
n


Then all the subsets of XX are Y_1,\ldots,Y_{2^n}, Y_1 \cup {a},\ldots,Y_{2^n} \cup {a}Y
1

,…,Y
2
n


,Y
1

∪{a},…,Y
2
n


∪{a} (i.e., the subsets of YY together with the subsets of YY with aa added to each one). There are 2\cdot2^n = 2^{n+1}2⋅2
n=2
n+1 sets in this list. This establishes the theorem for n+1n+1. Hence, by induction, it is true for all nn.

  • Valid
  • Invalid

Q3. True or false? If pp is a prime number, then \sqrt{p}
p is irrational. (Before entering your answer, you should either construct a proof of truth or find a counter-example, so you are sure. After you have completed the problem set, you should write up your proof or counter-example and share it with your study group for feedback. You can assume that if pp is prime, then whenever pp divides a product abab, pp divides at least one of a, ba,b. ) [3 points]

  • True
  • False

Q4. Evaluate this purported proof

according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box. An answer within 4 points of the instructor’s grade counts as correct. [3 points]

You should read the website section “Using the rubric” and watch the associated short explanatory video before attempting this question.

Enter answer here

Q5. This purported theorem is obviously false:

Select the line number of the (incorrect) statement where the proof logically breaks down.

  • Line 1
  • Line 2
  • Line 3
  • Line 4
  • Line 5
  • Line 6
  • Line 7
  • Line 8
  • Line 9
  • Line 10
  • Line 11
  • Line 12
  • Line 13
  • Line 14
  • Line 15
  • Line 16
  • Line 17
  • Line 18
  • Line 19

Week 7: Introduction to Mathematical Thinking Coursera Quiz Answers

Quiz 1: Problem Set 7

Q1. Say which of the following statements are true. (Leave the box blank to indicate that it is false.)

  • 20|30020∣300
  • 17|3517∣35
  • 5|05∣0
  • 0|50∣5
  • 21|(-21)21∣(−21)

Q2. Say whether the following proof is valid or not.

Theorem. The square of any odd number is 1 more than a multiple of 8. (For example, 3^{2}= 9 = 8 + 1, 5^{2} = 25 = 3 \cdot 8 + 13
2
=9=8+1,5
2
=25=3⋅8+1.)

Proof: By the Division Theorem, any number can be expressed in one of the forms 4q,\ 4q+1,\ 4q+2,\ 4q+34q, 4q+1, 4q+2, 4q+3. So any odd number has one of the forms 4q+1, 4q+34q+1,4q+3. Squaring each of these gives:

(4q+1)2(4q+3)2==16q2+8q+116q2+24q+9==8(2q2+q)+18(2q2+3q+1)+1
(4q+1)
2

(4q+3)
2

​=

16q
2
+8q+1
16q
2
+24q+9
​=​

8(2q
2
+q)+1
8(2q
2
+3q+1)+1
​In both cases the result is one more than a multiple of 8. This proves the theorem.

  • Valid
  • Invalid

Q3. Say whether the following verification of the method of induction is valid or not.

Proof: We have to prove that if:

then (\forall n)A(n)(∀n)A(n).

We argue by contradiction. Suppose the conclusion is false. Then there will be a natural number nn such that \neg A(n)¬A(n). Let mm be the least such number. By the first condition, m>1m>1, so m=n+1m=n+1 for some nn. Since n \lt mn<m, A(n)A(n). Then by the second condition, A(n+1)A(n+1), i.e., A(m)A(m). This is a contradiction, and that proves the result.

  • Valid
  • Invalid

Q4. Evaluate this purported proof

according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box. An answer within 4 points of the instructor’s grade counts as correct.

You should read the website section “Using the rubric” and watch the associated short explanatory video before attempting this question.

Enter answer here


Q5. Evaluate this purported proof

according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box. An answer within 4 points of the instructor’s grade counts as correct.

Enter answer here


Q6. Evaluate this purported proof

according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box. An answer within 4 points of the instructor’s grade counts as correct.

Enter answer here

Week 8: Introduction to Mathematical Thinking Coursera Quiz Answers

Quiz 1: Problem Set 8

Q1. Say which of the following are true. (Leave the box empty to indicate that it’s false.)

