What are common errors in maths proofs?

Dividing by zero, multiplying inequalities by unknown signs, squaring without checking, and assuming a function is one-to-one.

How do you find the error in a proof?

Test each line with a simple value. The first line that stops being true is where the error sits.

Why is cancelling a common factor risky?

The factor might be zero, so cancelling it can lose a solution, as when x^2=3x is divided by x.

TMUA Spotting Errors in Proofs

These questions show a short argument with a flaw and ask where it goes wrong. The specification mentions errors such as cancelling a common factor without justification and concluding that equal sines imply equal angles.

Paper 2 topic. Specification codes: Err1, Err2, Prf4. Last checked 12 November 2026.

How to check a proof

Test each line by substituting a simple number. The first line that fails is usually the error. Also ask whether each step follows from the one before, or only looks plausible.

Common faults

Dividing by an expression that could be zero.
Multiplying an inequality by an expression of unknown sign.
Squaring both sides and treating the result as equivalent.
Assuming a function is one-to-one, such as $\sin A=\sin B\Rightarrow A=B$ .

True conclusion, bad proof

A proof can reach a true conclusion by invalid steps. Judge the steps, not the final statement.

Worked examples

Example 1

A student solves $x^2=3x$ like this: 'Divide by $x$ to get $x=3$ .' What is wrong?

Show the worked solution

Dividing by $x$ is only valid if $x\ne0$ . But $x=0$ also satisfies $x^2=3x$ , so one solution is lost. The correct solutions are $x=0$ and $x=3$ .

Answer: Division by $x$ loses $x=0$

Example 2

A student says: ' $\sin A=\sin B$ , so $A=B$ .' Give a counterexample.

Show the worked solution

Sine is many-to-one. For $A=30^\circ$ and $B=150^\circ$ , both sines equal $\tfrac12$ but $A\ne B$ .

Answer: $A=30^\circ,\ B=150^\circ$

Example 3

A student solves $x>\dfrac1x$ : 'Multiply by $x$ to get $x^2>1$ , so $x>1$ .' Find the error and give a correct solution.

Show the worked solution

Multiplying by $x$ is invalid when $x<0$ , and $x^2>1$ also allows $x<-1$ . Testing $x=-\tfrac12$ : $-\tfrac12>-2$ is true, so the claim misses solutions. The solution is $-1<x<0$ or $x>1$ .

Answer: $-1<x<0$ or $x>1$

Common mistakes

Spotting a wrong-looking step that is actually valid.
Blaming the last line when the first error is earlier.
Accepting a step because it gives a true result for one example.
Forgetting that squaring, cancelling or taking roots can add or lose solutions.

Test yourself on this topic with our free practice questions.

Quick answers

Practise it under timed conditions

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TMUA Spotting Errors in Proofs

How to check a proof

Common faults

True conclusion, bad proof

Worked examples

Common mistakes

Quick answers

Practise it under timed conditions

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