1.01a, 1.01b - Proof, logic and mathematical arguments | OCR A-Level Mathematics A

A proof turns a convincing pattern into a conclusion you can trust for every allowed case. In 1.01a you will build deductions and complete case arguments; in 1.01b you will use implication, equivalence and congruence precisely. Along the way, the language of expressions, equations and functions will help you say exactly what your argument establishes.

Expressions, equations and identities

Why is $n^2+n$ always even when $n$ is an integer? At $n=3$ it is $12$ , and at $n=-2$ it is $2$ . Those examples are encouraging; this lesson will turn them into a proof that also covers every integer you have not tried.

First, notice what the question says. The expression $n^2+n$ represents a value. The equation $n^2+n=12$ makes a claim that is true only for particular values of $n$ . You need this distinction before you can judge a proof.

You already know

Expand brackets by multiplying every term. Substitute a negative number inside brackets: if $x=-2$ , then $x^2=(-2)^2=4$ , whereas $-x^2=-4$ .

Here is the vocabulary you will use throughout A-level mathematics.

Expression, equation and identity

An expression combines numbers, letters and operations, such as $3x^2-4x+7$ . An equation states that two expressions are equal, such as $3x+7=19$ . An identity holds for every value in its stated domain, such as $3(x+2)\equiv3x+6$ for real $x$ . The symbol $\equiv$ emphasises an identity. By contrast, $3x+7=19$ holds only when $x=4$ .

The domain matters even for an identity: $x/x\equiv1$ requires $x\ne0$ , because division by zero is undefined. Expanding brackets establishes an identity; substituting one value only verifies one case.

In $3x^2-4x+7$ , the terms are $3x^2$ , $-4x$ and $7$ . A coefficient is the multiplying factor of a variable or its power: the coefficient of $x$ is $-4$ . An index is a power, so the index in $x^2$ is $2$ . A constant has a fixed value; the constant term here is $7$ . A letter can represent a constant, so letters are not automatically variables.

A variable can take different values in a domain. An unknown is a value to be found in a problem. A function assigns exactly one output to each input in its domain; $\mathrm f(x)=3x^2-4x+7$ defines a rule, and $\mathrm f(2)=11$ gives its output at input $2$ . The notation is not multiplication of $\mathrm f$ by $x$ .

Worked example 1: reading and using a function

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

Plan. Read the sign with the coefficient, then replace every $x$ by the input.

\begin{aligned} \text{coefficient of }x&=-4 && \text{include its sign} \\ \mathrm f(2)&=3(2)^2-4(2)+7 && \text{substitute} \\ &=12-8+7=11 && \text{powers before products} \end{aligned}

Why this step? Substituting into the whole rule keeps the coefficient and the input separate.

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

Check. The three term values are $12$ , $-8$ and $7$ ; they total $11$ .

Example 2 changes only the input's sign. Brackets now prevent a negative square from being mistaken for the negative of a square.

Worked example 2: a negative input

Question. For $\mathrm f(x)=3x^2-4x+7$ , find $\mathrm f(-2)$ .

Plan. Substitute $-2$ in brackets and evaluate each term before adding.

\begin{aligned} \mathrm f(-2)&=3(-2)^2-4(-2)+7 && \text{substitute} \\ &=12+8+7 && \text{two negative factors give positive} \\ &=27 && \text{add} \end{aligned}

Why this step? The square acts on the input $-2$ , not on the coefficient $3$ .

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

Check. The square term is unchanged from input $2$ ; the linear term changes from $-8$ to $8$ , so the output increases by $16$ , from $11$ to $27$ .

Function notation records a calculation; an identity records an equality that survives every admissible input. Neither vocabulary nor these basic algebra rules is supplied on the formula sheet.

Watch out

Wrong: the coefficient of $x$ in $3x^2-4x+7$ is $4$ .

Right: the term is $-4x$ , so its coefficient is $-4$ . Keep the sign attached to its term.

Try the same reading and substitution with a new rule.

From assumptions to a conclusion

A deduction starts from assumptions and uses steps that remain valid for every object covered by the claim. General letters do the work that a long list of numerical checks cannot.

For example, $4+7=11$ suggests that an even integer plus an odd integer is odd. We can represent any such pair without choosing their values.

Even and odd integers

An integer is even if it has the form $2k$ for an integer $k$ , and odd if it has the form $2k+1$ . Thus $0$ is even and $-3=2(-2)+1$ is odd. These definitions include negative integers; they do not apply to a non-integer such as $1/2$ .

Use separate parameters for independent numbers. Writing both odd integers as $2k+1$ assumes they are equal, which a question about any two odd integers does not allow.

Worked example 3: an even integer plus an odd integer

Question. Prove that the sum of any even integer and any odd integer is odd.

Plan. Use the definitions, add, then identify twice an integer plus one.

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

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

Why this step? The final form matches the definition of odd because $a+b$ is an integer.

Answer. The sum is odd for every allowed pair.

Check. $a=2$ , $b=3$ gives $4+7=11$ . This illustrates the argument; the general algebra proves it.

Example 4 changes the first integer from even to odd. One extra unit changes the parity of the sum.

Worked example 4: two odd integers

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

Plan. Represent the integers independently and factor their sum by $2$ .

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

\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? $a+b+1$ is an integer, so the sum is an integer multiple of $2$ .

Answer. The sum of any two odd integers is even.

Check. With $a=2$ , $b=3$ , the sum is $5+7=12$ , agreeing with $2(2+3+1)$ .

The surprising reach of this tiny argument is that it covers an odd integer with a million digits just as completely as it covers $5$ . The size never enters the reasoning.

Exam tip

For “Prove”, define your integer parameters and show why the final form has the required property. Numerical trials do not replace a general argument. Never assume the conclusion as one of your starting facts: that is circular reasoning.

