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