  • A set AA of reals can have at most one least upper bound.
  • If a set AA of reals has a lower bound, it has infinitely many lower bounds.
  • If a set AA of reals has both a lower bound and an upper bound, then it is finite.
  • 0 is the least upper bound of the set of negative integers, considered as a subset of the reals.

Q2. Which of the following say that bb is the greatest lower bound of a set AA of reals? (Leave the box empty to indicate that it does not say that.)

  • b \leq ab≤a for all a \in Aa∈A and if c \leq ac≤a for all a \in Aa∈A, then b \geq cb≥c.
  • b \leq ab≤a for all a \in Aa∈A and if c \leq ac≤a for all a \in Aa∈A, then b > cb>c.
  • b \lt ab<a for all a \in Aa∈A and if c \lt ac<a for all a \in Aa∈A, then b \geq cb≥c.
  • b \lt ab<a for all a \in Aa∈A and if c \leq ac≤a for all a \in Aa∈A, then b \geq cb≥c.
  • b \leq ab≤a for all a \in Aa∈A and if \epsilon > 0ϵ>0 there is an a \in Aa∈A such that a \lt b + \epsilona<b+ϵ.

Q3. Evaluate this purported proof

according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box. An answer within 4 points of the instructor’s grade counts as correct.

You should read the website section “Using the rubric” and watch the associated short explanatory video before attempting this question.

Enter answer here

Q4. Evaluate thus purported proof

according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box. An answer within 4 points of the instructor’s grade counts as correct.

Enter answer here

Q5. Evaluate this purported proof

according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box. An answer within 4 points of the instructor’s grade counts as correct.

Enter answer here

Week 9: Introduction to Mathematical Thinking Coursera Quiz Answers

Quiz 1: Evaluation Exercise 1

Q1. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Say whether the following is true or false and support your answer by a proof:

(∃m ∈ N )(∃n ∈ N )(3m + 5n = 12)(∃m∈N)(∃n∈N)(3m+5n=12)

ANSWER It’s true. Let m = 4, n = 0m=4,n=0. Then 3m + 5n = 123m+5n=12.

Enter answer here

Q2. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Say whether the following is true or false and support your answer by a proof: The sum of any five consecutive integers is divisible by 5 (without remainder).

ANSWER True. 1 + 2 + 3 + 4 + 5 = 15, which is divisible by 5.

Enter answer here

Q3. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Say whether the following is true or false and support your answer by a proof: For any integer nn, the number n^2 + n + 1n
2
+n+1 is odd.

ANSWER We prove it by induction.

For n = 1, n^2 + n + 1 = 1 + 1 + 1 = 3n=1,n
2
+n+1=1+1+1=3, which is odd.

Suppose n^2 + n + 1n
2
+n+1 is odd. Then

(n + 1)^2 + (n + 1) + 1 = n^2 + 2n + 1 + n + 1 + 1 = n^2 + 3n + 2 + 1 = (n + 1)(n + 2) + 1(n+1)
2
+(n+1)+1=n
2
+2n+1+n+1+1=n
2
+3n+2+1=(n+1)(n+2)+1

But one of (n + 1),(n + 2)(n+1),(n+2) must be even, so (n + 1)(n + 2)(n+1)(n+2) is even. Hence (n + 1)^2 + (n + 1) + 1(n+1)
2
+(n+1)+1 is odd. This proves the result by induction.

Enter answer here

Q4. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Prove that every odd natural number is of one of the forms 4n + 14n+1 or 4n + 3,4n+3, where nn is an integer.

ANSWER We prove it by induction. For n = 1, 4n + 1 = 5n=1,4n+1=5, which is odd.

If it’s true for nn, then 4(n + 1) + 1 = 4n + 4 + 1 = 4n + 54(n+1)+1=4n+4+1=4n+5 and 4(n + 1) + 3 = 4n + 4 + 3 = 4n + 74(n+1)+3=4n+4+3=4n+7, which are both odd. This proves the result by induction

Enter answer here

Q5. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Prove that for any integer nn, at least one of the integers nn, n + 2, n + 4n+2,n+4 is divisible by 3.

