TMUA: Errors in purported proofs
Paper 2: identifying errors in purported proofs.
- Section 2 (Paper 2) · Err1-Err2 Identifying errors in proofs · spec Err1-Err2
- Paper 2 only
- 12 practice questions
- No calculator
What the specification covers
- Dividing by something that may be zero (ab = ac does not give b = c)
- sin A = sin B does not give A = B; squaring both sides creates extra roots
- Assuming the result, or checking only one direction of an 'if and only if'
Key ideas
- Dividing by a quantity that might be zero loses solutions (from \(ab=ac\) you cannot conclude \(b=c\)).
- Squaring both sides can introduce solutions that do not satisfy the original equation.
- From \(x^2=4\) you get \(x=\pm2\); taking only the positive root loses a solution.
- Multiplying an inequality by an expression of unknown sign is invalid; \(\sin A=\sin B\) does not give \(A=B\).
Common mistakes
- The first invalid step is the answer, even if later steps are also wrong.
- A correct final answer does not make every step valid.
Exam tip
Check each step: does it follow from the previous one for every value allowed?
Worked example
Worked example
A student solves \(x^3=4x\) as follows.
(1) \(x^3=4x\)
(2) Divide both sides by \(x\): \(x^2=4\)
(3) Take square roots: \(x=2\)
Which of the following is true?
- A solution is lost at step (2) only.
- A solution is lost at step (3) only.
- Solutions are lost at both step (2) and step (3).
- The working is correct and complete.
- An incorrect solution is introduced at step (3).
Answer: C: Solutions are lost at both step (2) and step (3).
The full solution set is \(\{-2,0,2\}\). Dividing by \(x\) assumes \(x\ne0\), losing \(x=0\) at step (2). From \(x^2=4\), \(x=\pm2\), so step (3) loses \(x=-2\).
Practice questions
Try each question before opening the solution. No calculator: the TMUA does not allow one. Every answer here was re-checked independently by computer before it was published.
Question 1
A student solves \(\dfrac{1}{x-1}<2\) by multiplying both sides by \(x-1\) to get \(1<2(x-1)\), and concludes that the solution is \(x>\tfrac32\). Which of the following is true?
- The answer misses some solutions and includes no wrong values
- The answer misses some solutions and includes some wrong values
- The answer includes some values that are not solutions
- The answer is correct
- There are no solutions
Show the answer and solution
Answer: A: The answer misses some solutions and includes no wrong values
Multiplying by \(x-1\) is only valid when \(x-1>0\). For \(x<1\), the left side is negative, so \(\tfrac1{x-1}<2\) holds too. The true solution is \(x<1\) or \(x>\tfrac32\). Every \(x>\tfrac32\) is a genuine solution, so the student's answer misses \(x<1\) and includes nothing wrong.
Question 2
Claim: if \(a\) and \(b\) are real numbers with \(a>b\), then \(a^2>b^2\).
'Proof': (1) Since \(a>b\), multiplying by \(a\) gives \(a^2>ab\). (2) Since \(a>b\), multiplying by \(b\) gives \(ab>b^2\). (3) So \(a^2>ab>b^2\).
Which of the following is true?
- The claim is true but step (1) is invalid.
- The claim is true and the proof is correct.
- The claim is false, and the first invalid step is (2).
- The claim is false, and the first invalid step is (1).
- The claim is false, and the first invalid step is (3).
Show the answer and solution
Answer: D: The claim is false, and the first invalid step is (1).
The claim is false: \(a=1\), \(b=-2\) gives \(1<4\). Step (1) multiplies an inequality by \(a\), valid only for \(a>0\): with \(a=-1\), \(b=-2\) it would say \(1>2\). So step (1) is the first invalid step (step (3) correctly combines (1) and (2)).
Question 3
A student claims that \(2^n>n^2\) for every positive integer \(n\), having checked \(n=1\) and \(n=5\). What is the smallest counterexample?
- \(n = 2\)
- \(n = 4\)
- There is no counterexample
- \(n = 6\)
- \(n = 3\)
Show the answer and solution
Answer: A: \(n = 2\)
\(n=2\): \(4>4\) is false, so \(n=2\) is a counterexample (\(n=3\): \(8<9\) and \(n=4\): \(16=16\) fail too; \(n=1\) works).
Question 4
A student solves an equation as follows.
(1) \(x^3=9x\)
(2) Divide both sides by \(x\): \(x^2=9\)
(3) Take square roots: \(x=3\)
Which of the following is true?
- An incorrect solution is introduced at step (3).
- A solution is lost at step (3) only.
- Solutions are lost at both step (2) and step (3).
- A solution is lost at step (2) only.
- The working is correct and complete.
Show the answer and solution
Answer: C: Solutions are lost at both step (2) and step (3).
The full solution set is \(\{-3,0,3\}\). Dividing by \(x\) loses \(x=0\) at step (2); from \(x^2=9\), \(x=\pm3\), so step (3) loses \(x=-3\).
