Marking time

Proofs Workshop

The following question appeared in the 1990 M203 exam: Prove that 3n2 ≥ (n + 6) 2 for n ≥ 9. a)

Rephrase the question in formal logic terms

b)

Spend a few minutes thinking about how you would approach the proof.

Below we show attempts by 3 students to prove the result. How many marks (out of 10) would you give each of them? Justify your marking, and show what’s needed to get full marks.

Student 1:

Assume 3n 2 ≥ (n + 6) 2 Then ⇒ ⇒ ⇒ ⇒

3n 2 ≥ (n 2 + 12n + 36) 2n 2 − 6n − 36 ≥ 0 n 2 − 6n − 18 ≥ 0 (n − 3) 2 − 9 − 18 ≥ 0 (n − 3) 2 ≥ 27

Now 62 ≥ 27, but 52 is not, so the smallest possible value of n − 3 is 6, and n ≥ 9.

Student 2:

Suppose 3n 2 < (n + 6) 2 Then ⇒ ⇒ ⇒ ⇒

3 n < ( n + 6) n+6 6 3< = 1+ n n 6 3 −1 < n n 1 < 6 3 −1 6 n< < 9 since 3 −1

6 3 −1

≈ 8.196

We have shown that if 3n2 < (n + 6)2, then n < 9. So 3n2 ≥ (n + 6)2 for n ≥ 9.

Student 3: We define the function f (n) = 3n2 − (n + 6)2. Then the question becomes: Prove that f (n) ≥ 0 for n ≥ 9.

3n2 (n + 6)2

We have f (9) = 3 × 92 − 152 = 243 – 225 = 18 f ' (n) = 6n − 2(n + 6) = 4n − 12, so f ' (n) > 0 for n > 3, and f is increasing on (3, ∞). Since f (9) > 0 and increasing, we must have f (n) ≥ 0 for n ≥ 9.

Shirleen Stibbe

http://www.shirleenstibbe.co.uk

Marking time – Tutors' view

Proofs Workshop

Tutor's comments

The following question appeared in the 1990 M203 exam: 2

Prove that 3n ≥ (n + 6) a)

2

Question doesn't say whether n is real or integer Mark: 8/10 for examiner.

for n ≥ 9.

Prove that if n ≥ 9 then 3n2 ≥ (n + 6)2, or Prove: ( n ≥ 9) ⇒ (3n2 ≥ (n + 6)2 )

Rephrase the question in formal logic terms:

Below we show attempts by 3 students to prove the result.

Marking plan: 5 for method, 5 for accuracy.

How many marks (out of 10) would you give each of them? Justify your marking, and show what’s needed to get full marks. Student 1: 2

Assume 3n ≥ (n + 6) Then ⇒ ⇒ ⇒ ⇒

Method: Direct (but wrong way round) (0/5 method) Assumes n an integer Conclusion is wrong: if (n – 3)2 ≥ 27, then |n − 3| ≥ √27 so n ≥ 9 or n ≤ −3. Algebra is wrong: should be −12n in line 3. Slip (back on the right track in line 4) so no extra marks deducted. (2/5 accuracy) Total: 2/10 How to fix it: Can’t use double arrows since the converse is not true. So: Assume n ≥ 9, where n ∈ R Then (n − 3)2 ≥ 27 …..and go backwards.

2

3n 2 ≥ (n 2 + 12n + 36) 2n 2 − 6n − 36 ≥ 0 n 2 − 6n − 18 ≥ 0 (n − 3) 2 − 9 − 18 ≥ 0 (n − 3) 2 ≥ 27

Now 62 ≥ 27, but 52 is not, so the smallest possible value of n − 3 is 6, and n ≥ 9.

Student 2: Method: Contrapositive

Suppose 3n 2 < (n + 6) 2 Then ⇒ ⇒ ⇒ ⇒

3 n < ( n + 6) n+6 6 3< = 1+ n n 6 3 −1 < n n 1 < 6 3 −1 6 n< < 9 since 3 −1 2

6 3 −1

(5/5 method)

Needs n positive to get from the second to the third line. Didn’t state that it’s a contrapositive proof (3 ½ /5 accuracy) Total: 8 ½ /10 How to fix it: Could use modulus throughout to fix the positive n problem, i.e. √3|n| < |n + 6|, but the arithmetic gets messy.

≈ 8.196

Easier approach is to state at the start: I shall prove that if 3n2 < (n + 6)2, then n < 9 (contrapositive proof ). We may assume without loss of generality that n > 0, since if n ≤ 0, then n < 9.

2

We have shown that if 3n < (n + 6) , then n < 9. So 3n2 ≥ (n + 6)2 for n ≥ 9. Student 3: We define the function f (n) = 3n2 − (n + 6)2 Then the question becomes: Prove that f (n) ≥ 0 for n ≥ 9. 2

Method: Graphical 3n2

(n + 6)

2

(5/5 method)

Assumes the graph is differentiable, which needs n to be real (to make it continuous). (4/5 accuracy) Brownie point for drawing the picture

2

We have f (9) = 3 × 9 − 15 = 243 – 225 = 18 f ' (n) = 6n − 2(n + 6) = 4n − 12, so f ' (n) > 0 for n > 3, and f is increasing on (3, ∞).

+½ Total: 9 ½/10

How to fix it: Start by saying Define the function f (n) = 3n2 − (n + 6)2, n ∈ R which is differentiable everywhere (by the combination rules).

Since f (9) > 0 and increasing, we must have f (n) ≥ 0 for n ≥ 9.

Shirleen Stibbe

http://www.shirleenstibbe.co.uk