Identifying errors in proofs

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1Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student attempts to solve the equation \(\sqrt{x + 3} = x - 3\) for real \(x\).

ISquaring both sides: \(x + 3 = (x - 3)^2 = x^2 - 6x + 9\).
IISo \(x^2 - 7x + 6 = 0\).
IIISo \((x - 1)(x - 6) = 0\).
IVHence the solutions are \(x = 1\) and \(x = 6\).

Which of the following best describes the solution?

AThe solution is wrong, and the first error occurs on line IV.
BThe solution is completely correct.
CThe solution is wrong, and the first error occurs on line I.
DThe solution is wrong, and the first error occurs on line II.
EThe solution is wrong, and the first error occurs on line III.

Worked solution

Squaring can introduce extraneous solutions, so line IV should check them: \(x = 1\) gives \(\sqrt4 = 2 \neq -2\). Only \(x = 6\) works. Lines I-III are valid deductions.

Full method — how and why

Squaring an equation is not reversible: it can create solutions that do not satisfy the original. The error is the missing check, not the algebra.

Check line I: squaring both sides of \(\sqrt{x+3} = x-3\) gives \(x + 3 = x^2 - 6x + 9\). Valid.
\(A = B \Rightarrow A^2 = B^2\) is always a legitimate deduction - it is the converse that fails.
Check lines II and III: \(x^2 - 7x + 6 = 0\) factorises as \((x-1)(x-6) = 0\). Valid.
\(-1 \times -6 = 6\) and \(-1 - 6 = -7\) ✓.
Line IV asserts both are solutions. Substitute \(x = 1\) into the original: \(\sqrt4 = 2\), but \(x - 3 = -2\). \(2 \neq -2\), so \(x=1\) is extraneous.
A square root is non-negative, so the right-hand side must be \(\geq 0\), forcing \(x \geq 3\).
Substitute \(x = 6\): \(\sqrt9 = 3\) and \(6-3 = 3\) ✓. So only \(x = 6\) is a solution, and the first error is line IV.
Lines I-III produce candidates; the check is a required final step.

Where it goes wrong

Blaming line I - squaring is a valid implication, just not an equivalence.
Declaring the solution correct because the factorising is right; correct algebra can still give a wrong answer set.
Checking the candidates in the squared equation rather than the original - both pass there, which proves nothing.

Takeaway After squaring, always substitute back into the original equation; squaring can only add solutions, never lose them.

Answer A · Identifying errors in proofs (Err2) · question ERR-01

2Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

Here is a purported proof that \(1 = 2\).

ILet \(a = b\), with \(a \neq 0\).
IIThen \(a^2 = ab\).
IIISo \(a^2 - b^2 = ab - b^2\), i.e. \((a - b)(a + b) = b(a - b)\).
IVDividing both sides by \((a - b)\): \(a + b = b\).
VSo \(2b = b\), and hence \(2 = 1\).

On which line does the first error occur?

AIV
BII
CIII
DV
EI

Worked solution

Since \(a = b\), \(a - b = 0\) and dividing by zero is invalid.

Full method — how and why

The classic '\(1=2\)' fallacy. Somewhere a step is division by an expression that is secretly zero.

Line I sets \(a = b\) with \(a \neq 0\) - a legitimate supposition.
Nothing wrong yet.
Line II: multiplying \(a = b\) by \(a\) gives \(a^2 = ab\). Valid.
Multiplying both sides by the same quantity is always allowed.
Line III: subtracting \(b^2\) from both sides and factorising both sides. Valid.
\(a^2 - b^2 = (a-b)(a+b)\) and \(ab - b^2 = b(a-b)\) - correct algebra.
Line IV divides by \((a-b)\). But \(a = b\), so \(a - b = 0\): this is division by zero. First error.
Dividing by zero is undefined, and it is exactly what manufactures the false conclusion.

Where it goes wrong

Blaming line V, where the absurdity finally appears; the last line is where you notice the error, not where it happens.
Blaming line III - the factorisation is perfectly correct.
Being reassured by the stated condition \(a \neq 0\); it is \(a - b\), not \(a\), that vanishes.

Takeaway In every fake proof of this type, look for a division by a factor that the hypotheses force to be zero.

Answer A · Identifying errors in proofs (Err2) · question ERR-02

3Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student tries to solve \(\sin\theta = \sin 40^\circ\) for \(0^\circ \leq \theta < 360^\circ\).

ISince \(\sin\theta = \sin 40^\circ\), we have \(\theta = 40^\circ\).
IIThe sine function has period \(360^\circ\), so \(\theta = 40^\circ + 360^\circ k\).
IIIIn the given interval the only solution is \(\theta = 40^\circ\).

Which of the following is correct?

AThe first error is on line I: \(\sin\theta = \sin\alpha\) does not imply \(\theta = \alpha\); \(\theta = 140^\circ\) is also a solution.
BThe solution is completely correct.
CThe first error is on line II: the period of sine is \(180^\circ\).
DThe first error is on line III: \(\theta = 400^\circ\) should be included.
EThe first error is on line II: it should be \(\theta = 40^\circ + 180^\circ k\).

Worked solution

'If \(\sin\alpha = \sin\beta\) then \(\alpha = \beta\)' is a classic invalid deduction; \(\sin(180^\circ - 40^\circ) = \sin40^\circ\) too.

