Prove it

From one side to the other (identities and integer claims)

In 4 of 20 recent exams
Paper 1, no calculator
2 to 4 marks

How it is set in the exam

Paper 1 (no calculator) only, 2–4 marks per part: a Section A question on its own, or part (a) of a longer one, sometimes followed by a second claim about the same integers. Typical (4 of 20 exam versions, all on Paper 1 (no calculator)). The same layout is also the first part of many questions in other topics (Where proof hides).

Spot it in the exam

Exam phrases to spot (5)

“show that … ≡ …”
“prove that … for all $n \in \mathbb{Z}$ ”
“show that … is a multiple of $k$ ”
“prove that the sum of any $k$ consecutive integers is divisible by …”
“prove that … is never divisible by …”

In disguise (3)

a claim that one expression in consecutive integers is the same as another, with no mention of dividing: it is an identity, so end on the given form
algebraic side lengths of a right-angled triangle: Pythagoras is the identity to prove. Start from $a^2 + b^2$ (the shorter sides) and end on $c^2$ (the hypotenuse), or go the other way; never both at once
a "show that" identity that prepares a later part (a rewrite before integrating, factorising and cancelling before a log equation): it belongs to the later part's type, but these layout rules still decide its marks (Where proof hides)

Variants (4)

identity: algebraic fractions or squared fractions, worked from one side; a relation between consecutive integers, ending on the given form
divisible by $k$ : the sum of consecutive integers, written as $k \times$ (an integer)
not divisible by $k$ : $n^2 + (n + 1)^2 + \ldots$ for consecutive integers, written as $k \times$ (an integer) $+ r$
a multiple of $k$ : $(an + b)^2 - (cn + d)^2$ , two linear expressions in $n$ (expand both, or $A^2 - B^2$ )

Looks like it, but isn't (6)

"verify that …" at a given value: substitute into both sides and compare; no algebra for all values is needed (show that)
"solve", or "find" the values that make an equation true: it holds for some values only, so solve it (Solving quadratic equations)
a quadratic's roots or sign "for all $k$ " (or all $x$ ): the discriminant or vertex form, Find k from the number of roots or Find the vertex: vertex form, range, max/min value
a trigonometric identity: trig identities (topic 3), with these layout rules plus the trig moves
"show that the sequence is arithmetic" (or geometric): Given a formula in $n$ (for $u_n$ or $S_n$ )
a statement shown false by one example, proof by induction, proof by contradiction: AHL 1.15, not SL

Worked example

An identity with algebraic fractions, from one side

Paper 1 (no calculator)

Question

(a) Show that $\dfrac{x + 1}{x - 1} - \dfrac{x - 1}{x + 1} \equiv \dfrac{4x}{x^2 - 1}$ , for $x \ne \pm 1$ . [3]
(b) Hence solve $\dfrac{x + 1}{x - 1} - \dfrac{x - 1}{x + 1} = \dfrac{8}{3}$ . [3]

Try it yourself first, then compare.

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Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.

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Which method?

The first moves are the same for every claim: write it algebraically, start on one side, expand. What changes is where you stop.

an identity: $\equiv$ , or $=$ "for all $x$ "; also a relation between consecutive integers
End on
exactly the other side, in its given form
Then write
" $=$ RHS" (or " $=$ LHS" if you started on the right) as the last line; for a claim in words, also say what that last line is in the claim's words
a multiple of $k$ , divisible by $k$ (even is $k = 2$ )
End on
$k \times (\ldots)$ , where the bracket is an integer because $n$ is
Then write
a sentence naming the integer factor
never divisible by $k$ (odd means never divisible by $2$ )
End on
$k \times (\text{an integer}) + r$ , with $0 < r < k$
Then write
a sentence: $r$ is left over, so $k$ does not divide it

The claim	End on	Then write
an identity: $\equiv$ , or $=$ "for all $x$ "; also a relation between consecutive integers	exactly the other side, in its given form	" $=$ RHS" (or " $=$ LHS" if you started on the right) as the last line; for a claim in words, also say what that last line is in the claim's words
a multiple of $k$ , divisible by $k$ (even is $k = 2$ )	$k \times (\ldots)$ , where the bracket is an integer because $n$ is	a sentence naming the integer factor
never divisible by $k$ (odd means never divisible by $2$ )	$k \times (\text{an integer}) + r$ , with $0 < r < k$	a sentence: $r$ is left over, so $k$ does not divide it

For a difference of two squares, $A^2 - B^2$ , both routes score: expand both squares (with their middle terms), or factorise as $(A + B)(A - B)$ , which is quicker when $A - B$ is a number.

