Mathematics · AQA 8300 §A10
Algebraic Proof
- Write valid algebraic proofs with clear, logical conclusions
- Represent even, odd and consecutive integers algebraically
- Prove results about sums, differences and products of integers
- Construct counter-examples to disprove mathematical statements
- Prove geometric results and divisibility results using algebra
🔑 Core Concepts
What is Algebraic Proof?
A proof is a logical argument that shows a statement is always true, for every possible value. Unlike verification (checking a few examples), a proof uses algebra to cover all cases simultaneously. In GCSE exams, you are expected to state a clear conclusion after your algebra.
Definition — Algebraic Proof
An algebraic proof starts with general expressions (using variables to represent all integers), manipulates them with valid algebraic steps, and ends with a conclusion that confirms the result is always true.
Common Mistake
Showing a result is true for three or four specific numbers is NOT a proof. You must use algebraic variables that represent all integers.
Exam Tip
Every proof must end with an explicit conclusion sentence: e.g. "Therefore the sum of two odd numbers is always even." Examiners award a mark specifically for this statement.
Algebraic Representations of Integers
The foundation of all integer proofs is expressing numbers in general form. Let $n$ be any integer. The table below summarises every key representation:
Core Integer Representations
$$\text{Even integer} = 2n$$
$n$ = any integergives: 0, 2, 4, 6, …
Odd Integer
$$\text{Odd integer} = 2n+1$$
$n$ = any integergives: 1, 3, 5, 7, …
Consecutive Integers
$$n, \quad n+1, \quad n+2$$
$n$ = first integere.g. 5, 6, 7
Consecutive Even Integers
$$2n, \quad 2n+2$$
$n$ = any integere.g. 4, 6 or 10, 12
Consecutive Odd Integers
$$2n+1, \quad 2n+3$$
differ by 2e.g. 7, 9 or 11, 13
Multiple of k
$$kn \quad (k \text{ fixed}, n \in \mathbb{Z})$$
e.g. multiple of 3: $3n$multiple of 5: $5n$
Memory Trick
"2n is even, 2n+1 is odd" — even numbers are always divisible by 2, so they equal $2 \times \text{something}$. Add 1 to get odd.
Proving Sums, Differences and Products
Most GCSE proof questions ask you to prove something about adding, subtracting or multiplying integers. The strategy is always the same:
Write general forms (e.g. $2m$, $2n+1$)
→
Perform the operation (add, multiply, etc.)
→
Simplify and factorise
→
Identify the form (even? odd? multiple of 3?)
→
State conclusion
Worked Example — Sum of Two Even Numbers
Let the two even numbers be $2m$ and $2n$. Then $2m + 2n = 2(m+n)$. Since $m+n$ is an integer, $2(m+n)$ is even. Therefore the sum of two even numbers is always even.
Important — Use Different Variables
When working with two different even (or odd) numbers, use $2m$ and $2n$, not $2n$ and $2n$ (which would mean they are the same number). This is one of the most penalised errors at Grade 9.
Definition — Product of Two Odd Numbers
Let the odd numbers be $2m+1$ and $2n+1$. Product $= (2m+1)(2n+1) = 4mn + 2m + 2n + 1 = 2(2mn+m+n)+1$. This has the form $2k+1$ where $k = 2mn+m+n$ is an integer, so it is odd.
Proving Algebraic Identities
An identity is an equation true for all values of the variable, written with $\equiv$. To prove an identity you manipulate one side (usually the more complex side) until it equals the other. You must not work on both sides simultaneously.
Exam Tip — Proving Identities
Always start with LHS = ... and expand/factorise until you reach the RHS. State "LHS = RHS, therefore proven." Never rearrange both sides at once — that assumes the result is already true.
Common Mistake — Identity Proofs
Writing things like $x^2 - 1 = (x-1)(x+1)$ and calling it "proven" without showing the manipulation is insufficient. You must show each algebraic step clearly.
Expressions Always Positive — Completing the Square
To prove that a quadratic expression is always positive (or always negative), complete the square to write it in the form $a(x+p)^2 + q$. Since $(x+p)^2 \geq 0$ for all real $x$, you can then conclude about the sign.