Full method — how and why

\(\sin\) is not one-to-one, so \(\sin\theta = \sin\alpha\) has more solutions than \(\theta = \alpha\).

Line I claims \(\sin\theta = \sin 40^\circ \Rightarrow \theta = 40^\circ\). This is the error.
Sine takes the same value twice per revolution: \(\sin(180^\circ - \alpha) = \sin\alpha\).
So \(\theta = 140^\circ\) also satisfies the equation.
\(\sin 140^\circ = \sin 40^\circ\) - check on the graph or with the symmetry of the unit circle.
Lines II and III are then correct statements in themselves (sine does have period \(360^\circ\)), but they are built on the incomplete line I.
The question asks for the first error, which is line I.
The full solution set in \(0^\circ \leq \theta < 360^\circ\) is \(\theta = 40^\circ\) and \(\theta = 140^\circ\).
Only two, because the sine is positive in the first and second quadrants.

Where it goes wrong

Claiming the period of sine is \(180^\circ\) - that is tangent.
Claiming \(\theta = 400^\circ\) was missed - it is outside the stated interval.
Declaring the work correct because each individual line looks plausible.

Takeaway \(\sin\theta = \sin\alpha\) gives \(\theta = \alpha + 360^\circ k\) or \(\theta = 180^\circ - \alpha + 360^\circ k\). Never cancel a sine.

Answer A · Identifying errors in proofs (Err2) · question ERR-03

4Identifying errors in proofs · spec Err1 · Paper 2 style · ★☆☆

Claim: for all real \(x\), if \(x^2 > 9\) then \(x > 3\).

ISuppose \(x^2 > 9\).
IITaking square roots of both sides, \(x > 3\).
IIIThis proves the claim.

Which of the following is true?

AThe claim is false; the first error is on line II, since \(x = -4\) satisfies \(x^2 > 9\) but not \(x > 3\).
BThe claim and proof are both correct.
CThe claim is true but line II should say \(|x| > 3\) and then deduce \(x > 3\).
DThe claim is false; the first error is on line I.
EThe claim is true; the first error is on line III.

Worked solution

\(x^2 > 9\) gives \(|x| > 3\), i.e. \(x > 3\) or \(x < -3\). The claim itself is false, and line II is where the invalid deduction happens.

Full method — how and why

Taking square roots of an inequality gives a modulus, not a bare inequality. Here the claim itself is false.

Test the claim: \(x = -4\) gives \(x^2 = 16 > 9\), but \(x = -4 \not> 3\). The claim is false.
Settle the truth of the claim first; that tells you an error must exist somewhere.
Line I is just the supposition - no error.
You are allowed to suppose the hypothesis.
Line II says 'taking square roots, \(x > 3\)'. What \(x^2 > 9\) actually gives is \(|x| > 3\), i.e. \(x>3\) or \(x<-3\). First error.
\(\sqrt{x^2} = |x|\), not \(x\).
The offered repair ('say \(|x|>3\) and then deduce \(x>3\)') is no repair at all, since \(|x|>3\) does not give \(x>3\).
That option admits the modulus but then makes the same invalid step.

Where it goes wrong

Accepting the option that says the claim is true - it is not.
Blaming line III; by then the damage is done.
Solving \(x^2 > 9\) as \(-3 < x < 3\), which is the solution of \(x^2 < 9\).

Takeaway \(x^2 > k^2 \iff |x| > k \iff x>k\) or \(x<-k\). Squaring and rooting both lose sign information.

Answer A · Identifying errors in proofs (Err1) · question ERR-04

5Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student proves that the equation \(x^2 + 2x + 3 = 0\) has no real solutions.

IComplete the square: \(x^2 + 2x + 3 = (x + 1)^2 + 2\).
II\((x + 1)^2 \geq 0\) for all real \(x\).
IIISo \((x + 1)^2 + 2 \geq 2 > 0\) for all real \(x\).
IVHence the equation has no real solutions.

Which of the following is correct?

AThe proof is completely correct.
BThe first error is on line I.
CThe first error is on line II: \((x + 1)^2\) can be negative.
DThe first error is on line III.
EThe first error is on line IV: a positive expression can still equal zero.

Worked solution

Every step is valid.

Full method — how and why

Another valid proof. Completing the square and bounding below is a standard, correct technique.

Line I: \((x+1)^2 = x^2+2x+1\), so \(x^2+2x+3 = (x+1)^2 + 2\) ✓.
Half the coefficient of \(x\) is \(1\); correct by \(3 - 1 = 2\).
Line II: any real square is \(\geq 0\) ✓.
True for every real \(x\).
Line III: adding \(2\) gives \(\geq 2\), and \(2 > 0\) ✓.
Adding a constant to both sides of an inequality is always valid.
Line IV: an expression that is always at least \(2\) can never equal \(0\), so there are no real roots ✓.
That is the definition of having no real solutions. (Check: discriminant \(4 - 12 = -8 < 0\) ✓.)

Where it goes wrong

Claiming \((x+1)^2\) can be negative - not for real \(x\).
Claiming 'a positive expression can still equal zero' - it cannot; that is what positive means.
Assuming there must be an error.

Takeaway Completing the square gives a rigorous lower bound; if the minimum is positive, there are no real roots.

