Mp1, Ma1 - Proof, logic and mathematical arguments | OCR A-Level Mathematics B (MEI)

Good mathematical writing makes a claim precise and then explains why it follows. Mp1 develops deduction and proof by exhaustion; Ma1 gives you the vocabulary and notation to express those arguments clearly. You will also read a short argument with a newly defined symbol and reconstruct what it says, a skill that matters whenever mathematics arrives as a passage rather than a familiar exercise.

Expressions, equations and identities

Can three consecutive integers ever have a product that is not divisible by $3$ ? Try $4\times5\times6=120$ , then $(-2)\times(-1)\times0=0$ . Both are divisible by $3$ ; the challenge is to explain why every possible starting integer works.

Before proving anything, decide exactly what the symbols mean. In $n(n+1)(n+2)$ , $n$ is a variable ranging over integers. In the equation $3x+7=19$ , $x$ is an unknown to be found. One letter can play either role; the question tells you which.

You already know

Use brackets around a negative input before raising it to a power. Expand a product by multiplying each term, and collect like terms with their signs attached.

The following distinctions stop a line of algebra from claiming more than it establishes.

Expression, equation and identity

An expression represents a value, for example $3x^2-4x+7$ . An equation asserts equality, for example $3x+7=19$ , true here only at $x=4$ . An identity is true for every allowed input, for example $3(x+2)\equiv3x+6$ for real $x$ . A stated domain is part of the claim: $x/x\equiv1$ applies only when $x\ne0$ .

An expression's terms are its added pieces, including their signs. The terms of $3x^2-4x+7$ are $3x^2$ , $-4x$ and $7$ . A coefficient multiplies a variable or its power, so the coefficient of $x$ is $-4$ . An index is a power: the index in $x^2$ is $2$ . A constant has a fixed value; the constant term here is $7$ , though a letter may also denote a fixed constant.

A function gives exactly one output for each input in its domain. The rule $\mathrm f(x)=3x^2-4x+7$ defines a function; $\mathrm f(2)$ means the output at $2$ , not $\mathrm f$ multiplied by $2$ .

Worked example 1: vocabulary that controls a calculation

Question. For $\mathrm f(x)=3x^2-4x+7$ , state the coefficient of $x$ and evaluate $\mathrm f(2)$ .

Plan. Identify the whole linear term, then substitute the input into every occurrence of $x$ .

\begin{aligned} \text{coefficient of }x&=-4 && \text{from the term }-4x \\ \mathrm f(2)&=3(2)^2-4(2)+7 && \text{use the rule} \\ &=12-8+7=11 && \text{evaluate and add} \end{aligned}

Why this step? The coefficient includes its sign; dropping that sign would change the function you are evaluating.

Answer. The coefficient is $-4$ , and $\mathrm f(2)=11$ .

Check. Subtract the constant from the output: $11-7=4$ . The two variable terms give $3(2)^2-4(2)=12-8=4$ , agreeing with that remainder and keeping the coefficient's negative sign.

For the second example only the sign of the input changes. Watch what happens to the square term and the linear term separately.

Worked example 2: the same function at a negative input

Question. Find $\mathrm f(-2)$ for $\mathrm f(x)=3x^2-4x+7$ .

Plan. Keep $-2$ in brackets until its square and products are evaluated.

\begin{aligned} \mathrm f(-2)&=3(-2)^2-4(-2)+7 && \text{substitute the whole input} \\ &=3(4)+8+7 && \text{square first} \\ &=27 && \text{combine the terms} \end{aligned}

Why this step? $(-2)^2$ is $4$ ; the minus sign belongs inside the square.

Answer. $\mathrm f(-2)=27$ .

Check. Changing $2$ to $-2$ leaves $3x^2$ unchanged and increases $-4x$ by $16$ . Thus $11+16=27$ .

The basic vocabulary and these elementary algebra rules are not printed on the formula sheet. Use the words to describe what your working actually does.

Watch out

Wrong: “ $\mathrm f(x)$ is $\mathrm f$ times $x$ .”

Right: $\mathrm f(x)$ is a function's output at input $x$ . Substitution changes the input inside the complete rule.

Now use the same care with another function.