Completing the Square for Proof
$$ax^2 + bx + c = a\!\left(x + \frac{b}{2a}\right)^{\!2} + c - \frac{b^2}{4a}$$
$(x+p)^2 \geq 0$ alwaysif constant term $q > 0$ and $a > 0$: always positive
Worked Example — Prove $x^2 + 4x + 7 > 0$ for all $x$
Complete the square: $x^2 + 4x + 7 = (x+2)^2 - 4 + 7 = (x+2)^2 + 3$.Since $(x+2)^2 \geq 0$ for all real $x$, we have $(x+2)^2 + 3 \geq 3 > 0$.
Therefore $x^2 + 4x + 7 > 0$ for all real values of $x$.
Divisibility Proofs
These proofs require showing an expression is always divisible by a given integer $k$. The key technique is to factorise the result so that $k$ appears as a factor.
Definition — Divisibility
An integer $N$ is divisible by $k$ if and only if $N = k \times (\text{integer})$, i.e. $k$ is a factor of $N$. In a proof, you must explicitly show this factorised form.
Exam Tip — Divisibility Proofs
After expanding and collecting terms, try factorising out the divisor $k$. If the result is $k \times (\text{integer expression})$, you are done. Don't forget to confirm the bracket contains an integer.
Disproving Statements — Counter-Examples
To disprove a statement you only need a single counter-example: one specific value that shows the statement is false. The counter-example must be a valid input for the statement and must produce a result that contradicts the claim.
Definition — Counter-Example
A counter-example is a specific case that proves a general statement false. You only need one counter-example to disprove a universal claim.
Example — Disproving a Statement
Claim: "The square of any odd number is always a multiple of 3."Counter-example: $n = 1$, odd. $1^2 = 1$, and $1$ is not a multiple of 3. Statement is false.
Common Mistake — Counter-Examples
Don't just say "it doesn't work for some numbers" — you must give a specific numerical example with a clear calculation showing the claim fails.
Memory Trick
"To PROVE — use algebra for ALL. To DISPROVE — find ONE counter-example." The word "disprove" sounds like "dis one" — just one example needed!
Proof Structure and Valid Proofs
A valid algebraic proof must contain all of: (1) general algebraic expressions set up correctly, (2) correct algebraic manipulation, (3) a factorised form identifying the property, and (4) a clear concluding sentence. Missing any of these typically loses at least one mark.
Exam Tip — Proof Checklist
Before writing your final answer, check: Did I define my variables? Did I use different letters for different unknowns? Did I expand and simplify fully? Did I state a conclusion?
🗺️ Visual Notes
Algebraic
Proof
Proof
Integer Representations
- Even: $2n$
- Odd: $2n+1$
- Consecutive: $n, n+1, n+2$
- Multiple of $k$: $kn$
Proof Techniques
- Expand brackets
- Collect like terms
- Factorise result
- State conclusion
Types of Proof
- Sum/product of integers
- Divisibility proofs
- Always positive (CTS)
- Geometric results
Disproving
- Only need ONE counter-example
- Give specific number
- Show calculation
- Explain why it fails
Common Errors
- Using $2n$ for both even numbers
- No conclusion sentence
- Checking examples only
- Algebraic errors in expansion
Grade 9 Techniques
- Multi-step with squares/cubes
- Completing the square
- Proofs about consecutive squares
- Geometric expressions
Comparison: Proof vs. Verification
| Feature | Verification (Not a Proof) | Algebraic Proof |
|---|---|---|
| Scope | Tests specific numbers only | Covers ALL integers |
| Method | Substitute e.g. $n=2,4,6$ | Use variable $n$ for all |
| Validity | Does NOT prove the result | Constitutes a valid proof |
| Exam credit | 0 marks for proof question | Full marks if correct |
| Example | "Works for 2+4=6, 6+8=14" | "$2m+2n = 2(m+n)$, even" |
Integer Type Comparison
| Integer Type | General Form | Example ($n=3$) | Key Property |
|---|---|---|---|
| Even | $2n$ | $6$ | Divisible by 2 |
| Odd | $2n+1$ | $7$ | Not divisible by 2 |
| Consecutive (3) | $n, n+1, n+2$ | $3,4,5$ | Each differs by 1 |
| Consec. even | $2n, 2n+2$ | $6,8$ | Differ by 2, both even |
| Consec. odd | $2n+1, 2n+3$ | $7,9$ | Differ by 2, both odd |
| Multiple of 3 | $3n$ | $9$ | $3$ is a factor |
Proof Process Decision Tree
Read the claim
→
Prove or Disprove?
