Sometimes tested Paper 2 (Mathematical Reasoning)

Identifying Errors in Proofs: Finding the Invalid Step (TMUA Paper 2 Reasoning)

A purported proof can look convincing yet fail at a single step. TMUA Paper 2 hands you an offered argument and asks you to pinpoint where the reasoning breaks: dividing by a possible zero, dropping a ± case, or confusing a statement with its converse.

Part of the TMUA syllabus — revision for the Test of Mathematics for University Admission (TMUA), the UAT-UK maths test used by Cambridge, Oxford, Imperial, UCL, LSE, Warwick and Durham. No calculator; multiple choice.

Key points

  • Your job is to find the FIRST invalid step, not to judge whether the final conclusion is true. A correct-looking conclusion reached by broken reasoning still makes the proof flawed.
  • Read each line as a claimed implication: does this line genuinely follow from the previous one for EVERY allowed value? A single counterexample to one step kills that step.
  • Dividing both sides by an expression that could be zero is invalid. Before cancelling any factor, ask 'could this be zero?'
    if ab = ac then b = c fails when a = 0
  • Taking square roots drops the negative case: x2 = 9 gives x = ± 3, not just x = 3. Losing the ± discards genuine solutions.
  • A statement and its converse are different claims: 'if it rains then the ground is wet' does NOT give 'if the ground is wet then it rained'. Reusing the converse as if proved is a classic flaw.
  • Assuming the conclusion (circular reasoning) means a step quietly uses the very thing you are meant to prove, which makes the whole argument worthless.
  • Many-to-one functions break naive cancelling: sin A = sin B does NOT give A = B. The same trap applies to squaring, modulus and cosine.
    e.g. sin 30 = sin 150, in degrees
  • In multiple choice, test each candidate line with a quick numerical counterexample; the step that breaks under even one legal value is your answer.

Definitions

Purported (offered) proof
An argument presented as a proof. TMUA does not promise it is valid, so you must check whether each line truly follows from the ones before it.
Counterexample
One specific case that satisfies a step's assumptions but not its claimed conclusion, which proves that step is not a valid deduction.
Converse
The statement made by swapping the 'if' and 'then' parts. Its truth is independent of the original, so it can never be assumed for free.
Invalid step
A line that does not follow from earlier lines for all permitted values, even when the final conclusion happens to be true.

Worked examples

1

A student offers this 'proof' that 2 = 1. Assume a = b, where a and b are equal and nonzero. Multiply by a, then subtract b2, then factorise, then divide by (a - b), reaching a + b = b, and since a = b this gives 2b = b so 2 = 1. Find the single step that is not a valid deduction.

  1. 1

    Multiply the given equation by a:

    a2 = ab
  2. 2

    Subtract b2 from both sides:

    a2 - b2 = ab - b2
  3. 3

    Factorise each side (a true identity):

    (a-b)(a+b) = b(a-b)
  4. 4

    With a and b equal, the shared factor is zero:

    a - b = 0
  5. 5

    Dividing by (a - b) divides by zero, so this step is invalid:

    a + b = b

Answer: The invalid step is dividing both sides by (a - b). Since a = b we have a - b = 0, so that division is illegal; every earlier line is fine.

2

An offered proof claims that if x2 = y2 then x = y, justified by 'taking the square root of both sides'. Show by a counterexample that the step is invalid, and state the correct conclusion.

  1. 1

    A square root returns a size, not a sign:

    √(x2) = |x|
  2. 2

    Applied to both sides, |x| = |y| unpacks into two cases:

    x = y or x = -y
  3. 3

    Try a counterexample:

    x = 3, y = -3
  4. 4

    Both squares match but the values differ:

    32 = 9 = (-3)2, yet x ≠ y
  5. 5

    The invalid move was keeping only the positive root; correct conclusion:

    x = ± y

Answer: Invalid: taking square roots kept only the positive case. A counterexample is x = 3, y = -3, and the correct deduction is x = y or x = -y.