ANSWER Given mm, by the Division Theorem, m = 4n + qm=4n+q, where 0 ≤ q < 40≤q<4. If we divide nn by 3, either it divides evenly or it leaves a remainder of 1 or 2. So 3 has to divide one of n, n + 2, n + 4.n,n+2,n+4.

Enter answer here

Q6. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION A classic unsolved problem in number theory asks if there are infinitely many pairs of ‘twin primes’, pairs of primes separated by 2, such as 3 and 5, 11 and 13, or 71 and 73. Prove that the only prime triple (i.e. three primes, each 2 from the next) is 3, 5, 7.

ANSWER Suppose p, qp,q is a pair of twin primes, where p > 5p>5. We show that it is impossible to extend p, qp,q to be a prime triple Let N = p.q + 1N=p.q+1. Then, either NN is prime or else there is a prime rr such that r|Nr∣N. It follows that there is no prime that can be added to give a prime triple.

Enter answer here

Q7. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Prove that for any natural number n:

2 + 2^2
2

  • 2^3
    3
  • . . . + 2n = 2^{n+1}
    n+1
    − 2

ANSWER For n = 1n=1, the identity reduces to 2 = 22 − 22=22−2, which is true.

Assume it hold for nn. Then, adding 2n+12n+1 to both sides of the identity,

2 + 2^2 + 2^3 + . . . + 2^n + 2^{n+1} = 2^{n+1} − 2 + 2^{n+1} = 2.2^{n+1} − 2 = 2^{n+2} − 22+2
2
+2
3
+…+2
n
+2
n+1
=2
n+1
−2+2
n+1
=2.2
n+1
−2=2
n+2
−2

This is the identity at n + 1n+1. That completes the proof.

Enter answer here

Q8. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Prove (from the definition of a limit of a sequence) that if the sequence {a_n}_{n=1}^\infty{a
n

}
n=1


tends to limit LL as n \rightarrow \inftyn→∞, then for any fixed number M > 0M>0, the sequence {Ma_n}_{n=1}^\infty{Ma
n

}
n=1


tends to the limit MLML.

ANSWER By the assumption, we can find an NN such that

n \geq N \rightarrow | a_n = L | < \epsilon /Mn≥N→∣a
n=L∣<ϵ/M

Then,

n \geq N \Rightarrow |Ma_n – M L|= M. |a_n – L| < M. \epsilon / M = \epsilonn≥N⇒∣Ma
n

−ML∣=M.∣a
n

−L∣<M.ϵ/M=ϵ

which shows that {Ma_nMa
n

}_{n=1}^\infty
n=1


tends to the limit M LML

Enter answer here

Q9. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Given a collection A_nA
n

, n = 1, 2, \ldotsn=1,2,… of intervals of the real line, their intersection is defined to be

\bigcap_{n=1}^\infty A_n = {x|(\forall n)(x \in A_n)}⋂
n=1


A
n

={x∣(∀n)(x∈A
n

)}.

Give an example of a family of intervals A_nA
n

, n = 1, 2, \ldotsn=1,2,…, such that A_{n+1} \subset A_nA
n+1

⊂A
n

for all nn and \bigcap_{n=1}^\infty A_n = \emptyset⋂
n=1


A
n

=∅

Prove that your example has the stated property.

ANSWER Let A_n = ( \frac{1}{n+1} , \frac{1}{n} )A
n

=(
n+1
1

,
n
1

).

For any x > 0x>0, we can find an mm such that 1/m < x1/m<x, and then x \notin ( \frac{1}{m+1} , \frac{1}{m} )x∈ /(m+11,m1).

Hence \bigcap_{n=1}^\infty A_n = \emptyset⋂
n=1


A
n=∅

Enter answer here

Q10. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Give an example of a family of intervals A_n
n

, n = 1, 2, \ldotsn=1,2,…, such that A_{n+1} \subset A_nA
n+1

⊂A
n

for all nn and \bigcap_{n=1}^\infty A_n⋂
n=1


A
n

consists of a single real number.