Answer A · Identifying errors in proofs (Err2) · question ERR-09

6Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

Claim: if \(a > b\) then \(\dfrac{1}{a} < \dfrac{1}{b}\).

ISuppose \(a > b\).
IIDividing both sides by \(ab\): \(\dfrac{1}{b} > \dfrac{1}{a}\).
IIISo \(\dfrac1a < \dfrac1b\).

Which of the following is correct?

AThe claim is false in general and the first error is on line II, since \(ab\) may be negative or zero.
BThe claim and proof are correct.
CThe first error is on line III.
DThe claim is true but the first error is on line I.
EThe claim is false and the first error is on line III.

Worked solution

\(a = 1\), \(b = -1\): \(a > b\) but \(\frac1a = 1 > -1 = \frac1b\). Dividing by \(ab\) without knowing its sign is invalid (and \(b = 0\) would be division by zero).

Full method — how and why

The same unknown-sign trap as ERR-05, now with division, plus a false claim.

Test the claim: \(a = 1\), \(b = -1\). Then \(a > b\) but \(\frac1a = 1\) and \(\frac1b = -1\), so \(\frac1a > \frac1b\). Claim false.
Opposite signs are the natural test for a reciprocal claim.
Line I is just the supposition - fine.
No deduction yet.
Line II divides by \(ab\) and keeps the direction. That needs \(ab > 0\), which is not given; \(ab\) could be negative (as in the counterexample) or zero. First error.
Dividing by a negative reverses the inequality, and dividing by zero is undefined.
The claim is true if \(a\) and \(b\) have the same sign - which is the condition that was silently assumed.
Naming the missing hypothesis is the cleanest way to describe this kind of error.

Where it goes wrong

Accepting the argument because it 'works for positive numbers' - the claim is stated for all reals.
Blaming line III, which is a trivial restatement of line II.

Takeaway Reciprocals reverse order only within a fixed sign: for \(a,b>0\) (or both negative), \(a>b \Rightarrow \frac1a<\frac1b\).

Answer A · Identifying errors in proofs (Err2) · question ERR-10

7Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student solves \(x^2 = 4x\).

IDivide both sides by \(x\): \(x = 4\).
IISo the only solution is \(x = 4\).

Which of the following is correct?

AThe first error is on line I: dividing by \(x\) discards the solution \(x = 0\).
BThe solution is completely correct.
CThe first error is on line II.
DThe first error is on line I: \(x^2 \div x = 2x\).
EThe first error is on line I: it should give \(x = 2\).

Worked solution

Factorise instead: \(x(x - 4) = 0\) gives \(x = 0\) or \(x = 4\).

Full method — how and why

Dividing an equation by a variable destroys the root where that variable is zero.

Line I divides \(x^2 = 4x\) by \(x\). That is only legitimate when \(x \neq 0\) - and \(x = 0\) does satisfy the equation. First error.
\(0^2 = 4 \times 0\) ✓, so \(x=0\) is a genuine solution that the division removes.
The correct method: bring everything to one side and factorise. \(x^2 - 4x = 0\), so \(x(x-4) = 0\).
Factorising preserves every root, because a product is zero exactly when a factor is.
So \(x = 0\) or \(x = 4\) - two solutions.
Line II's 'only solution' is wrong as a consequence of line I.

Where it goes wrong

Blaming line II - it faithfully reports what line I produced; the first error is the division.
Thinking \(x^2 \div x = 2x\); it is \(x\).
Trying to 'fix' it by remembering to add \(x = 0\) afterwards - correct in this case, but factorising is the reliable habit.

Takeaway Never divide an equation by a variable expression. Factorise instead: division loses roots, squaring adds them.

Answer A · Identifying errors in proofs (Err2) · question ERR-12

8Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

Claim: if \(x\) and \(y\) are real with \(x^2 = y^2\), then \(x = y\).

Which of the following is a counterexample to the claim?

A\(x = -1\), \(y = -1\)
B\(x = 1\), \(y = 2\)
C\(x = 2\), \(y = -2\)
D\(x = 0\), \(y = 0\)
E\(x = 2\), \(y = 2\)

Worked solution

A counterexample must satisfy the hypothesis (\(x^2 = y^2\)) and fail the conclusion (\(x \neq y\)).

Full method — how and why

A counterexample must satisfy the hypothesis and break the conclusion. Test both halves on each candidate.

Hypothesis: \(x^2 = y^2\). Conclusion to break: need \(x \neq y\).
Both conditions, or it is not a counterexample.
\(x=2, y=-2\): \(4 = 4\) ✓ hypothesis, and \(2 \neq -2\) ✓ conclusion fails. Counterexample.
Equal squares with opposite signs is exactly the gap in the claim.
\(x=2,y=2\) and \(x=0,y=0\) and \(x=-1,y=-1\): hypothesis holds but so does the conclusion - these are examples of the claim.
Confirming instances never refute.
\(x=1,y=2\): \(1 \neq 4\), so the hypothesis fails and the case is irrelevant.
A case outside the hypothesis says nothing.

Where it goes wrong

Picking a pair where the hypothesis fails and thinking the different values refute the claim.
Picking a pair where both hold, which supports rather than refutes.

Takeaway \(x^2=y^2 \Rightarrow x = \pm y\). A counterexample satisfies the 'if' and fails the 'then'.

