Mathematical argument, language and notation

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Proof and mathematical argument are not a separate corner of the course.

This unit is examined on all three papers: sometimes as a question of its own (a counter example, a proof by exhaustion, a proof by contradiction), and far more often inside questions on algebra, calculus, mechanics and statistics.

Every Show that, every Prove, every "Fully justify your answer" and every Explain is marked on the skills in this unit.

This first section is about the raw material of every argument: the words, the symbols and the ways of connecting statements.

You need to be able to "Construct and present mathematical arguments through appropriate use of diagrams; sketching graphs; logical deduction; precise statements involving correct use of symbols and connecting language, including: constant, coefficient, expression, equation, function, identity, index, term, variable."

You also need to "Understand and use mathematical language and syntax as set out in the content."

Why precise words matter

In everyday speech "equation", "expression" and "formula" are used loosely.

In mathematics each word names a different kind of object, and an argument that mixes them up is not a correct argument.

"Solve the expression $x^2 - 5x$ " makes no sense, because an expression has no equals sign and so nothing to solve.

"Show that this equation is true" is ambiguous: an equation is true only for some values.

Using the right word shows that you know what you are doing, and it stops you from doing the wrong thing.

Word

Meaning

Example

variable

a letter that can take different values

$x$ in $y = 3x + 2$

constant

a quantity that does not change in the problem

$2$ , $\pi$ , or $c$ in $y = mx + c$

term

one part of an expression, separated by $+$ or $-$

$3x^2$ , $-5x$ and $7$ in $3x^2 - 5x + 7$

coefficient

the number (or constant) multiplying a variable in a term

$3$ is the coefficient of $x^2$ in $3x^2 - 5x + 7$

index

a power: the small raised number

$2$ in $x^2$ ; $\tfrac{1}{2}$ in $x^{\frac{1}{2}}$

expression

a collection of terms with no equals sign

$3x^2 - 5x + 7$

equation

a statement that two expressions are equal, true only for particular values of the variable

$3x^2 - 5x + 7 = 9$

identity

a statement that two expressions are equal for every value of the variable, written with $\equiv$

$(x + 1)^2 \equiv x^2 + 2x + 1$

function

a rule that gives exactly one output for each input in its domain

$\mathrm{f}(x) = 3x^2 - 5x + 7$

The distinction between an equation and an identity is the one that earns and loses the most marks.

An equation asks a question ("for which $x$ is this true?")

and you answer it by solving.

An identity makes a claim ("this is true for every $x$ ") and you justify it by proving, usually by turning one side into the other.

The symbol $\equiv$ means "is identical to", and you should use it whenever you mean "for all values".

If you are told to prove an identity, you must never treat it as an equation and solve it: there is nothing to find.

Whether a letter is a variable or a constant depends on the problem.

In $y = mx + c$ for one particular line, $m$ and $c$ are constants and $x$ and $y$ are variables.

In a question about "the family of lines through $(0, 4)$ ", $m$ changes from one line to the next, so it is behaving as a variable (a parameter).

When a question says "where $k$ is a constant", it is telling you that $k$ is one fixed number, even though you do not yet know it.

Connecting language: implies, is implied by, is equivalent to

A proof is a chain of statements, and each link must be justified.

The symbols you need to read and write are these, with the meanings given in the notation list you are expected to know:

Symbol

Meaning

$p \Rightarrow q$

$p$ implies $q$ (if $p$ then $q$ )

$p \Leftarrow q$

$p$ is implied by $q$ (if $q$ then $p$ )

$p \Leftrightarrow q$

$p$ implies and is implied by $q$ ( $p$ is equivalent to $q$ )

$\therefore$

therefore

$\because$

because

The key fact is that an implication can be true in one direction only.

" $x = 3 \Rightarrow x^2 = 9$ " is true.

The reverse, " $x^2 = 9 \Rightarrow x = 3$ ", is false, because $x = -3$ also gives $x^2 = 9$ .

So $x = 3 \Leftrightarrow x^2 = 9$ is false.

This is exactly why squaring both sides of an equation can create solutions that do not work: squaring is an implication in one direction only, and you must check your answers in the original equation.

When every step of a solution is reversible (adding the same number to both sides, multiplying by a non-zero number), the steps are equivalences and the final answer is guaranteed.

When a step is not reversible (squaring, multiplying by an expression that might be zero), the chain only runs forwards and you must check.

Writing $\Rightarrow$ between lines is a good habit, but only if it is the true relationship.

Writing $\Leftrightarrow$ between two lines that are not equivalent is a false statement, and in a proof a false statement costs the reasoning mark.