→
If PROVE: set up general forms
→
Expand, simplify, factorise
→
Conclude with words
If DISPROVE:
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Find one specific counter-example
→
Show calculation contradicts claim
→
State: "Statement is false"
✏️ Worked Examples
Simple — Grade 4–5
Note: different letters $m$ and $n$ because these can be different odd numbers.
Therefore the sum of two odd numbers is always even.
Prove that the sum of two odd numbers is always even.
1
Set up general expressions
Let the two odd numbers be $2m + 1$ and $2n + 1$, where $m$ and $n$ are integers.Note: different letters $m$ and $n$ because these can be different odd numbers.
2
Add the expressions
$$(2m+1) + (2n+1) = 2m + 2n + 2$$
3
Factorise
$$= 2(m + n + 1)$$
4
Identify the form and conclude
Since $m + n + 1$ is an integer, $2(m+n+1)$ is of the form $2k$, which is even.Therefore the sum of two odd numbers is always even.
✓ Sum = $2(m+n+1)$, which is always even.
Medium — Grade 6–7
Case 2: $n = 2k+1$ (odd): $n(n+1) = (2k+1)(2k+2) = 2(2k+1)(k+1)$, even.
In both cases the product is even.
Prove that the product of two consecutive integers is always even.
1
Set up consecutive integers
Let the consecutive integers be $n$ and $n+1$, where $n$ is an integer.
2
Multiply
$$n(n+1) = n^2 + n$$
3
Factorise differently
$$= n(n+1)$$
Now argue from the factorised form: of any two consecutive integers $n$ and $n+1$, one must be even (either $n$ is even or $n+1$ is even — they alternate). The product of an even number and any integer is even.
4
Formal algebraic version (alternative)
Case 1: $n = 2k$ (even): $n(n+1) = 2k(2k+1) = 2 \times k(2k+1)$, even.Case 2: $n = 2k+1$ (odd): $n(n+1) = (2k+1)(2k+2) = 2(2k+1)(k+1)$, even.
In both cases the product is even.
✓ Product = $n(n+1)$. One of $n$, $n+1$ is always even, so the product is always even.
Grade 9
Prove that the difference between the squares of any two consecutive odd numbers is always a multiple of 8.
1
Set up consecutive odd integers
Let the consecutive odd numbers be $2n+1$ and $2n+3$, where $n$ is an integer.
2
Square each expression
$$(2n+3)^2 = 4n^2 + 12n + 9$$
$$(2n+1)^2 = 4n^2 + 4n + 1$$
3
Find the difference (larger minus smaller)
$$(2n+3)^2 - (2n+1)^2 = (4n^2+12n+9) - (4n^2+4n+1)$$
$$= 8n + 8$$
4
Factorise to reveal divisibility
$$= 8(n + 1)$$
Since $n+1$ is an integer, $8(n+1)$ is a multiple of 8.
5
Conclusion
Therefore the difference between the squares of any two consecutive odd numbers is always a multiple of 8.
✓ $(2n+3)^2 - (2n+1)^2 = 8(n+1)$, a multiple of 8 for all integers $n$.
❓ Exam Questions
Question 11 mark
Give a counter-example to disprove the statement: "The square of any prime number is always odd."
Counter-example: $p = 2$ (which is prime). $2^2 = 4$, which is even, not odd.
Mark scheme: Any correct prime that gives an even square — only $p=2$ works. Must show the calculation. [1]
Mark scheme: Any correct prime that gives an even square — only $p=2$ works. Must show the calculation. [1]
Question 22 marks
Prove that the sum of any three consecutive integers is always a multiple of 3.
Let the three consecutive integers be $n$, $n+1$, $n+2$.
