TMUA Paper 2: logic and proof
Paper 2 (Mathematical Reasoning) is 20 multiple-choice questions in 75 minutes. It uses all the maths of Paper 1, and adds Section 2 of the specification: the logic of arguments, mathematical proof and identifying errors in proofs. A Level courses teach little of this, so it is where focused practice pays off most.
What Section 2 covers
| Spec code | Content | Our notes |
|---|---|---|
| Arg1 | True/false; and, inclusive or, not; if A then B; A if B; A only if B; A if and only if B; A statement is equivalent to its contrapositive, not its converse; No symbolic notation or truth tables are expected | Converse and contrapositive |
| Arg2 | A is sufficient for B: A ⇒ B; A is necessary for B: B ⇒ A; Testing each direction with examples and counterexamples | Necessary and sufficient conditions |
| Arg3-Arg4 | for all, for some (at least one), there exists; Negating: not (for all x, P) = there exists x with not P; Negating and/or statements (De Morgan in words) | Quantifiers and negation |
| Prf1 | Chains of deduction; Cases such as even/odd or sign of x; Proof by contradiction (e.g. irrationality, infinitely many primes); One counterexample disproves a 'for all' claim | Types of proof and counterexamples |
| Prf2-Prf5 | Deducing what must, could or cannot be true; Conjectures from small cases, then justification; Putting the lines of a proof in a valid order; Longer chains of reasoning | Deductions, conjectures and ordering proofs |
| Err1-Err2 | 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' | Errors in purported proofs |
The specification notes that candidates will not be expected to use symbolic logic notation or to complete formal truth tables: questions are written in words.
How to prepare
- Learn the vocabulary exactly: 'only if', 'necessary', 'sufficient', 'for all', 'there exists', the converse and the contrapositive.
- Practise translating every statement into 'if ... then ...' form before judging it.
- Work through proof-error questions by checking each step on its own: does it follow for every value allowed?
- Sit timed Paper 2s: our paper simulator and the official past papers.
Examples
Example 1
What is the contrapositive of the statement: 'If \(n^2\) is odd, then \(n\) is odd'? (\(n\) is an integer.)
- If \(n\) is even, then \(n^2\) is odd.
- \(n^2\) is odd only if \(n\) is even.
- If \(n\) is odd, then \(n^2\) is odd.
- If \(n^2\) is even, then \(n\) is even.
- If \(n\) is even, then \(n^2\) is even.
Show the answer and solution
Answer: E: If \(n\) is even, then \(n^2\) is even.
The contrapositive of 'if A then B' is 'if not B then not A'. Here A is '\(n^2\) is odd' and B is '\(n\) is odd', so it is 'if \(n\) is even (not odd), then \(n^2\) is even'. Option 2 is the converse; option 3 is the converse's contrapositive (the inverse).
Example 2
For a real number \(x\), the condition \(x^2<4\) is ______ for \(x<2\). Which phrase fills the gap?
- necessary but not sufficient
- necessary and sufficient
- neither necessary nor sufficient
- sufficient but not necessary
Show the answer and solution
Answer: D: sufficient but not necessary
If \(x^2<4\) then \(-2
Example 3
Which of the following is the negation of the statement: 'For every real number \(x\) there is an integer \(n\) with \(n>x\)'?
- There is a real number \(x\) and an integer \(n\) with \(n\le x\).
- For every real number \(x\), every integer \(n\) satisfies \(n\le x\).
- For every real number \(x\) there is an integer \(n\) with \(n\le x\).
- There is a real number \(x\) such that every integer \(n\) satisfies \(n\le x\).
- There is an integer \(n\) such that every real number \(x\) satisfies \(n\le x\).
Show the answer and solution
Answer: D: There is a real number \(x\) such that every integer \(n\) satisfies \(n\le x\).
'Not (for every \(x\), P(\(x\)))' is 'there is an \(x\) with not P(\(x\))'. And 'not (there is an \(n\) with \(n>x\))' is 'every \(n\) has \(n\le x\)'. So the negation is 'there is a real \(x\) such that every integer \(n\) satisfies \(n\le x\)'. (The original statement is true, so its negation is false.)
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