Common mistakes

  • ×Judging the conclusion instead of the steps: a true conclusion can still be reached by an invalid argument, and that argument is still flawed.
  • ×Cancelling a common factor from both sides without first asking whether that factor could be zero.
  • ×Writing √(x2) = x instead of |x|, so the negative solution silently disappears.
  • ×Treating a statement as if its converse were automatically proved as well.
  • ×Stopping at the first line that merely looks unusual; the invalid step is the one that fails a counterexample, not the one that looks odd.

No-calculator tips

  • ✓Test any suspect step with small numbers (0, 1, -1, 2); one counterexample proves a step invalid instantly.
  • ✓Scan for the 'danger moves': a division, a square root, a cancellation, or an 'if⋯then' being used backwards.
  • ✓Zero is the usual trap - whenever a factor is cancelled, substitute the value that makes it zero.
  • ✓Work line by line and stop at the FIRST broken deduction; the later lines no longer matter once one step fails.
  • ✓Eliminate answer options by confirming the earlier lines are genuinely valid, narrowing down to the one broken step.

Test yourself

Original practice questions, no calculator. Work each out before revealing the answer.

Q1.A student presents this 'proof' that 2 = 1. Start from a = b, where a and b are nonzero. Step 2: a2 = ab. Step 3: a2 - b2 = ab - b2. Step 4: (a - b)(a + b) = b(a - b). Step 5: a + b = b. Step 6: 2b = b, so 2 = 1. At which step does the argument FIRST become invalid?

  • A. Step 2
  • B. Step 3
  • C. Step 4
  • D. Step 5
  • E. Step 6
Show answer

Answer: D — Step 5

Going from Step 4 to Step 5 divides both sides by (a - b), but a = b makes a - b = 0, so this is division by zero. Every step up to and including Step 4 is valid; the false conclusion in Step 6 is only a symptom.

Q2.To solve (x - 5)2 = 9, a student takes the square root of both sides to get x - 5 = 3, and concludes that x = 8 is the only solution. Which statement correctly identifies the flaw?

  • A. No flaw; x = 8 is the only solution
  • B. The square root gives x - 5 = ± 3, so x = 8 or x = 2; the negative root was dropped
  • C. It should give x - 5 = ± 9, so x = 14 or x = -4
  • D. √(9) is not a real number, so the equation has no solutions
  • E. x - 5 = 3 gives x = 2, so the answer is wrong by an arithmetic slip
Show answer

Answer: B — The square root gives x - 5 = ± 3, so x = 8 or x = 2; the negative root was dropped

If u2 = 9 then u = ± 3, so x - 5 = 3 or x - 5 = -3, giving x = 8 or x = 2; keeping only the positive root loses the solution x = 2. Option 2 wrongly forgets to square-root the 9, and option 4 mis-solves x - 5 = 3.

Q3.For a, b ≥ 0 a student 'proves' a + b ≥ 2*√(ab) as follows. Step 1: Assume a + b ≥ 2*√(ab). Step 2: Then (a + b)2 ≥ 4ab. Step 3: So a2 + 2ab + b2 ≥ 4ab. Step 4: So a2 - 2ab + b2 ≥ 0. Step 5: So (a - b)2 ≥ 0, which is true. Step 6: Therefore a + b ≥ 2*√(ab). Which statement best describes the logical error?

  • A. There is no error; assuming the statement and reaching a true conclusion proves it
  • B. Step 2 is invalid because you cannot square both sides of an inequality
  • C. Step 5 is false because (a - b)2 can be negative for some real a and b
  • D. The proof assumes the very thing to be proved and reasons forward to a known truth; it is only valid if every step is reversible
  • E. The proof fails because it divides by zero when forming √(ab)
Show answer

Answer: D — The proof assumes the very thing to be proved and reasons forward to a known truth; it is only valid if every step is reversible

The argument starts by assuming the conclusion and derives the true statement (a - b)2 ≥ 0, which only shows 'conclusion implies something true' - not the reverse. It becomes a valid proof only when the reversible steps are run in the opposite order, starting from (a - b)2 ≥ 0.

Read this topic in the official UAT-UK TMUA content specification →

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