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Direct proof and proof by cases

1.[1p]

How should a direct proof of "for every integer n, if n is odd then n2 is odd" begin?

Correct

The answer is: Let n be an odd integer.

The answer is: Let n be an odd integer.

The answer is: Let n be an odd integer.

2.[2p]

Which of these statements are true?

Select all that apply

Correct

The answer is: 3∣12, −4∣12, 7∣0, 1∣13

Correct
Correct

The answer is: 3∣12, −4∣12, 7∣0, 1∣13

Correct

3.[2p]

Put the lines of this proof that if a∣b and b∣c then a∣c in order.

  1. By definition, b=ak and c=bm for some integers k and m.

  2. Let a, b and c be integers with a∣b and b∣c.

  3. Substituting, c=(ak)m=a(km).

  4. Since k and m are integers, so is km.

  5. So c is a times an integer, and by definition a∣c.

Show the answer

a, b, c, d, e

4.[1p]

A proof of |x+y|≤|x|+|y| that treats only the case x≥0,y≥0 and the case x<0,y<0 is complete.

The answer is: False

Correct

5.[2p]

The proof that every odd square is one more than a multiple of 8 writes n=2k+1 and gets n2=4(k2+k)+1. What finishes it?

Correct

The answer is: The earlier result that k2+k is even for every integer k

The answer is: The earlier result that k2+k is even for every integer k

The answer is: The earlier result that k2+k is even for every integer k

6.[3p]

Match each claim to the method its proof in the lesson used.

  • If a∣b and b∣c then a∣c

  • n2+n is even for every integer n

  • |x+y|≤|x|+|y| for all reals

  • For every k there are k consecutive non-primes

  • Some irrational a and b have ab rational

  • cases on signs

  • a non-constructive existence proof

  • a constructive existence proof

  • a direct proof from the definition

  • cases on parity

Show the answer

If a∣b and b∣c then a∣c: a direct proof from the definition n2+n is even for every integer n: cases on parity |x+y|≤|x|+|y| for all reals: cases on signs For every k there are k consecutive non-primes: a constructive existence proof Some irrational a and b have ab rational: a non-constructive existence proof

7.[3p]

The proof that some irrational a and b give ab rational considers 22. Which of these are true of it?

Select all that apply

Correct
Correct

The answer is: It shows that one of two pairs works without saying which, It uses the fact that every real number is rational or irrational, It relies on 2 being irrational

The answer is: It shows that one of two pairs works without saying which, It uses the fact that every real number is rational or irrational, It relies on 2 being irrational

Correct

8.[2p]

The constructive proof takes N=(k+1)! and the numbers N+2 to N+(k+1). For k=5, which run of non-primes does it produce?

Correct

The answer is: 722 to 726

The answer is: 722 to 726

The answer is: 722 to 726

9.[2p]

Why does a direct proof of "if n2 is even then n is even" stall?

Correct

The answer is: The hypothesis n2=2k gives n=±2k, which is not visibly twice an integer

The answer is: The hypothesis n2=2k gives n=±2k, which is not visibly twice an integer

The answer is: The hypothesis n2=2k gives n=±2k, which is not visibly twice an integer

10.[1p]

What is the remainder given by the division algorithm when −23 is divided by 7?

CorrectNot quite: 5

11.[2p]

In the uniqueness proof of the division algorithm, why does 0≤b(q−q′)<b, with b>0, force q=q′?

Correct

The answer is: The only integer m with 0≤bm<b is m=0

The answer is: The only integer m with 0≤bm<b is m=0

The answer is: The only integer m with 0≤bm<b is m=0

12.[2p]

A student proves x2+y2≥2xy by writing: "x2+y2≥2xy, so x2−2xy+y2≥0, so (x−y)2≥0, which is true." What is wrong?

Correct

The answer is: It reasons from the claim to a true statement, which proves nothing about the claim

The answer is: It reasons from the claim to a true statement, which proves nothing about the claim

The answer is: It reasons from the claim to a true statement, which proves nothing about the claim

13.[3p]

Which of these pairs of proofs together establish P⟺Q?

Select all that apply

Correct
Correct
Correct

The answer is: A proof of P⇒Q and a proof of Q⇒P, A proof of P⇒Q and a proof of ¬P⇒¬Q, A proof of Q⇒P and a proof of ¬Q⇒¬P

The answer is: A proof of P⇒Q and a proof of Q⇒P, A proof of P⇒Q and a proof of ¬P⇒¬Q, A proof of Q⇒P and a proof of ¬Q⇒¬P

14.[1p]

"For all real numbers x and y, y−x≥0" can be proved by saying "without loss of generality x≤y".

The answer is: False

Correct