Sum $= n + (n+1) + (n+2) = 3n + 3 = 3(n+1)$.
Since $n+1$ is an integer, $3(n+1)$ is a multiple of 3.
Therefore the sum of any three consecutive integers is always a multiple of 3. ✓
Mark scheme: Correct setup [1] + factorised form with conclusion [1]
Sum $= n + (n+1) + (n+2) = 3n + 3 = 3(n+1)$.
Since $n+1$ is an integer, $3(n+1)$ is a multiple of 3.
Therefore the sum of any three consecutive integers is always a multiple of 3. ✓
Mark scheme: Correct setup [1] + factorised form with conclusion [1]
Question 33 marks
Prove that $(n+3)^2 - (n-3)^2 \equiv 12n$ for all values of $n$.
LHS $= (n+3)^2 - (n-3)^2$
$= (n^2 + 6n + 9) - (n^2 - 6n + 9)$
$= n^2 + 6n + 9 - n^2 + 6n - 9$
$= 12n$
$= $ RHS
Therefore LHS $\equiv$ RHS. ✓
Mark scheme: Expand both brackets correctly [1] + collect like terms [1] + reach $12n$ with conclusion [1]
$= (n^2 + 6n + 9) - (n^2 - 6n + 9)$
$= n^2 + 6n + 9 - n^2 + 6n - 9$
$= 12n$
$= $ RHS
Therefore LHS $\equiv$ RHS. ✓
Mark scheme: Expand both brackets correctly [1] + collect like terms [1] + reach $12n$ with conclusion [1]
Question 44 marks
$n$ is a positive integer. Prove that $n^2 + n + 1$ is always odd.
Method — by cases:
Case 1: $n$ is even. Let $n = 2k$. Then $n^2 + n + 1 = 4k^2 + 2k + 1 = 2(2k^2 + k) + 1$. This is odd. ✓
Case 2: $n$ is odd. Let $n = 2k+1$. Then $n^2 + n + 1 = (2k+1)^2 + (2k+1) + 1 = 4k^2+4k+1+2k+1+1 = 4k^2+6k+3 = 2(2k^2+3k+1)+1$. This is odd. ✓
In both cases $n^2+n+1$ is odd, so the result holds for all positive integers $n$.
Alternative: $n^2+n+1 = n(n+1)+1$. Product of consecutive integers $n(n+1)$ is always even (one of them is even), so $n(n+1)+1$ is even $+$ 1 = odd. ✓
Mark scheme: Attempt at general form [1] + correct expansion [1] + identify odd form [1] + conclusion [1]
Case 1: $n$ is even. Let $n = 2k$. Then $n^2 + n + 1 = 4k^2 + 2k + 1 = 2(2k^2 + k) + 1$. This is odd. ✓
Case 2: $n$ is odd. Let $n = 2k+1$. Then $n^2 + n + 1 = (2k+1)^2 + (2k+1) + 1 = 4k^2+4k+1+2k+1+1 = 4k^2+6k+3 = 2(2k^2+3k+1)+1$. This is odd. ✓
In both cases $n^2+n+1$ is odd, so the result holds for all positive integers $n$.
Alternative: $n^2+n+1 = n(n+1)+1$. Product of consecutive integers $n(n+1)$ is always even (one of them is even), so $n(n+1)+1$ is even $+$ 1 = odd. ✓
Mark scheme: Attempt at general form [1] + correct expansion [1] + identify odd form [1] + conclusion [1]
Question 54 marks
Prove that $x^2 - 6x + 11 > 0$ for all real values of $x$.
Complete the square:
$x^2 - 6x + 11 = (x - 3)^2 - 9 + 11 = (x-3)^2 + 2$
Since $(x-3)^2 \geq 0$ for all real $x$, we have $(x-3)^2 + 2 \geq 2 > 0$.
Therefore $x^2 - 6x + 11 > 0$ for all real values of $x$. ✓
Mark scheme: Attempt to complete the square [1] + correct form $(x-3)^2 + 2$ [1] + $(x-3)^2 \geq 0$ stated [1] + final conclusion [1]
$x^2 - 6x + 11 = (x - 3)^2 - 9 + 11 = (x-3)^2 + 2$
Since $(x-3)^2 \geq 0$ for all real $x$, we have $(x-3)^2 + 2 \geq 2 > 0$.