From assumptions to a conclusion

A deduction explains why a conclusion follows for arbitrary values satisfying the assumptions. A numerical example can check your algebra, but cannot do the job of a general argument.

An even integer plus an odd integer gives an odd result: $4+7=11$ . To reach every allowed pair, replace the chosen values by general forms.

Parity

An integer is even when it equals $2k$ for some integer $k$ , and odd when it equals $2k+1$ . Every integer has exactly one of these parities. For example, $0=2(0)$ is even and $-3=2(-2)+1$ is odd. A non-integer such as $1/2$ is neither even nor odd.

Independent numbers need independent parameters. The forms $2a+1$ and $2b+1$ , with $a,b\in\mathbb Z$ , allow the numbers to be equal or different. Using $2a+1$ twice would prove only a special case.

Worked example 3: deduction from definitions

Question. Prove that an even integer plus an odd integer is odd.

Plan. Add their general forms, then match the result to the definition of odd.

Let the numbers be $2a$ and $2b+1$ , with $a,b\in\mathbb Z$ .

\begin{aligned} 2a+(2b+1)&=2a+2b+1 && \text{add the expressions} \\ &=2(a+b)+1 && \text{factor the even part} \end{aligned}

Why this step? $a+b$ is an integer, so this has the form $2k+1$ required by the definition.

Answer. The sum is odd for any such pair of integers.

Check. $a=2$ , $b=3$ gives $4+7=11$ .

Change the even input to an odd one in example 4. Nothing about the size or sign of either integer needs to change in the reasoning.

Worked example 4: changing one parity

Question. Prove that the sum of two odd integers is even.

Plan. Use independent general odd forms and collect an integer multiple of $2$ .

For $a,b\in\mathbb Z$ ,

\begin{aligned} (2a+1)+(2b+1)&=2a+2b+2 && \text{collect terms} \\ &=2(a+b+1) && \text{factor out }2 \end{aligned}

Why this step? The integer bracket makes divisibility by $2$ explicit.

Answer. The sum is even for every pair of odd integers.

Check. $a=2$ , $b=3$ gives $5+7=12$ , as the general formula predicts.

A short proof can cover numbers far beyond a calculator's display: the same argument applies to an odd integer with a million digits. Its size never needs to be known.

Exam tip

A “Prove” answer needs a valid general route. Define the parameters, show the algebra, and link the final integer form to the claimed property. MEI proof marks reward that reasoning; several successful numerical trials cannot replace it.

A deduction may use definitions, established facts and valid algebraic steps. Dividing by a possibly zero expression or assuming the desired conclusion breaks that chain. These proof rules are not supplied on the formula sheet. Try changing the operation instead of the parity.

Implication, equivalence and congruence

A connecting symbol tells the reader which direction your reasoning supports. That is a mathematical claim of its own.

For real $x$ , knowing $x=3$ tells you that $x^2=9$ . Knowing $x^2=9$ leaves two possible values: $3$ and $-3$ .

Implication and equivalence

$P\Rightarrow Q$ means “if $P$ , then $Q$ ”. The converse is $Q\Rightarrow P$ , or $P\Leftarrow Q$ . The statement $P\Leftrightarrow Q$ means both implications hold: “ $P$ if and only if $Q$ ”. A sufficient condition guarantees the conclusion; a necessary condition must hold whenever the starting statement holds.

Thus $x=3$ is sufficient for $x^2=9$ , and $x^2=9$ is necessary for $x=3$ . The converse fails at $x=-3$ . Over the reals the complete equivalence is

x^2=9\Leftrightarrow x=3\text{ or }x=-3

If the domain is restricted to non-negative $x$ , the negative candidate disappears, and $x^2=9\Leftrightarrow x=3$ becomes true. Squaring an equation can lose sign information, so test any candidates against its starting conditions.

The symbol $\equiv$ also needs its context. In $(x+1)^2\equiv x^2+2x+1$ it denotes an identity. In $17\equiv2\pmod5$ it denotes congruence: $17-2=15$ is a multiple of $5$ , so the integers have the same remainder. It does not assert ordinary equality. In geometry, congruent triangles have equal corresponding lengths and angles, hence the same shape and size.

These distinctions let you read an argument without silently reversing an implication. None is a supplied formula-sheet rule.

