# A Level Maths: The mathematical process of proof

*Hints and tips - five minute read*

**Steven Walker, OCR Maths Subject Advisor**

Following on from the series of blogs about studying mathematics away from the classroom, I will now turn my attention to issues seen with questions on the individual topics of A Level Maths, beginning with proof.

### DfE criteria for maths

Formal proof was not particularly a key feature of the legacy specifications, but it is in the reformed A Level Maths criteria.

The AS content includes: an introduction to the language and structure of proof, proof by deduction, proof by exhaustion and disproof by counter example.

At A Level this initial content is developed further and proof by contradiction is introduced.

### Further Maths

Proof by induction is covered in Further Maths. Each year a few candidates will answer an A Level Maths proof question with a full proof by induction despite this being “off specification”. Obviously, correct solutions will gain full credit but there is no inherent advantage to using techniques taught in Further Maths in the A Level Maths exam.

### The language of proof

Students will have been introduced to proof at GCSE, although the work at that level would have been more of a ‘show that’ demonstration rather than the more formal approach needed at AS and A Level. Careful algebraic manipulation is needed when approaching proof questions, with the most common errors seen when multiplying negative terms.

The direction of connectives ( → , ← and ↔ ) often causes problems.2* ^{n}* – 1 is not prime if

*n*is a positive even integer greater than 2 so

2

*– 1 is not prime ←*

^{n}*n*is a positive even integer greater than 2.

Given that *x* is a positive integer, then *x*^{2} is a positive integer so

*x* is a positive integer → *x*^{2} is a positive integer.

*ABC* is a right angle triangle and *a*^{2} + *b*^{2} = *c*^{2} so

*ABC* is a right angle triangle ↔ *a*^{2} + *b*^{2} = *c*^{2}.

### Deductive proof

This is where students must prove a statement is true starting from some known facts. It is often written as a left hand side (LHS) expression equal to a right hand side (RHS) expression. A common context is trigonometric identities, but questions could be set on any GCSE prior knowledge.

A common mistake is to start manipulating both sides of the equation, but candidates should only work with either the LHS or the RHS. Generally, the expectation is to manipulate the LHS into the RHS form but there is no rule that states this is the only route.

### Proof by exhaustion

A single numerical example will not prove something is true, neither will multiple numerical examples unless every possible number can be checked. Numerical proof will only be feasible if there is a clear, manageable, range of numbers to be checked.

For example:

Prove that there are no perfect squares between 90 and 99.

Here it is straight forward to show that √90, √91, √92, √93, √94, √95, √96, √97, √98, √99 all give irrational results.

Proof by exhaustion can also be used algebraically, provided that all numeric values can be clearly represented.

For example:

Prove that all cube numbers are either a multiple of 9, or one more or one less than a multiple of 9.

Here it is important to explicitly state that all integers (*n*) can be written as either (3*a* – 1), (3*a*) or (3*a* + 1).

Therefore

If *n* = (3*a* – 1) then *n*^{3} = 9 (3*a*^{3} – 3*a*^{2} + *a*) – 1

If *n* = (3*a*) then *n*^{3} = 9 (3*a*^{3})

If *n* = (3*a* + 1) then *n*^{3} = 9 (3*a*^{3} + 3*a*^{2} + *a*) + 1

In this type of question, the most common mistake seen is to not exhaust all the possible values.

### Counter example

It only takes one example to disprove a general statement.

For example:

The statement ‘all square numbers are even’ can be disproved by 3^{2} = 9.

### Contradiction

A common misconception is to misread this as counter example. Proof by contradiction requires candidates to make an assumption that can subsequently be proved to be impossible.

For example:

Prove that √7 is irrational

Assume that if √7 is rational then √7 = *p* / *q* where *p* and *q* are integers with no common factors.

However, √7 = *p* / *q* gives 7*q*^{2} = *p*^{2} so p must be a multiple of 7 (*p* = 7*k*)

But this means that 7*q*^{2}= 49*k*^{2} so *q* must be a multiple of 7.

This is a contradiction since this has shown that *p* and *q* both have a factor of 7 when the original statement assumed that *p* and *q* have no common factors.

### OCR Support

For a range of exam questions see the practice and past papers on Interchange and the qualification website (for Mathematics A H230/H240 or for Mathematics B(MEI) H630/H640). For notes on common mistakes and misconceptions see the examiner reports.

ExamBuilder is a useful tool for quickly sourcing specific topic questions. Simply filter on ‘proof’ to locate all the questions tagged to proof along with the respective mark schemes and the examiner report comments. The following Check In tests and Delivery Guides on proof are available from the planning and teaching section of the qualification pages.

Mathematics A H240 | Mathematics B (MEI) H640 |

Check In Test: 1.01 Proof | Check In Test: Proof |

Delivery Guide: 1.01 Proof | Delivery Guide: Proof |

## Stay connected

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If you have any questions, email us at maths@ocr.org.uk, call us on 01223 553998 or Tweet us @OCR_Maths. You can also sign up for email updates to receive information about resources and support.

### About the author

#### Steven Walker, OCR Maths Subject Advisor

Steven joined OCR in 2014 during the major qualification reform period and now primarily focuses on supporting the Level 3 maths qualifications. Steven originally studied engineering before completing a PGCE in secondary mathematics. He began his teaching career with VSO in Malawi and has taught maths in both the UK and overseas. He is currently balancing his ‘work from home’ commitments with supporting his daughter with her year one lessons.