Prove that your example has the stated property.

ANSWER Let A_n = (−1/n, +1/n)A
n=(−1/n,+1/n).

For any n, 0 \in A_n,n,0∈A
n

, so 0 \in \bigcap_{n=1}^\infty A_n0∈⋂
n=1


A
n
On the other hand, if x \ne 0x


=0, then there is an mm such that 1/m < |x|1/m<∣x∣, and for that m, x \notin A_mm,x∈
/

A
m

, so x \notin \bigcap_{n=1}^\infty A_nx∈
/


n=1


A
n

.

Hence \bigcap_{n=1}^\infty A_n = {0}⋂
n=1


A n={0}.

Enter answer here

Quiz 2: Evaluation Exercise 2

Q1. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Say whether the following is true or false and support your answer by a proof:

(∃m ∈ N )(∃n ∈ N )(3m + 5n = 12)(∃m∈N)(∃n∈N)(3m+5n=12)

ANSWER It’s false. We need only look at values of mm from 1 to 3 (since 3×4 = 12, which already
gives the right-hand side) and values of nn from 1 to 2 (since 5 × 3 = 15 ≥ 12). If you calculate 3m + 5n3m+5n for the six possible pairs in this range, you find that the answer is never 12. This proves
the result.

Enter answer here

Q2. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Say whether the following is true or false and support your answer by a proof: The sum of any five
consecutive integers is divisible by 5 (without remainder).

ANSWER False. Let n, n + 1, n + 2, n + 3, n + 4n,n+1,n+2,n+3,n+4 be any five consecutive integers. Then

n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 5n + 1 + 2 + 3 + 4 = 5n + 8 = 5(n + 1) + 3n+(n+1)+(n+2)+(n+3)+(n+4)=5n+1+2+3+4=5n+8=5(n+1)+3

which is not a multiple of 5 since in the Division Theorem it leaves a remainder of 3.

Enter answer here

Q3. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Say whether the following is true or false and support your answer by a proof: For any integer nn,
the number n^2 + n + 1n
2
+n+1 is odd.

ANSWER For any nn, n^2 +n+ 1 = n(n+ 1) + 1n
2
+n+1=n(n+1)+1. But n(n+ 1)n(n+1) is always even (since one of n, n+ 1n,n+1 is even and the other odd). Hence n(n + 1)n(n+1) is always odd, as claimed.

Enter answer here

Q4. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Prove that every odd natural number is of one of the forms 4n + 14n+1 or 4n + 34n+3, where nn is an integer.

ANSWER This is not true. For example, if n = −1n=−1, which is an integer, then 4n + 1 = −34n+1=−3 and 4n + 3 = −14n+3=−1. But −3 and −1 are not natural numbers.

Enter answer here

Q5. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Prove that for any integer nn, at least one of the integers n, n + 2, n + 4n,n+2,n+4 is divisible by 3

ANSWER nn can be expressed in one of the forms 3q, 3q + 1, 3q + 23q,3q+1,3q+2, for some qq.
In the first case, nn is divisible by 3.
In the second case n + 2 = 3q + 3 = 3(q + 1n+2=3q+3=3(q+1), so n + 2n+2 is divisible by 3.
In the third case n + 4 = 3q + 6 = 3(q + 2)n+4=3q+6=3(q+2), so n + 4n+4 is divisible by 3.

Enter answer here

Q6. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION A classic unsolved problem in number theory asks if there are infinitely many pairs of ‘twin primes’,
pairs of primes separated by 2, such as 3 and 5, 11 and 13, or 71 and 73. Prove that the only prime
triple (i.e. three primes, each 2 from the next) is 3, 5, 7

ANSWER Let n, n + 2, n + 4n,n+2,n+4 be any three successive natural numbers, where n > 3n>3. I show that
3 divides one of these numbers. If 3 does not divide nn, then by the Division Theorem, n = 3q + 1n=3q+1 or n = 3q+ 2n=3q+2, for some qq. In the first case, n+ 2 = 3q+ 3n+2=3q+3, so 3|n3∣n, and in the second case n+ 4 = 3q+ 644n+4=3q+644,
so again 3|n3∣n. Thus 3 must divide one of the three numbers, which means they cannot all be prime.