Question 5
A student solves an equation as follows.
(1) \((x-3)(x+1)=3(x-3)\)
(2) Divide both sides by \((x-3)\): \(x+1=3\)
(3) So \(x=2\)
Which of the following is true?
- The working is correct and complete.
- A solution is lost at step (2) only.
- Solutions are lost at both step (2) and step (3).
- An incorrect solution is introduced at step (3).
- A solution is lost at step (3) only.
Show the answer and solution
Answer: B: A solution is lost at step (2) only.
The full solution set is \(\{3,2\}\). Dividing by \(x-3\) assumes \(x\ne3\), losing \(x=3\) at step (2); step (3) is correct.
Question 6
A student solves an equation as follows.
(1) \((x-4)(x+5)=3(x-4)\)
(2) Divide both sides by \((x-4)\): \(x+5=3\)
(3) So \(x=-2\)
Which of the following is true?
- A solution is lost at step (2) only.
- Solutions are lost at both step (2) and step (3).
- A solution is lost at step (3) only.
- The working is correct and complete.
- An incorrect solution is introduced at step (3).
Show the answer and solution
Answer: A: A solution is lost at step (2) only.
The full solution set is \(\{4,-2\}\). Dividing by \(x-4\) assumes \(x\ne4\), losing \(x=4\) at step (2); step (3) is correct.
Question 7
A student solves an equation as follows.
(1) \((x-1)(x+4)=2(x-1)\)
(2) Divide both sides by \((x-1)\): \(x+4=2\)
(3) So \(x=-2\)
Which of the following is true?
- A solution is lost at step (2) only.
- Solutions are lost at both step (2) and step (3).
- An incorrect solution is introduced at step (3).
- The working is correct and complete.
- A solution is lost at step (3) only.
Show the answer and solution
Answer: A: A solution is lost at step (2) only.
The full solution set is \(\{1,-2\}\). Dividing by \(x-1\) assumes \(x\ne1\), losing \(x=1\) at step (2); step (3) is correct.
Question 8
A student solves an equation as follows.
(1) \(x^3=16x\)
(2) Divide both sides by \(x\): \(x^2=16\)
(3) Take square roots: \(x=4\)
Which of the following is true?
- An incorrect solution is introduced at step (3).
- The working is correct and complete.
- A solution is lost at step (3) only.
- Solutions are lost at both step (2) and step (3).
- A solution is lost at step (2) only.
Show the answer and solution
Answer: D: Solutions are lost at both step (2) and step (3).
The full solution set is \(\{-4,0,4\}\). Dividing by \(x\) loses \(x=0\) at step (2); from \(x^2=16\), \(x=\pm4\), so step (3) loses \(x=-4\).
Question 9
A student solves an equation as follows.
(1) \(x^3=36x\)
(2) Divide both sides by \(x\): \(x^2=36\)
(3) Take square roots: \(x=6\)
Which of the following is true?
- A solution is lost at step (2) only.
- The working is correct and complete.
- A solution is lost at step (3) only.
- An incorrect solution is introduced at step (3).
- Solutions are lost at both step (2) and step (3).
Show the answer and solution
Answer: E: Solutions are lost at both step (2) and step (3).
The full solution set is \(\{-6,0,6\}\). Dividing by \(x\) loses \(x=0\) at step (2); from \(x^2=36\), \(x=\pm6\), so step (3) loses \(x=-6\).
Question 10
A student solves an equation as follows.
(1) \((x-2)(x+3)=6(x-2)\)
(2) Divide both sides by \((x-2)\): \(x+3=6\)
(3) So \(x=3\)
Which of the following is true?
- Solutions are lost at both step (2) and step (3).
- The working is correct and complete.
- A solution is lost at step (2) only.
- A solution is lost at step (3) only.
- An incorrect solution is introduced at step (3).
Show the answer and solution
Answer: C: A solution is lost at step (2) only.
The full solution set is \(\{2,3\}\). Dividing by \(x-2\) assumes \(x\ne2\), losing \(x=2\) at step (2); step (3) is correct.
Question 11
A student solves an equation as follows.
(1) \((x-2)(x+3)=3(x-2)\)
(2) Divide both sides by \((x-2)\): \(x+3=3\)
(3) So \(x=0\)
Which of the following is true?
- An incorrect solution is introduced at step (3).
- The working is correct and complete.
- Solutions are lost at both step (2) and step (3).
- A solution is lost at step (3) only.
- A solution is lost at step (2) only.
Show the answer and solution
Answer: E: A solution is lost at step (2) only.
The full solution set is \(\{2,0\}\). Dividing by \(x-2\) assumes \(x\ne2\), losing \(x=2\) at step (2); step (3) is correct.
Keep going
- Previous topic: Deductions, conjectures and ordering proofs
- Next topic: Indices and surds
- All TMUA topics · Timed TMUA paper simulator · Official TMUA past papers and specimen papers · TMUA Paper 2: logic and proof
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