Answer C · Identifying errors in proofs (Err2) · question ERR-14

9Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student argues: 'Every prime number is odd, because if it were even it would be divisible by \(2\) and so not prime.' Which of the following is correct?

AThe argument is wrong because \(2\) is even and prime: being divisible by \(2\) does not stop \(2\) itself being prime.
BThe argument is correct.
CThe argument is wrong because some odd numbers are not prime.
DThe argument is wrong because \(1\) is odd and not prime.
EThe argument is wrong because \(0\) is even.

Worked solution

A number divisible by \(2\) is not prime unless it equals \(2\). The conclusion 'every prime is odd' is false.

Full method — how and why

An argument that proves 'almost' the right thing. Find the single case it mishandles.

The reasoning is: even \(\Rightarrow\) divisible by \(2\) \(\Rightarrow\) not prime.
Trace the chain and test each link.
The second link fails for one number: \(2\) is divisible by \(2\) and is prime, because divisibility by itself is allowed.
A prime has exactly two divisors, \(1\) and itself; for \(2\) those are \(1\) and \(2\).
So the conclusion 'every prime is odd' is false - \(2\) is an even prime.
A single counterexample refutes it.
The correct statement is 'every prime greater than \(2\) is odd', and the argument does establish that.
Identifying the corrected claim shows precisely what the flaw was.

Where it goes wrong

Objecting that some odd numbers are not prime - true, but irrelevant; the claim is not the converse.
Objecting via \(1\) or \(0\) - neither is a counterexample to 'every prime is odd'.

Takeaway Check whether an argument's chain breaks at a boundary case. \(2\) is the even prime that breaks this one.

Answer A · Identifying errors in proofs (Err2) · question ERR-16

10Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student solves \(2\cos^2 x = \cos x\) for \(0 \leq x < 2\pi\).

IDivide both sides by \(\cos x\): \(2\cos x = 1\).
IISo \(\cos x = \dfrac12\).
IIISo \(x = \dfrac{\pi}{3}\) or \(x = \dfrac{5\pi}{3}\).

How many solutions has the student missed, and where is the first error?

ATwo solutions missed; the first error is on line I.
BNo solutions missed; the working is correct.
COne solution missed; the first error is on line III.
DTwo solutions missed; the first error is on line III.
EOne solution missed; the first error is on line I.

Worked solution

Dividing by \(\cos x\) loses \(\cos x = 0\), i.e. \(x = \frac\pi2\) and \(x = \frac{3\pi}{2}\): two solutions.

Full method — how and why

Dividing a trigonometric equation by \(\cos x\) loses the solutions where \(\cos x = 0\).

Line I divides by \(\cos x\), valid only if \(\cos x \neq 0\). First error.
The equation \(2\cos^2x = \cos x\) is satisfied whenever \(\cos x = 0\), so those solutions are discarded.
The correct method: \(2\cos^2x - \cos x = 0\), so \(\cos x(2\cos x - 1) = 0\).
Factorise rather than divide - it keeps both branches.
Branch \(\cos x = 0\): in \(0 \leq x < 2\pi\) that is \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\) - two missed solutions.
Cosine vanishes at the top and bottom of the unit circle.
Branch \(\cos x = \frac12\): \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\) - the two the student found.
Cosine is positive in the first and fourth quadrants, so \(x = \frac\pi3\) and \(2\pi - \frac\pi3\). Four solutions in total.

Where it goes wrong

Saying one solution was missed; \(\cos x = 0\) has two solutions in a full revolution.
Blaming line III - the student's own two values are correct for their (incomplete) equation.
Declaring the working correct because the two values found do satisfy the original equation.

Takeaway Never divide a trigonometric equation by a trig function - factorise, and count the solutions of each branch across the whole interval.

Answer A · Identifying errors in proofs (Err2) · question ERR-18

11Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student solves \(x^2 - 5x + 6 > 0\) as follows.

I\(x^2 - 5x + 6 = (x - 2)(x - 3)\).
IISo \((x - 2)(x - 3) > 0\).
IIISo \(x - 2 > 0\) and \(x - 3 > 0\).
IVSo \(x > 3\).

Which of the following is correct?

AThe first error is on line III: a product is positive when both factors are positive or both are negative, so \(x < 2\) is missed.
BThe solution is completely correct.
CThe first error is on line I.
DThe first error is on line IV: it should be \(x > 2\).
EThe first error is on line II.

Worked solution

The full solution set is \(x < 2\) or \(x > 3\).

Full method — how and why

A quadratic inequality solved as if it were an equation-style 'both factors positive'. Find where the missing branch is lost.

Lines I and II are fine: \(x^2-5x+6 = (x-2)(x-3)\), so the inequality is \((x-2)(x-3)>0\).
Check: \(-2\times-3 = 6\) and \(-2-3=-5\) ✓.
Line III says both factors must be positive. But a product is positive when both are positive or both are negative. First error.
Two negatives multiply to a positive - the case the student omits.
The missing branch: \(x-2<0\) and \(x-3<0\), i.e. \(x<2\).
Both factors negative requires the smaller bound.
Full solution: \(x<2\) or \(x>3\). (Sketch check: the parabola is above the axis outside its roots.)
A quick sketch of an upward parabola confirms the two outer intervals.