Steps

Integer claims: write the integers algebraically.
"Let $n \in \mathbb{Z}$ ", then the integers as $n$ , $n + 1$ , … ( $2n$ for even, $2n + 1$ for odd), or the expressions in $n$ that the question gives.
Choose one side:
the one with work to do, or, for an integer claim, the expression the claim is about. Write it as the first line. Never work on both sides.
Expand every square with its middle term
( $(A + B)^2 = A^2 + 2AB + B^2$ ), put fractions over a common denominator, or factorise a difference of two squares. Put a bracket around anything that is subtracted.
Simplify to the ending in the table:
the given form, $k \times (\ldots)$ , or $k \times (\ldots) + r$ . If the target is described in words (a perfect square, the product of two consecutive integers), write what it looks like in algebra first ( $(\ldots)^2$ , $m(m + 1)$ ) and aim for that; numerical cases often show it.
Integer claims: finish with the sentence,
naming the integer factor, and for "never", saying why the leftover $r$ stops $k$ dividing it.
Check on scrap paper
with one value (not $0$ ): the expression you started from and your last line must give the same number. Testing the given result alone checks nothing you wrote. The check is not part of the proof.

Common mistakes

Squaring term by term
$(5n + 3)^2$ is not $25n^2 + 9$ . Markschemes give it no method mark, and every later mark depends on that one. Write the middle term: $25n^2 + 30n + 9$ .
A minus sign that stops at the first term
$-(n + 3)^2 = -\left(n^2 + 6n + 9\right) = -n^2 - 6n - 9$ . Expand inside a bracket first, then remove the bracket.
Subtracting the wrong way round
"the square of the larger minus the square of the smaller" fixes the order, and the reverse order gives the negative. Markschemes check the order.

More mistakes (9)

Working on both sides,
or moving terms across the $=$ , in an identity: capped at on a 3-mark proof, or the final mark withheld. Start on one side and stay there.
Substituting a value into the given result
working backwards, no marks. Writing the whole statement first and rearranging it down to $0 = 0$ counts as working on both sides (above). Starting from the right-hand side alone is fine.
Proving with examples
$n = 1, 2, 3$ is evidence. The claim is about every integer, so $n$ must stay a letter.
Stopping at $24n^2 + 24n$
the claim needs $24 \times$ (something), and a sentence saying the something is an integer. The reasoning mark needs the factor written out and the algebra before it correct.
Stopping one line short of the given form
if the given side is $(n + 3)^2$ , then $n^2 + 6n + 9$ is not yet it. Write $= (n + 3)^2 =$ RHS; markschemes have tied the last mark to reaching it (rule 4 of the layout).
Calling a leftover that is too big the remainder
$4n + 6$ does not leave remainder $6$ on division by $4$ . Take out another $4$ : $4(n + 1) + 2$ , remainder $2$ . (" $4n$ is a multiple of $4$ and $6$ is not" is also a valid reason, and markschemes accept this kind of reason; "remainder $6$ " is not.)
One letter for two different numbers
two odd numbers that need not be consecutive are $2m + 1$ and $2n + 1$ .
Consecutive odd numbers as $n$ , $n + 2$ , …
unless you say $n$ is odd, these can all be even ( $0 + 2 + 4 + 6 = 12$ , not a multiple of $8$ ). Write $2n + 1$ , $2n + 3$ , …, so the numbers are odd whatever $n$ is and the factor appears directly: $8(n + 2)$ .
More working after " $=$ RHS"
that contradicts the given result: the last mark goes.

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