Precise statements: "for all", "there exists", "if and only if"

Many claims have the form "for every $x$ , something is true".

To prove such a claim you need an argument that works for every $x$ at once: checking examples never does it, however many you check.

To disprove it you need only one value of $x$ for which it fails: a counter example.

The opposite kind of claim, "there is an $x$ for which something is true", is proved by one example and disproved only by a general argument.

Knowing which kind of claim you face tells you which kind of answer is wanted.

"If and only if" is the words for $\Leftrightarrow$ : " $n$ is even if and only if $n^2$ is even" is two claims at once, " $n$ even $\Rightarrow n^2$ even" and " $n^2$ even $\Rightarrow n$ even", and a full proof must deal with both.

Arguments with diagrams and sketched graphs

A sketch can be part of a proof, provided you say in words what the sketch shows and why it is enough.

The classic use is counting solutions.

The number of real solutions of $\mathrm{f}(x) = \mathrm{g}(x)$ is the number of points where the graphs $y = \mathrm{f}(x)$ and $y = \mathrm{g}(x)$ meet.

A cubic equation you cannot factorise can often be rearranged into two graphs you can sketch easily, and the sketch tells you how many real roots there are.

The diagram below shows $y = \dfrac{1}{x}$ and $y = 4 - x^2$ on the same axes.

They meet three times, at A, B and C. Worked example 3 turns that picture into an argument about the equation $x^3 - 4x + 1 = 0$ .

The curve y = 1/x, solid, in two branches, and the parabola y = 4 − x², dashed, with its maximum at (0, 4). They cross at three points: A, where x is a little less than −2; B, where x is a little more than 0; and C, where x is between 1 and 2.

A sketch argument needs more than a picture.

You must say why the curves cross where they do (for example, which curve is higher at which values of $x$ ), and you must account for every region of the graph, including where one curve is negative.

A sketch with no words is not an argument.

The language of the questions

The papers use the same words in the same way every year, and each one tells you what kind of answer earns the marks.

The question says

What it means for your answer

Find, Calculate, Determine

Work it out, with the method shown

Show that

The answer is printed: give every step that reaches it, then a concluding sentence

Prove

Give a complete argument that covers every case

State, Write down

No working is needed

Explain

A precise reason in a full sentence, naming the quantities involved

Hence

Use the result you have just found or been given; another method scores nothing

Hence, or otherwise

The result just found is the intended route, but any correct method earns full marks

exact

Keep surds, fractions, $\pi$ , $\ln$ and $\mathrm{e}$ ; a decimal earns nothing

in the form ...

Your final answer must match that form exactly

Fully justify your answer

The last mark is for the reason, written out

"Show that" questions print the answer so that you can carry on to the next part even if you cannot do this one.

That is also why every step counts: the answer is already on the page, so the marks are all for the working that reaches it.

Key terms

Expression: a collection of terms combined by operations, with no equals sign.

Equation: a statement that two expressions are equal, true only for particular values of the variable.

Identity: a statement, written with $\equiv$ , that two expressions are equal for every value of the variable.

Coefficient: the number or constant that multiplies a variable in a term.

$p \Rightarrow q$ : $p$ implies $q$ (if $p$ then $q$ ).

$p \Leftrightarrow q$ : $p$ implies and is implied by $q$ ( $p$ is equivalent to $q$ ).

How the marks work

M marks are for a correct method; A marks are for accuracy and need the M mark before them; B marks stand alone.

R marks are for reasoning: completing an argument or proof, or making a deduction. E marks are for an explanation in words.

An F suffix (A1F, B1F, R1F, E1F) means follow-through: credit for correct work on your own earlier wrong value.

AG marks a printed answer that you must reach by correct working. CAO is correct answer only; CSO correct solution only; AWRT "answers which round to"; ISW "ignore subsequent working"; PI "possibly implied" by later work.

Where a question names a method ("Use the factor theorem", "use proof by exhaustion", "Hence"), that method must be seen. A correct answer with no working earns full marks only when a calculator gives it directly; a wrong answer with no working scores nothing.

The last mark of a Show that or Prove is usually an R1: every step shown, brackets right, and a concluding sentence that repeats the claim.