Enter answer here

Q7. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Prove that for any natural number nn:

2 + 2^2 + 2^3 + . . . + 2^n = 2^{n+1} − 22+2
2
+2
3
+…+2
n
=2
n+1
−2

ANSWER We prove the result by induction. For n = 1n=1, the identity reduces to 2 = 2^2 − 22=2
2
−2, which
is true. Assume it hold for nn.

Then, 2 + 2^2 + 2^3 + . . . + 2^n + 2^{n+1} = 2^{n+1} − 2 + 2^{n+1} = 2.2^{n+1} − 2 = 2^{n+2} − 22+2
2
+2
3
+…+2
n
+2
n+1
=2
n+1
−2+2
n+1
=2.2
n+1
−2=2
n+2
−2

This is the identity at n + 1n+1. The result follows by induction.

Enter answer here

Q8. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Prove (from the definition of a limit of a sequence) that if the sequence {a_n}_{n=1}^\infty{a
n

}
n=1


tends to limit LL as n → ∞n→∞, then for any fixed number M > 0M>0, the sequence {Ma_n}_{n=1}^\infty{Ma
n

}
n=1


tends to the limit MLML.

ANSWER Pick \epsilonϵ > 0. Since {a_na
n

}{^∞{n=1}} n=1 ∞ ​ tends to limit L L as n → ∞n→∞, there is an NN such that a_na n ​ is within a distance of \epsilon/Mϵ/M of LL whenever n>Nn>N. For any such nn, Ma_nMa n ​ is within a distance M(\epsilon/M)M(ϵ/M) = \epsilonϵ of MLML. Hence {Ma_nMa n ​ }{^\infty{n=1}}
n=1


tends to MLML as nn tend to \infty∞

Enter answer here

Q9. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Given a collection A_nA
n

, nn = 1, 2, . . . of intervals of the real line, their intersection is defined to be \bigcap_{n=1}^\infty ⋂
n=1


A_nA
n

= {x |(∀n)(x ∈ An)}x∣(∀n)(x∈An). Give an example of a family of intervals A_n, n = 1, 2A
n

,n=1,2, . . ., such that A_{n+1} ⊂A
n+1

⊂ A_nA
n

for all nn and

\bigcap_{n=1}^\infty⋂
n=1


A_n = ∅A
n

=∅

Prove that your example has the stated property.

ANSWER Take the sequence (0, 1), (0, 1/2). (0, 1/4), (0, 1/8), . . . That is, A_nA
n

= (0, 1/2^{n−1}
n−1
).
Since {1/2^n
n
}^∞{n=1} n=1 ∞ ​ tends to 0 as n → ∞n→∞, \bigcap{n=1}^\infty⋂
n=1


A_n = \emptysetA
n

=∅

Enter answer here

Q10. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Give an example of a family of intervals A_n, n = 1, 2A
n

,n=1,2, . . ., such that A_{n+1} ⊂ A_nA
n+1

⊂A
n

for all nn and \bigcap_{n=1}^\infty⋂
n=1


A_nA
n

consists of a single real number. Prove that your example has the stated
property.

ANSWER Take A_n = [0, 1/2^n]A
n

=[0,1/2
n
]. Then 0 ∈ A_n0∈A
n

for all nn, so 0 ∈ \bigcap_{n=1}^\infty⋂
n=1


A_nA
n

. By the same argument
as in question 9 above, it follows that \bigcap_{n=1}^\infty⋂
n=1


A_nA
n

= {\emptyset∅}

Enter answer here

Quiz 3: Evaluation Exercise 3

Q1. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Say whether the following is true or false and support your answer by a proof:

(∃m ∈ N )(∃n ∈ N )(3m + 5n = 12)(∃m∈N)(∃n∈N)(3m+5n=12)

ANSWER It’s false. If n ≥ 2n≥2, then for any m, 3m + 5n ≥ 13m,3m+5n≥13, so we need only show that there is
no m such that 3m + 5 = 123m+5=12, i.e. no m such that 3m = 73m=7. This is immediate.