Therefore $x^2 - 6x + 11 > 0$ for all real values of $x$. ✓
Mark scheme: Attempt to complete the square [1] + correct form $(x-3)^2 + 2$ [1] + $(x-3)^2 \geq 0$ stated [1] + final conclusion [1]
Question 66 marks
$p$ and $q$ are consecutive even integers, with $q = p + 2$. Prove that $q^2 - p^2$ is always a multiple of 8 but is not always a multiple of 16. You must show working and give an example to justify the second part.
Part 1 — Prove multiple of 8:
Let $p = 2n$ (even), so $q = 2n+2$.
$q^2 - p^2 = (2n+2)^2 - (2n)^2 = 4n^2+8n+4 - 4n^2 = 8n+4 = 4(2n+1)$
Hmm — this is $4(2n+1)$. Since $2n+1$ is odd, this is $4 \times \text{odd}$, which is a multiple of 4 but the odd factor prevents it always being a multiple of 8. Let me recheck the factorisation:
$8n + 4 = 4(2n+1)$. For $n=0$: $4 \times 1 = 4$. This is NOT a multiple of 8.
Note: The claim as stated is actually FALSE in general — $q^2 - p^2 = 4(2n+1)$ which is always a multiple of 4, but NOT always a multiple of 8. For example $p=0, q=2$: $4-0=4$, not divisible by 8.
Corrected proof (as the question intends): Prove $q^2 - p^2$ is always a multiple of 4 and show it's not always a multiple of 8.
$q^2 - p^2 = 4(2n+1)$. Since $2n+1$ is always odd, $4(2n+1)$ is always a multiple of 4. [3]
To show it's NOT always a multiple of 8: take $n=0$, $p=0$, $q=2$: $q^2-p^2=4$. Since $4 \div 8$ is not an integer, this is not a multiple of 8. [3]
Mark scheme: Setup with $p=2n, q=2n+2$ [1] + correct expansion and simplification [1] + factorised form $4(2n+1)$ [1] + conclusion multiple of 4 [1] + valid counter-example given [1] + clear explanation why not always multiple of 8 [1]
Let $p = 2n$ (even), so $q = 2n+2$.
$q^2 - p^2 = (2n+2)^2 - (2n)^2 = 4n^2+8n+4 - 4n^2 = 8n+4 = 4(2n+1)$
Hmm — this is $4(2n+1)$. Since $2n+1$ is odd, this is $4 \times \text{odd}$, which is a multiple of 4 but the odd factor prevents it always being a multiple of 8. Let me recheck the factorisation:
$8n + 4 = 4(2n+1)$. For $n=0$: $4 \times 1 = 4$. This is NOT a multiple of 8.
Note: The claim as stated is actually FALSE in general — $q^2 - p^2 = 4(2n+1)$ which is always a multiple of 4, but NOT always a multiple of 8. For example $p=0, q=2$: $4-0=4$, not divisible by 8.
Corrected proof (as the question intends): Prove $q^2 - p^2$ is always a multiple of 4 and show it's not always a multiple of 8.
$q^2 - p^2 = 4(2n+1)$. Since $2n+1$ is always odd, $4(2n+1)$ is always a multiple of 4. [3]
To show it's NOT always a multiple of 8: take $n=0$, $p=0$, $q=2$: $q^2-p^2=4$. Since $4 \div 8$ is not an integer, this is not a multiple of 8. [3]
Mark scheme: Setup with $p=2n, q=2n+2$ [1] + correct expansion and simplification [1] + factorised form $4(2n+1)$ [1] + conclusion multiple of 4 [1] + valid counter-example given [1] + clear explanation why not always multiple of 8 [1]
⭐ Grade 9 Model Answers
Model Answer — Consecutive Odd Squares (Grade 9 Proof)
Question: Prove that the difference between the squares of any two consecutive odd numbers is always a multiple of 8.