Worked example 1: an identity, then an equation

(a) Show that $(2x - 1)^2 - (x - 2)^2 \equiv 3(x^2 - 1)$ . [3]

(b) Hence solve the equation $(2x - 1)^2 - (x - 2)^2 = 9$ . [2]

(a) $(2x - 1)^2 = 4x^2 - 4x + 1$ and $(x - 2)^2 = x^2 - 4x + 4$ M1

$(2x - 1)^2 - (x - 2)^2 = 4x^2 - 4x + 1 - x^2 + 4x - 4 = 3x^2 - 3$ A1

$= 3(x^2 - 1)$ , so $(2x - 1)^2 - (x - 2)^2 \equiv 3(x^2 - 1)$ for all values of $x$ .

(b) Using part (a), the equation is $3(x^2 - 1) = 9$ , so $x^2 = 4$ M1

$x = 2$ or $x = -2$ A1

Answer: (b) $x = \pm 2$ .

Part (a) is an identity, true for every $x$ ; part (b) is an equation, true for two values only.

"Hence" means part (a) must be used: expanding from scratch is allowed only after "Hence, or otherwise".

The bracket in $-(x - 2)^2$ matters: dropping it gives $-x^2 - 4x + 4$ , the commonest slip here.

Worked example 2: choosing the right implication

In each statement, $x$ is a real number.

Write down the symbol $\Rightarrow$ , $\Leftarrow$ or $\Leftrightarrow$ that makes the statement correct and complete.

(a) $x = 3 \;\square\; x^2 = 9$ [1]

(b) $x^3 = -8 \;\square\; x = -2$ [1]

(a) $\Rightarrow$ : if $x = 3$ then $x^2 = 9$ , but $x^2 = 9$ also allows $x = -3$ . B1

(b) $\Leftrightarrow$ : a real number has exactly one real cube root, so each statement gives the other. B1

Answer: (a) $\Rightarrow$ (b) $\Leftrightarrow$ (c) $\Leftarrow$ .

Test each direction separately and look for one value that breaks it: a single failure shows the arrow cannot point that way.

Worked example 3: a sketch used in an argument

Show that the equation $x^3 - 4x + 1 = 0$ has exactly three real roots.

You may use the diagram above. [4]

$x = 0$ is not a root, because $0 - 0 + 1 = 1 \neq 0$ , so I may divide by $x$ :

$x^3 - 4x + 1 = 0 \Leftrightarrow x^2 - 4 + \dfrac{1}{x} = 0 \Leftrightarrow \dfrac{1}{x} = 4 - x^2$ B1

So the real roots are the $x$ -coordinates of the points where $y = \dfrac{1}{x}$ meets $y = 4 - x^2$ .

For $x > 0$ : as $x \to 0$ , $\dfrac{1}{x}$ is above $4 - x^2$ ; at $x = 1$ , $\dfrac{1}{x} = 1 < 3 = 4 - x^2$ ; at $x = 2$ , $\dfrac{1}{x} = 0.5 > 0 = 4 - x^2$ .

The curves change order twice, so they cross at least twice, at B and C. For $-2 \leqslant x < 0$ : $4 - x^2 \geqslant 0 > \dfrac{1}{x}$ , so they do not meet.

At $x = -2$ the parabola is above ( $0 > -0.5$ ) and at $x = -3$ it is below ( $-5 < -\frac{1}{3}$ ), so they cross at A. A1

That gives at least three real roots, and a cubic equation has at most three, so the equation $x^3 - 4x + 1 = 0$ has exactly three real roots.

Answer: shown.

The roots are about $-2.11$ , $0.254$ and $1.86$ , but the question does not ask for them: the argument is about how many, not where.

The first line, ruling out $x = 0$ , matters: dividing by something that might be zero is a step you must justify.

Worked example 4: using the words precisely

The curve $C$ has equation $y = 2x^3 + ax^2 - 7x + 6$ , where $a$ is a constant.

The point $(1, 5)$ lies on $C$ .

(a) Find the value of $a$ . [2]

(b) State the coefficient of $x$ and the constant term in the equation of $C$ . [1]

(a) Substituting $x = 1$ , $y = 5$ : $5 = 2 + a - 7 + 6$ M1

$a = 4$ A1

(b) The coefficient of $x$ is $-7$ and the constant term is $6$ . B1

Answer: (a) $a = 4$ (b) $-7$ and $6$ .

The coefficient includes its sign: writing $7$ for the coefficient of $x$ in $-7x$ loses the mark.

"Where $a$ is a constant" tells you $a$ is one fixed number, so one point on the curve is enough to find it.

Exam tip

End every Show that and Prove with a sentence that repeats the claim in the question's own words ("so the equation has exactly three real roots", "for all values of $x$ ").

The missing concluding statement is the most common lost mark on all three papers, and the wording can almost always be lifted straight from the question.