Every operation also has conditions. You may divide only by a nonzero quantity. Definitions and proof methods are not on the formula sheet; you need to build the argument yourself. Now change addition to subtraction.

Implication, equivalence and congruence

The direction of a statement matters: knowing an input fixes its square, but knowing its square can leave a choice of signs.

For real $x$ , $x=3$ guarantees $x^2=9$ . Yet $x^2=9$ does not guarantee $x=3$ , because $x=-3$ also works.

Implication and equivalence

$P\Rightarrow Q$ means “if $P$ , then $Q$ ”. Here $P$ is sufficient for $Q$ , and $Q$ is necessary for $P$ . The converse is $Q\Rightarrow P$ , also written $P\Leftarrow Q$ . Equivalence, $P\Leftrightarrow Q$ , means both directions hold: “ $P$ if and only if $Q$ ”, or “iff”.

Thus $x=3\Rightarrow x^2=9$ is true over the reals, while $x=3\Leftrightarrow x^2=9$ is false. A true equivalence is

x^2=9\Leftrightarrow x=3\text{ or }x=-3 \qquad (x\in\mathbb R)

The “or” includes every possible solution. Restricting the domain to $x\ge0$ would instead make $x=3\Leftrightarrow x^2=9$ true. This is why domains belong in an argument, not in the reader's guesswork.

The symbol $\equiv$ has meanings supplied by context. In algebra, $(x+1)^2\equiv x^2+2x+1$ is an identity. For integers, $17\equiv2\pmod5$ is a congruence: their difference $15$ is divisible by $5$ , so they have the same remainder on division by $5$ . It does not say $17=2$ . Congruent triangles have equal corresponding sides and angles: the same shape and size.

Squaring can therefore preserve a forward implication while losing a reverse one. Check candidates against the equation you started with; do not replace every equality sign by an equivalence arrow automatically. These logical meanings are not supplied on the formula sheet.

Key idea

For an equivalence, check both directions under the stated domain. One valid direction earns an implication, not an “if and only if”.

Here a missing sign is enough to break the claimed equivalence.

Covering every possible case

Proof by exhaustion works when your cases cover the whole domain. For integers, even and odd are two general cases, each containing infinitely many values.

There are other useful complete splits. An integer leaves remainder $0$ , $1$ or $2$ on division by $3$ . This proves that $n(n+1)(n+2)$ is divisible by $3$ :

Form of $n$ , with $k\in\mathbb Z$	A factor divisible by $3$
$n=3k$	$n=3k$
$n=3k+1$	$n+2=3(k+1)$
$n=3k+2$	$n+1=3(k+1)$

Every row makes the product divisible by $3$ , and every integer belongs to a row. Negative integers are included: $-1=3(-1)+2$ .

Proof by exhaustion

A proof by exhaustion establishes a statement in every case of a complete set of possibilities. Checking all inputs in a finite domain works; for an infinite domain, each of finitely many case arguments must itself be general. Checking one even integer and one odd integer does not cover all integers.

Return to the opening expression. Parity gives just the case split we need.

Worked example 5: the opening question

Question. Use exhaustion to prove that $n^2+n$ is even for every integer $n$ .

Plan. Substitute the two general parity forms and factor out $2$ in each case.

For $n=2k$ , with $k\in\mathbb Z$ ,

n^2+n=4k^2+2k=2(2k^2+k) \qquad \text{an even integer}

For $n=2k+1$ , with $k\in\mathbb Z$ ,

n^2+n=4k^2+6k+2=2(2k^2+3k+1) \qquad \text{an even integer}

Why this step? Each bracket is an integer, so each form proves evenness for its entire case.

Answer. Every integer is even or odd; both cases give an even result, so $n^2+n$ is always even.

Check. $n=3$ gives $12$ , and $n=-2$ gives $2$ , matching the opening examples.

Example 6 changes just the constant term by adding $3$ . The case split stays the same, but the final forms must now establish oddness.

Worked example 6: adding an odd constant

Question. Use exhaustion to prove that $n^2+n+3$ is odd for every integer $n$ .

Plan. Use the two parity cases and write each result as twice an integer plus one.

If $n=2k$ ,

n^2+n+3=4k^2+2k+3=2(2k^2+k+1)+1

If $n=2k+1$ ,

n^2+n+3=4k^2+6k+5=2(2k^2+3k+2)+1

In both cases $k$ is an integer, so the bracket is an integer.

Why this step? The final $+1$ identifies an odd integer; merely writing an expanded quadratic would leave that point unexplained.

Answer. Both exhaustive cases are odd, so the result holds for every integer $n$ .

Check. At $n=2$ , the expression is $4+2+3=9$ , which is odd.

You can also read $n^2+n$ as $n(n+1)$ : one of two consecutive integers is even. That short argument is the same complete parity split in words. No proof principle here is supplied on the formula sheet.

Watch out

Wrong: “It works for $n=2$ and $n=3$ , so both parity cases are proved.”

Right: use $n=2k$ and $n=2k+1$ , with arbitrary integer $k$ . They cover whole cases rather than one value from each.

Change the added constant again, keeping the general reasoning.

Quick Recap

An equation holds at its solutions; an identity holds throughout its stated domain.
Define your variables and use separate parameters for independent integers.
$P\Rightarrow Q$ needs one direction; $P\Leftrightarrow Q$ needs both.
Exhaustion proves a claim only when the cases cover the entire domain.

The opening expression $n^2+n=n(n+1)$ is always even because one of two consecutive integers is even; the two general parity cases prove this for every integer.

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1.01a, 1.01b - Proof, logic and mathematical arguments

Expressions, equations and identities

From assumptions to a conclusion

Implication, equivalence and congruence

Covering every possible case

Quick Recap

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