Where it goes wrong

Blaming line IV - it correctly follows from line III's (incomplete) conditions.
Giving the answer as \(20\).
Solving factor-by-factor at all; sketching the parabola is faster and safer.

Takeaway \((x-p)(x-q)>0\) holds outside the roots, \(<0\) between them. Sketch rather than case-split.

Answer A · Identifying errors in proofs (Err2) · question X2-25

12Identifying errors in proofs · spec Err2 · Paper 2 style · ★★☆

A student solves the inequality \(\dfrac{x + 1}{x - 2} < 3\).

IMultiply both sides by \((x - 2)\): \(x + 1 < 3x - 6\).
IISo \(7 < 2x\).
IIISo \(x > \dfrac{7}{2}\).

Which of the following is correct?

AThe first error is on line I: multiplying by \(x - 2\) is only valid as written when \(x - 2 > 0\).
BThe solution is completely correct.
CThe first error is on line II: it should be \(7 < 4x\).
DThe first error is on line III: it should be \(x < \frac72\).
EThe first error is on line I: you cannot multiply an inequality by an expression.

Worked solution

For \(x < 2\) the multiplier is negative and the inequality reverses; indeed \(x = 0\) gives \(-\frac12 < 3\), a solution missed by the student. The full solution set is \(x < 2\) or \(x > \frac72\).

Full method — how and why

Multiplying an inequality by an expression whose sign is unknown. The error is the unstated assumption.

Line I multiplies by \((x-2)\) and keeps the direction of the inequality.
That is only valid if \(x - 2 > 0\); if \(x < 2\) the inequality must reverse.
So line I is the first error: it silently assumes \(x > 2\).
The line is not wrong for all \(x\), but it is wrong as an unconditional step - which is what makes it an error.
Consequence: the student misses a whole branch. Test \(x = 0\): \(\frac{0+1}{0-2} = -\frac12 < 3\) ✓ - a genuine solution not in \(x > \frac72\).
A single test value exposes the omission.
The correct solution set is \(x < 2\) or \(x > \frac72\).
For \(2 < x < \frac72\) the fraction exceeds \(3\); at \(x = 2\) it is undefined.

Where it goes wrong

Saying you can never multiply an inequality by an expression - you can, if you case-split on its sign, or multiply by \((x-2)^2\), which is always positive.
Blaming line II or III; the arithmetic there is right, given line I.
Forgetting that \(x = 2\) must be excluded from any solution set.

Takeaway Never multiply an inequality by an expression of unknown sign: split into cases, or multiply by its square.

Answer A · Identifying errors in proofs (Err2) · question ERR-05

13Identifying errors in proofs · spec Err2 · Paper 2 style · ★★☆

A student writes the following argument.

I\(0.1 > 0.01\).
IITaking logarithms to base \(10\) of both sides: \(-1 > -2\).
IIIMultiplying both sides by \(\log_{10}\frac12\), which is negative, reverses the inequality: \(-\log_{10}\frac12 < -2\log_{10}\frac12\).

Which of the following is correct?

AAll three lines are correct.
BLine II is wrong: taking logarithms reverses an inequality.
CLine III is wrong: \(\log_{10}\frac12\) is positive.
DLine II is wrong: \(\log_{10} 0.1 = 1\).
ELine I is wrong.

Worked solution

Logarithms are increasing functions, so line II is valid; \(\log_{10}\frac12 < 0\), so reversing the inequality in line III is correct.

Full method — how and why

Sometimes the argument is simply valid. Check each step against the rules rather than assuming a trap.

Line I: \(0.1 > 0.01\) ✓.
\(\frac{1}{10} > \frac{1}{100}\).
Line II: \(\log_{10}\) is an increasing function, so it preserves the inequality: \(\log_{10}0.1 = -1\) and \(\log_{10}0.01 = -2\), and indeed \(-1 > -2\) ✓.
Applying an increasing function to both sides keeps the direction; \(\log_{10} 10^{-1} = -1\).
Line III: \(\log_{10}\frac12 < 0\) because \(\frac12 < 1\). Multiplying an inequality by a negative number reverses it, which is exactly what is written ✓.
\(-1 > -2\) becomes \(-1 \times c < -2 \times c\) for \(c<0\).
So all three lines are correct.
The question is testing whether you can certify a valid argument, not only spot faults.

Where it goes wrong

Assuming logarithms reverse inequalities - they do not; \(\log\) is increasing on its domain.
Thinking \(\log_{10}\frac12\) is positive; logs of numbers below \(1\) are negative.
Misreading \(\log_{10}0.1\) as \(1\) rather than \(-1\).
Hunting for an error simply because the question implies one might exist.

Takeaway Increasing functions preserve inequalities; multiplying by a negative reverses them. \(\log_b x < 0\) exactly when \(01\)).

Answer A · Identifying errors in proofs (Err2) · question ERR-07

14Identifying errors in proofs · spec Err1 · Paper 2 style · ★★☆

Claim: if \(n\) is an integer and \(n^2\) is divisible by \(4\), then \(n\) is divisible by \(4\).

ISuppose \(n^2\) is divisible by \(4\).
IIThen \(n^2 = 4k\) for some integer \(k\).
IIISo \(n = 2\sqrt{k}\), and since \(n\) is an integer \(\sqrt k\) is an integer.
IVHence \(n\) is even, so \(n = 2m\), and \(n^2 = 4m^2\) is divisible by \(4\), so \(n\) is divisible by \(4\).