Common mistakes

Treating an identity as an equation. Asked to show $A \equiv B$ , a common slip is to set $A = B$ and "solve", getting $0 = 0$ or a value of $x$ . It is tempting because the two look alike on the page; but an identity is proved by turning one side into the other, and there is nothing to solve.

Arrows pointing the wrong way. Writing $x^2 = 9 \Leftrightarrow x = 3$ is false. Check both directions before you write $\Leftrightarrow$ .

Reaching a printed answer through wrong working. If your algebra does not produce the printed result, do not change signs or terms to force it. Incorrect working that arrives at the given answer earns nothing for the R1.

"It" and "they" in explanations. Name the graphs, equations or quantities you are comparing: "the curve $y = \frac{1}{x}$ is above the parabola", not "it is above it".

Ignoring "Hence". A correct answer by a different route after "Hence" scores nothing; use the result you were just given, even if you did not manage to prove it.

A decimal where "exact" is asked. $\sqrt{2}$ is exact; $1.414$ is not, and earns no accuracy mark.

Check your understanding

1. Which of these is an identity?

$2(x + 3) = 2x + 3 \qquad (x - 2)^2 = x^2 - 4 \qquad x^2 - 4 = (x - 2)(x + 2) \qquad x^2 = 4$

Circle your answer. [1]

2. $x$ is a real number. Which statement is correct?

☐ $x^2 > 4 \Rightarrow x > 2$

☐ $x > 2 \Rightarrow x^2 > 4$

☐ $x^2 > 4 \Leftrightarrow x > 2$

☐ $x > 2 \Leftarrow x^2 > 4$

Tick (✓) one box. [1]

3. (a) Show that $(3x + 2)^2 - (3x - 2)^2 \equiv 24x$ . [2]

(b) Hence find the value of $302^2 - 298^2$ . [2]

4. Priya says that $x^2 - 5x + 6 = 0$ is an identity, because it is true when $x = 2$ and when $x = 3$ .

Explain why Priya is wrong. [1]

5. The expression $5x^4 - 3x^2 + 2x - 9$ is given.

(a) Write down the coefficient of $x^2$ . [1]

(b) Write down the number of terms in the expression. [1]

6. By considering the graphs of $y = x^2 - 1$ and $y = \dfrac{3}{x}$ , show that the equation $x^3 - x - 3 = 0$ has exactly one real root. [4]

Answers

1. $x^2 - 4 = (x - 2)(x + 2)$ B1: it is true for every $x$ (difference of two squares). $2(x + 3) = 2x + 3$ forgets to multiply the 3; $(x - 2)^2 = x^2 - 4$ drops the middle term $-4x$ ; $x^2 = 4$ is an equation, true only for $x = \pm 2$ .

2. $x > 2 \Rightarrow x^2 > 4$ R1. The first and fourth say the same thing and fail at $x = -3$ ; the third is false for the same reason.

3. (a) $(3x + 2)^2 - (3x - 2)^2 = (9x^2 + 12x + 4) - (9x^2 - 12x + 4)$ M1; $= 24x$ , so $(3x + 2)^2 - (3x - 2)^2 \equiv 24x$ for all $x$ R1.
(b) $302 = 3 \times 100 + 2$ and $298 = 3 \times 100 - 2$ , so take $x = 100$ M1; $302^2 - 298^2 = 24 \times 100 = 2400$ A1.

4. An identity must be true for every value of $x$ , but $x^2 - 5x + 6 = 0$ is false for other values (for example $x = 0$ gives $6 = 0$ ), so it is an equation with two solutions E1.

5. (a) $-3$ B1 (b) $4$ B1.

6. $x = 0$ is not a root ( $-3 \neq 0$ ), so $x^3 - x - 3 = 0 \Leftrightarrow x^2 - 1 = \dfrac{3}{x}$ B1. The roots are where the graphs meet M1. For $x > 0$ : near $x = 0$ , $\frac{3}{x}$ is far above $x^2 - 1$ , and for large $x$ the parabola is above, so they cross; since $\frac{3}{x}$ is decreasing and $x^2 - 1$ is increasing for $x > 0$ , they cross exactly once. For $x < 0$ : $\frac{3}{x} < 0$ , which needs $x^2 - 1 < 0$ , so $-1 < x < 0$ ; there $\frac{3}{x} < -3$ but $x^2 - 1 \geq -1$ , so they do not meet A1. The graphs meet once, so the equation $x^3 - x - 3 = 0$ has exactly one real root R1.

Mathematical argument, language and notation

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