Key idea

An “if and only if” statement must survive both directions in its stated domain. Squaring alone does not establish its reverse direction.

Decide what the domain lets you conclude here.

Covering every possible case

A complete split into cases can turn an infinite set of inputs into a manageable proof. The crucial word is complete.

For parity, every integer is $2k$ or $2k+1$ . For division by $3$ , every integer is $3k$ , $3k+1$ or $3k+2$ , with integer $k$ . Negative inputs are included: $-1=3(-1)+2$ .

Proof by exhaustion

Proof by exhaustion proves a claim in each case of a complete set of possibilities. For a finite domain you may check every individual input. For an infinite domain, a finite split works only when each case argument itself covers all inputs in that case.

Here is a short mathematical passage. Read its definition before using its conclusion.

Three-integer blocks

For an integer $t$ , define a block value by $C(t)=t(t+1)(t+2)$ . The letters inside the brackets are inputs, so $C(t+1)$ is the product of the three integers starting at $t+1$ . To establish divisibility of every block value by $3$ , sort the possible starting integers by their remainders on division by $3$ . Each class gives a factor divisible by $3$ . The conclusion concerns all integer inputs, so it can subsequently be used at a shifted integer input as well.

The passage gives a definition and a strategy; the following working supplies the missing justification.

Worked example 5: reconstructing the block argument

Question. Use exhaustion to prove that $C(t)$ is divisible by $3$ for every integer $t$ .

Plan. Use all three possible remainders and locate a multiple of $3$ in each product.

For $k\in\mathbb Z$ ,

\begin{array}{c|c|c} t & \text{factor divisible by }3 & \text{reason} \\ \hline 3k & t=3k & \text{first factor} \\ 3k+1 & t+2=3(k+1) & \text{third factor} \\ 3k+2 & t+1=3(k+1) & \text{second factor} \end{array}

The other factors are integers, so a multiple of $3$ in each product makes that whole product divisible by $3$ .

Why this step? A row represents an entire remainder class, not just one successful trial.

Answer. The three cases cover every integer, so $C(t)$ is always divisible by $3$ .

Check. $C(4)=4\times5\times6=120$ , while $C(-2)=0$ ; both agree with the general conclusion.

Example 6 changes only the input from $t$ to $t+1$ . Using a supplied conclusion requires checking that its conditions still hold.

Worked example 6: using a result at a shifted input

Question. Hence prove that $(t+1)(t+2)(t+3)$ is divisible by $3$ for every integer $t$ .

Plan. Identify this product as a block value and check its input is an integer.

\begin{aligned} u&=t+1\in\mathbb Z && \text{integer plus integer} \\ C(u)&=u(u+1)(u+2) && \text{use the definition} \\ &=(t+1)(t+2)(t+3) && \text{substitute }u=t+1 \end{aligned}

Since every $C(u)$ with integer $u$ is divisible by $3$ , this product is too.

Why this step? “Hence” asks you to use the established result, and $u\in\mathbb Z$ checks the exact condition that allows it.

Answer. The shifted product is divisible by $3$ for every integer $t$ .

Check. At $t=1$ , it equals $2\times3\times4=24$ , a multiple of $3$ .

The same idea explains the parity pattern. If $n=2k$ , then $n^2+n+1=2(2k^2+k)+1$ . If $n=2k+1$ , it equals $2(2k^2+3k+1)+1$ . Both are odd forms with integer brackets, and these two cases cover every integer. A proof can be short without leaving cases unstated.

Watch out

Wrong: “Check $t=0$ , $t=1$ and $t=2$ ; all remainders are proved.”

Right: use $t=3k$ , $t=3k+1$ and $t=3k+2$ . Three sample values do not cover their infinite classes.

Definitions, deduction and exhaustion are not on the formula sheet. Practise reading the defined input and applying the general result once more.

Quick Recap

A function value is an output; an equation and an identity make different equality claims.
Deduction uses stated assumptions and valid steps for arbitrary allowed values.
Implication has a direction; equivalence requires both directions.
Exhaustion needs a complete set of general cases; a supplied passage's result can be reused only when its conditions hold.

Three consecutive integers always have a product divisible by $3$ : in every possible remainder class, one of the three factors is a multiple of $3$ .

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Mp1, Ma1 - Proof, logic and mathematical arguments