Which of the following is true?

AThe claim is false (\(n = 2\) is a counterexample) and the first invalid step is on line IV.
BThe claim and proof are correct.
CThe claim is true but the first error is on line III.
DThe claim is false and the first error is on line II.
EThe claim is false and the first error is on line I.

Worked solution

\(n = 2\): \(n^2 = 4\) is divisible by \(4\) but \(n\) is not. Lines I-III are correct deductions (\(n\) even). Line IV concludes '\(n\) divisible by \(4\)' from '\(n^2\) divisible by \(4\)', which is exactly the unproved claim - circular and false.

Full method — how and why

A circular argument: the last line assumes the very thing being proved. Also, the claim is false.

Test the claim: \(n = 2\) gives \(n^2 = 4\), divisible by \(4\), but \(2\) is not. Claim false.
Establish this first; an error must exist.
Lines I-III: from \(n^2 = 4k\), \(n^2\) is even so \(n\) is even. The algebra \(n = 2\sqrt k\) with \(\sqrt k\) an integer is correct given \(n\) is an integer. Valid.
These lines legitimately establish that \(n\) is even.
Line IV: it writes \(n = 2m\), computes \(n^2 = 4m^2\) (true but irrelevant), and then concludes '\(n\) is divisible by \(4\)'. That conclusion does not follow from anything. First error.
Showing \(n^2\) is divisible by \(4\) was the hypothesis; restating it is not progress. Divisibility of \(n\) by \(4\) would require \(m\) even, which is never shown.
Compare with the working proof for \(8 \mid n^2 \Rightarrow 4 \mid n\): there the extra factor forces \(m\) even. Here there is no spare factor.
Seeing where the analogous proof gets its leverage shows exactly what is missing.

Where it goes wrong

Accepting line IV because every individual equation in it is true - the inference is what fails.
Blaming line III; introducing \(\sqrt k\) is unusual but not invalid here.
Accepting the claim because \(4 \mid n \Rightarrow 4 \mid n^2\) (the converse) is true.

Takeaway A step that restates the hypothesis and then asserts the conclusion is circular. \(4\mid n^2\) gives only \(2 \mid n\).

Answer A · Identifying errors in proofs (Err1) · question ERR-08

15Identifying errors in proofs · spec Err1 · Paper 2 style · ★★☆

Claim: \(\displaystyle\int_{-1}^{1} \frac{1}{x^2}\,dx = -2\).

IAn antiderivative of \(x^{-2}\) is \(-x^{-1}\).
IISo the integral equals \(\left[-\frac1x\right]_{-1}^{1} = -1 - 1 = -2\).
IIIHence the value is \(-2\).

Which of the following is correct?

AThe first error is on line II: the integrand is undefined at \(x = 0\), which lies inside the interval, so the fundamental theorem cannot be applied.
BThe proof is completely correct.
CThe first error is on line I.
DThe first error is on line II: \(-1 - 1 = 0\).
EThe first error is on line III: the value should be \(+2\).

Worked solution

The integrand is positive everywhere it is defined, so a negative answer is a warning sign; the function is not defined on the whole interval and the integral does not exist in the ordinary sense.

Full method — how and why

The fundamental theorem of calculus needs the integrand to be defined (and continuous) throughout the interval.

Spot the warning sign: \(\frac{1}{x^2} > 0\) wherever it is defined, yet the 'answer' is negative.
An integral of a positive function over an interval cannot be negative - the sign alone signals an error.
Line I is fine: \(\frac{d}{dx}\left(-x^{-1}\right) = x^{-2}\) ✓.
Differentiate: \(-(-1)x^{-2} = x^{-2}\).
Line II applies the fundamental theorem across \([-1,1]\). But \(\frac{1}{x^2}\) is undefined at \(x = 0\), which lies inside the interval. First error.
The antiderivative is not continuous on the whole interval, so the theorem does not apply.
The arithmetic in line II is itself right (\(-\frac11 - \left(-\frac{1}{-1}\right) = -1 - 1 = -2\)); the fault is applying the theorem at all.
Correct arithmetic on an invalid application still gives nonsense.

Where it goes wrong

Trying to 'fix' the sign to \(+2\) - the integral does not exist in the ordinary sense, so no finite value is right.
Blaming line I; the antiderivative is correct.
Ignoring the discontinuity because the limits themselves are perfectly ordinary numbers.

Takeaway Before applying \(\int_a^b f = F(b)-F(a)\), check \(f\) is defined on all of \([a,b]\); and a positive integrand can never give a negative integral.

Answer A · Identifying errors in proofs (Err1) · question ERR-11

16Identifying errors in proofs · spec Err1 · Paper 2 style · ★★☆

Claim: for all positive real \(x\), \(x + \dfrac{1}{x} \geq 2\).

ISuppose \(x + \dfrac1x \geq 2\).
IIMultiply by \(x > 0\): \(x^2 + 1 \geq 2x\).
IIISo \(x^2 - 2x + 1 \geq 0\), i.e. \((x - 1)^2 \geq 0\), which is true.
IVHence the claim is proved.

Which of the following best describes the argument?