Enter answer here

Q2. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Say whether the following is true or false and support your answer by a proof: The sum of any five
consecutive integers is divisible by 5 (without remainder)

ANSWER True. Let n, n + 1, n + 2, n + 3, n + 4n,n+1,n+2,n+3,n+4 be any five consecutive integers. Then

n + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 5n + 1 + 2 + 3 + 4 = 5n + 10 = 5(n + 2)n+(n+1)+(n+2)+(n+3)+(n+4)=5n+1+2+3+4=5n+10=5(n+2)

which proves the result.

Enter answer here

Q3. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Say whether the following is true or false and support your answer by a proof: For any integer nn,
the number n^2 + n + 1n
2
+n+1 is odd.

ANSWER True. Consider the two case n even and n odd separately.
If nn is even, say n = 2kn=2k, then

n^2 + n + 1 = 4k^2 + 2k + 1 = 2(2k^2 + k) + 1n
2
+n+1=4k
2
+2k+1=2(2k
2
+k)+1

which is odd.
If nn is odd, say n = 2k + 1n=2k+1, then

n^2+n+1 = (2k+1)^2+(2k+1)+1 = 4k^2+4k+1+2k+1+1 = 4k^2+6k+2+1 = 2(2k^2+3k+1)+1n
2
+n+1=(2k+1)
2
+(2k+1)+1=4k
2
+4k+1+2k+1+1=4k
2
+6k+2+1=2(2k
2
+3k+1)+1

which is odd.
In both cases, n^2 + n + 1n
2
+n+1 is odd.

Enter answer here

Q4. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Prove that every odd natural number is of one of the forms 4n + 1 or 4n + 34n+1or4n+3, where n is an integer.

ANSWER Let m be a natural number. By the Division Theorem, there are unique numbers nn, rr
such that m = 4n + rm=4n+r, where 0 ≤ r < 40≤r<4. Thus m is one of 4n, 4n + 1, 4n + 2, 4n + 34n,4n+1,4n+2,4n+3. Since 4n4n and 4n + 24n+2 are even, if mm is odd, the only possibilities are 4n + 14n+1 and 4n + 34n+3.

Enter answer here

Q5. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Prove that for any integer nn, at least one of the integers nn, n + 2n+2, n + 4n+4 is divisible by 33.

ANSWER By the Division Theorem, nn can be expressed in one of the forms 3q, 3q + 1, 3q + 23q,3q+1,3q+2, for
some qq. In the first case, nn is divisible by 33. In the second case n + 2 = 3q + 3 = 3(q + 1)n+2=3q+3=3(q+1), so n + 2n+2 is divisible by 33. In the third case n + 4 = 3q + 6 = 3(q + 2),n+4=3q+6=3(q+2), so n + 4n+4 is divisible by 33.

Enter answer here

Q6. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION A classic unsolved problem in number theory asks if there are infinitely many pairs of ‘twin primes’,
pairs of primes separated by 2, such as 3 and 5, 11 and 13, or 71 and 73. Prove that the only prime
triple (i.e. three primes, each 2 from the next) is 3, 5, 7

ANSWER Consider any three numbers of the form nn, n + 2n+2, n + 4n+4, where n > 3n>3. By the answer
to the previous question, one of these numbers is divisible by 33, and hence is not prime.

Enter answer here

Q7. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Prove that for any natural number nn: 2 + 2^2 + 2^3 + . . . + 2^n = 2^{n+1} − 22+2
2
+2
3
+…+2
n
=2
n+1
−2

ANSWER Let S = 2 + 2^2 + 2^3 + . . . + 2^nS=2+2
2
+2
3
+…+2
n
. Then 2S = 2^2 + 2^3 + 2^4 + . . . + 2^n + 2^{n+1}2S=2
2
+2
3
+2
4
+…+2
n
+2
n+1
. Subtracting
the first identity from the second gives 2S − S = 2^{n+1} − 22S−S=2
n+1
−2. But 2S − S = S2S−S=S, so this establishes the
stated identity.

