TTMUA Lab
TMUA 2020 · Paper 2 · Question 8 of 20

TMUA 2020 Paper 2 Question 8

Proof and counterexample — Judging a disproof by counterexample. Try it first; the answer and a full worked solution are below.

TMUA 2020 · Paper 2Proof and counterexampleJudging a disproof by counterexample5 options
A student is asked to prove whether the following statement (×) is true or false:

(×) For all real numbers a and b, |a+b|<|a|+|b|

The student's proof is as follows:

Statement (×) is false. A counterexample is a= 3, b= 4, as |3+4|= 7 and |3|+|4|= 7, but 7 < 7 is false.

Which of the following best describes the student's proof?

  1. AThe statement (×) is true, and the student's proof is not correct.
  2. BThe statement (×) is false, but the student's proof is not correct: the counterexample is not valid.
  3. CThe statement (×) is false, but the student's proof is not correct: the student needs to give all the values of a and b where |a+b|<|a|+|b| is false.
  4. DThe statement (×) is false, but the student's proof is not correct: the student should have instead stated that for all real numbers a and b, |a+b||a|+|b|.
  5. EThe statement (×) is false, and the student's proof is fully correct.
Show the answer and worked solution
answer · E
  1. AThe statement (×) is true, and the student's proof is not correct.
  2. BThe statement (×) is false, but the student's proof is not correct: the counterexample is not valid.
  3. CThe statement (×) is false, but the student's proof is not correct: the student needs to give all the values of a and b where |a+b|<|a|+|b| is false.
  4. DThe statement (×) is false, but the student's proof is not correct: the student should have instead stated that for all real numbers a and b, |a+b||a|+|b|.
  5. EThe statement (×) is false, and the student's proof is fully correct.
To disprove a "for all" statement you need exactly one pair of values that makes it fail, and the student has one: with a= 3 and b= 4 both sides are 7, so the strict inequality |a+b|<|a|+|b| does not hold. That is all a disproof requires. Listing every failing pair is not needed, and stating the correct inequality |a+b||a|+|b| would be a different (true) claim rather than a repair of this proof. So the statement is false and the proof stands.