AThe argument assumes what it is trying to prove (line I); it would be valid if the steps were reversed, starting from \((x - 1)^2 \geq 0\).
BThe argument is completely correct as written.
CThe first error is on line II: multiplying by \(x\) reverses the inequality.
DThe first error is on line III.
EThe claim is false.

Worked solution

Starting from the conclusion and reaching a truth does not prove the conclusion unless every step is reversible and the argument is presented in the forward direction. Here the steps are reversible, so the repair is to write it backwards.

Full method — how and why

Assuming the conclusion. The chain of steps is valid but it runs the wrong way.

Line I supposes \(x + \frac1x \geq 2\) - which is exactly the claim to be proved. First error.
A proof may suppose the hypothesis (\(x>0\)), never the conclusion.
Lines II and III are then correct algebra, ending at \((x-1)^2 \geq 0\), a truth.
Deriving a true statement from an assumption proves nothing: false statements can also imply true ones.
The repair: run it backwards. Start from \((x-1)^2 \geq 0\) (known true), expand to \(x^2 - 2x + 1 \geq 0\), so \(x^2 + 1 \geq 2x\), and divide by \(x > 0\) to get \(x + \frac1x \geq 2\).
Every step is reversible here - the division by \(x\) is safe because \(x>0\) - so the reversal is valid.
So the best description is: it assumes what it is proving, and would be valid reversed.
Both halves matter: the flaw, and the fact that it is repairable.

Where it goes wrong

Calling the argument correct because each line follows from the previous one - direction is what matters.
Calling the claim false; it is true, as the reversed proof shows.
Thinking multiplying by \(x\) reverses the inequality - \(x>0\) is given.

Takeaway Working backwards is how you find a proof; to present one, start from a known truth and check every step reverses.

Answer A · Identifying errors in proofs (Err1) · question ERR-13

17Identifying errors in proofs · spec Err1 · Paper 2 style · ★★☆

Claim: the sum of the first \(n\) odd numbers is \(n^2\).

IThe first odd number is \(1 = 1^2\), so the claim holds for \(n = 1\).
IISuppose the claim holds for some \(n\), i.e. \(1 + 3 + \cdots + (2n - 1) = n^2\).
IIIAdding the next odd number, \(2n + 1\), gives \(n^2 + 2n + 1 = (n + 1)^2\).
IVSo the claim holds for \(n + 1\), and hence for all positive integers \(n\).

Which of the following is correct?

AThe proof is completely correct.
BThe first error is on line II: you cannot assume the claim.
CThe first error is on line III: the next odd number is \(2n - 1\).
DThe first error is on line IV: it only shows the claim for \(n + 1\).
EThe first error is on line I.

Worked solution

This is a valid proof by induction: a base case, and a step showing that truth for \(n\) implies truth for \(n + 1\).

Full method — how and why

A correct proof by induction. Check the base case, the inductive hypothesis, the step and the conclusion in turn.

Line I: base case \(n=1\) gives \(1 = 1^2\) ✓.
An induction needs a case to start from.
Line II: assuming the result for some \(n\) is the inductive hypothesis - not circular, because it is a supposition inside the step, not a global assumption.
This is the point most often misidentified as an error.
Line III: the next odd number after \(2n-1\) is \(2n+1\) ✓, and \(n^2 + 2n + 1 = (n+1)^2\) ✓.
Check the term: odd numbers go up in twos, so \((2n-1)+2 = 2n+1\).
Line IV: base case plus step gives the result for every positive integer ✓. The proof is correct.
\(n=1\) gives \(n=2\), which gives \(n=3\), and so on without end.

Where it goes wrong

Calling line II circular - assuming for one \(n\) in order to deduce \(n+1\) is exactly how induction works.
Claiming the next odd number is \(2n-1\); that is the \(n\)th one, already included.
Claiming line IV proves only the \(n+1\) case - the chain from the base case gives all \(n\).

Takeaway Induction \(=\) base case \(+\) (truth at \(n\) \(\Rightarrow\) truth at \(n+1\)). The inductive hypothesis is not circular reasoning.

Answer A · Identifying errors in proofs (Err1) · question ERR-15

18Identifying errors in proofs · spec Err1 · Paper 2 style · ★★☆

Claim: if \(f'(a) = 0\) and \(f''(a) = 0\) then \(f\) has a point of inflexion at \(x = a\).

Which of the following shows the claim is false?

A\(f(x) = x^3 - 3x\) at \(a = 1\)
B\(f(x) = x\) at \(a = 0\)
C\(f(x) = x^2\) at \(a = 0\)
D\(f(x) = x^3\) at \(a = 0\)
E\(f(x) = x^4\) at \(a = 0\)

Worked solution

\(x^4\) has \(f'(0) = f''(0) = 0\) but a minimum at \(0\), not an inflexion. (\(x^3\) does have an inflexion; \(x^2\) has \(f''(0) \neq 0\).)

Full method — how and why

To disprove a claim about a test for inflexion, produce a function satisfying the conditions but failing the conclusion.