Enter answer here

Q8. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Prove (from the definition of a limit of a sequence) that if the sequence {a_n}_{n=1}^\infty{a
n

}
n=1


tends to limit LL as n → ∞n→∞, then for any fixed number M > 0M>0, the sequence {Ma_n}_{n=1}^\infty{Ma
n

}
n=1


tends to the limit MLML

ANSWER Let \epsilonϵ > 0 be given. By the assumption, we can find an NN such that

n \ge N \Rightarrow |a_n – L| \lt \epsilon / Mn≥N⇒∣a
n

−L∣<ϵ/M

Then,

n \ge N \Rightarrow |Ma_n – M L| = M. |a_n – L| < M.\epsilon/M = \epsilonn≥N⇒∣Ma
n

−ML∣=M.∣a
n

−L∣<M.ϵ/M=ϵ

which shows that {Ma_nMa
n

}{^\infty_{n=1}}
n=1


tends to the limit MLML

Enter answer here

Q9. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Given a collection A_n, n = 1, 2, . . .A
n

,n=1,2,… of intervals of the real line, their intersection is defined to be

\bigcap_{n=1}^\infty A_n⋂
n=1


A
n

= {x |(∀n)(x ∈ An)x∣(∀n)(x∈An)}

Give an example of a family of intervals A_n, n = 1, 2, . . .A
n

,n=1,2,…, such that A_{n+1} ⊂ A_nA
n+1

⊂A
n

for all nn and

\bigcap_{n=1}^\infty A_n = ∅⋂
n=1


A
n

=∅

Prove that your example has the stated property.

ANSWER Let A_n = (0, 1/n)A
n

=(0,1/n). Clearly, \bigcap_{n=1}^\infty A_n⋂
n=1


A
n

⊆ A1 = (0, 1)⊆A1=(0,1). Hence any element of the
intersection must be a member of (0, 1)(0,1). But if x ∈ (0, 1)x∈(0,1), we can find a natural number nn such
that 1/n < x1/n<x. Then x \notin A_nx∈
/

A
n

, so x \notinx∈
/

\bigcap_{n=1}^\infty A_n⋂
n=1


A
n

. Thus \bigcap_{n=1}^\infty A_n⋂
n=1


A
n

= \emptyset∅.

Enter answer here

Q10. Grade this answer according to the course rubric. Enter your grade (which should be a whole number between 0 and 24, inclusive) in the box below.

QUESTION Give an example of a family of intervals A_n, n = 1, 2, . . .A
n

,n=1,2,…, such that A_{n+1} ⊂ A_nA
n+1

⊂A
n

for all nn and \bigcap_{n=1}^\infty A_n⋂
n=1


A
n

consists of a single real number. Prove that your example has the stated property.

ANSWER Let A_nA
n

= [0, 1/nn). Clearly, 0 ∈ \bigcap_{n=1}^\infty A_n0∈⋂
n=1


A
n

. But the same argument as above shows that
no other number is in the intersection. Hence \bigcap_{n=1}^\infty A_n⋂
n=1


A
n

= {00}.

We will Update These Answers Soon.

More About This Course

Learn how to think the way mathematicians do – a powerful cognitive process developed over thousands of years.

Mathematical thinking is not the same as doing mathematics – at least not as mathematics is typically presented in our school system. School math typically focuses on learning procedures to solve highly stereotyped problems.

Professional mathematicians think a certain way to solve real problems, problems that can arise from the everyday world, from science, or from within mathematics itself. The key to success in school math is to learn to think inside the box. In contrast, a key feature of mathematical thinking is thinking outside the box – a valuable ability in today’s world. This course helps to develop that crucial way of thinking.

SKILLS YOU WILL GAIN

  • Number Theory
  • Real Analysis
  • Mathematical Logic
  • Language

Conclusion

Hopefully, this article will be useful for you to find all the Week, final assessment, and Peer Graded Assessment Answers of the Introduction to Mathematical Thinking Quiz of Coursera and grab some premium knowledge with less effort. If this article really helped you in any way then make sure to share it with your friends on social media and let them also know about this amazing training. You can also check out our other course Answers. So, be with us guys we will share a lot more free courses and their exam/quiz solutions also, and follow our Techno-RJ Blog for more updates.

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