Need \(f'(a) = f''(a) = 0\) with no inflexion at \(a\).
Both hypotheses must hold, and the conclusion must fail.
Take \(f(x) = x^4\) at \(a=0\): \(f'(x) = 4x^3\) so \(f'(0)=0\) ✓; \(f''(x) = 12x^2\) so \(f''(0)=0\) ✓.
Both derivatives vanish - the hypotheses are satisfied.
But \(x^4\) has a minimum at \(0\) (it is positive on both sides), not a point of inflexion.
An inflexion requires the concavity to change sign; \(f'' = 12x^2 \geq 0\) on both sides, so it does not.
Check the others: \(x^3\) does have an inflexion at \(0\) (so it confirms, not refutes); \(x^2\) has \(f''(0) = 2 \neq 0\); \(x^3-3x\) at \(a=1\) has \(f'(1) = 0\) but \(f''(1) = 6 \neq 0\); \(f(x)=x\) has \(f'(0) = 1 \neq 0\).
Each fails one of the two hypotheses, or satisfies the conclusion.

Where it goes wrong

Choosing \(x^3\), the standard example of the claim rather than a counterexample.
Choosing a function where one of the two derivative conditions fails at the stated point.

Takeaway \(f''(a)=0\) does not guarantee an inflexion; the second derivative must change sign. \(x^4\) at \(0\) is the standard counterexample.

Answer E · Identifying errors in proofs (Err1) · question ERR-17

19Identifying errors in proofs · spec Err1 · Paper 2 style · ★★★

Claim: the function \(f(x) = x^3 - 3x^2 + 3x\) is increasing for all real \(x\).

I\(f'(x) = 3x^2 - 6x + 3 = 3(x - 1)^2\).
IISo \(f'(x) \geq 0\) for all \(x\), with equality only at \(x = 1\).
IIISince \(f'(x) > 0\) for all \(x\), \(f\) is increasing.

Which of the following is true?

ALine III overstates what was shown: \(f'(1) = 0\); but \(f\) is still increasing for all \(x\), so the claim is true and the proof needs a better justification.
BThe proof is completely correct.
CThe first error is on line I.
DThe first error is on line II: \(f'(x)\) can be negative.
EThe claim is false: \(f\) has a turning point at \(x = 1\).

Worked solution

\(f\) is (strictly) increasing because \(f'(x) \geq 0\) with equality only at an isolated point, so \(f(a) < f(b)\) whenever \(a < b\); but line III's assertion '\(f'(x) > 0\) for all \(x\)' is false at \(x = 1\). There is no turning point at \(x = 1\) (the derivative does not change sign).

Full method — how and why

A subtle one: the claim is true but one line asserts more than was proved. Distinguish a false conclusion from an unjustified step.

Check line I: \(f'(x) = 3x^2 - 6x + 3 = 3(x^2-2x+1) = 3(x-1)^2\). Correct.
Differentiate term by term, then factorise.
Check line II: a square is non-negative, so \(f'(x) \geq 0\), with equality only at \(x = 1\). Correct.
The word 'only' is right - \((x-1)^2 = 0\) just at \(x=1\).
Line III says '\(f'(x) > 0\) for all \(x\)' - false at \(x = 1\), where \(f'(1) = 0\). So line III overstates line II.
It asserts a strict inequality that was never established.
Is the claim itself true? Yes: \(f' \geq 0\) everywhere with equality at a single isolated point still gives \(f(a) < f(b)\) for \(a
A derivative vanishing at one point does not create a flat stretch; there is no turning point, because \(f'\) does not change sign.

Where it goes wrong

Concluding the claim is false because \(f'(1) = 0\) - a stationary point of inflexion is not a turning point.
Blaming line I or II, both of which are correct.
Calling the proof completely correct, ignoring the strictness error in line III.

Takeaway \(f' \geq 0\) with equality only at isolated points still means strictly increasing - but a proof must not claim strict positivity it has not shown.

Answer A · Identifying errors in proofs (Err1) · question ERR-06

How this topic has been examined

Real TMUA questions on Identifying errors in proofs, newest first. Links open the official paper at the question and the online marking page.

2023 · Paper 2Question 4student's attempt: consecutive odd primesquestion paper sit online

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2022 · Paper 2Question 17student proves the converse (cubic roots)question paper sit online

2021 · Paper 2Question 5first error: cos x = sqrt(1 - sin^2 x)question paper sit online

2021 · Paper 2Question 11student's attempt: a^2 - 4b^3question paper sit online

2020 · Paper 2Question 3divisibility argument with 9^n and 3^nquestion paper sit online

2020 · Paper 2Question 8proof about |a + b| < |a| + |b|question paper sit online

2020 · Paper 2Question 16student uses the FTC wronglyquestion paper sit online

2019 · Paper 2Question 6student solves cos x + sin x tan x = 2 sin x - 1question paper sit online

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October 2018 · Paper 2Question 9attempt to solve 4x sqrt(2x - 1) = 10x - 5question paper sit online

October 2018 · Paper 2Question 13attempted proof: tan > 0 implies sin + cos > 1question paper sit online

November 2017 · Paper 2Question 4student finds sin 2x from tan x = 3question paper sit online

November 2017 · Paper 2Question 9attempted proof of a^3 + b^3 = c^3 having no solutionsquestion paper sit online

November 2017 · Paper 2Question 18solving (1/2)^n < 1/10 with logsquestion paper sit online

Specimen papers · Paper 2Question 3attempt to solve sqrt(x + 5) = x + 3question paper sit online

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