Annotated Grade 9 Answer
Examiner note: Using $2n+1$ and $2n+3$ correctly captures any pair of consecutive odd numbers. Using $n$ and $n+2$ would be wrong — those aren't guaranteed odd. This earns the setup mark.
$(2n+1)^2 = 4n^2 + 4n + 1$
Examiner note: Both expansions must be correct. A sign error here is one of the most common mark losses.
Examiner note: Show the subtraction explicitly. Collect all terms carefully.
Examiner note: The factorisation revealing 8 as a factor is crucial. This is the "proof" step.
Examiner note: The conclusion sentence is explicitly required. "Hence proven" alone does not score this mark — you must reference "multiple of 8".
1
Set up — 1 mark
Let the consecutive odd numbers be $2n+1$ and $2n+3$, where $n \in \mathbb{Z}$.Examiner note: Using $2n+1$ and $2n+3$ correctly captures any pair of consecutive odd numbers. Using $n$ and $n+2$ would be wrong — those aren't guaranteed odd. This earns the setup mark.
2
Squaring — 1 mark each bracket
$(2n+3)^2 = 4n^2 + 12n + 9$$(2n+1)^2 = 4n^2 + 4n + 1$
Examiner note: Both expansions must be correct. A sign error here is one of the most common mark losses.
3
Subtract — 1 mark
$(2n+3)^2 - (2n+1)^2 = (4n^2+12n+9)-(4n^2+4n+1) = 8n+8$Examiner note: Show the subtraction explicitly. Collect all terms carefully.
4
Factorise — 1 mark
$8n + 8 = 8(n+1)$Examiner note: The factorisation revealing 8 as a factor is crucial. This is the "proof" step.
5
Conclusion — 1 mark
Since $n+1$ is an integer, $8(n+1)$ is a multiple of 8. Therefore the difference between the squares of any two consecutive odd numbers is always a multiple of 8. ✓Examiner note: The conclusion sentence is explicitly required. "Hence proven" alone does not score this mark — you must reference "multiple of 8".
Full marks: 5/5 — Setup, two correct expansions, correct subtraction, factorisation revealing ×8, explicit conclusion.
Why This Earns Grade 9
This question combines: choosing the correct algebraic form for odd numbers, expanding squares (a common error point), careful subtraction of polynomials, factorisation identifying the divisor, and a formal conclusion. Mastery of all these steps together is the hallmark of Grade 9 algebra.
📋 Revision Sheet
Key Definitions
| Proof | Shows a result true for ALL cases |
| Counter-example | One case proving a statement false |
| Identity (≡) | True for all values of variables |
| Divisibility | $N$ divisible by $k$ means $N=km$ |
| Consecutive | Differ by 1 (integers in order) |
Essential Formulae
Even: $2n$ | Odd: $2n+1$
Consec.: $n, n+1, n+2$
Consec. even: $2n, 2n+2$
Consec. odd: $2n+1, 2n+3$
Multiple of $k$: $kn$
CTS: $(x+p)^2 + q$ to prove always positive
$(a+b)^2 = a^2+2ab+b^2$
Memory Hooks
- "2n is EVEN — two's company"
- "2n+1 is ODD — add one more"
- "Different numbers → different letters"
- "To PROVE: algebra. To DISPROVE: one example"
- "Always conclude in words"
- "CTS for always positive: $(x+p)^2 \geq 0$"
- "Factorise to reveal the factor"
Exam Tips
- State your variable: "let $n$ be any integer"
- Different variables for different unknowns
- Show ALL algebraic steps — no skipping
- Final sentence must name the property
- For identities: work one side only
- For counter-examples: give one specific value
- Check expansions: $(a-b)^2 \neq a^2-b^2$
- Divisibility: write $k \times (\text{integer})$
🔄 Flashcards
Click a card to reveal the answer.
✗ Common Mistakes
Mistake 1 — Using the Same Variable for Two Different Numbers
What students do: Write $2n + 2n$ for the sum of two even numbers.Why marks are lost: $2n + 2n = 4n$, which implies both numbers are the same — the proof only covers one specific case, not all pairs of even numbers.
How to avoid: Always use different letters: $2m + 2n = 2(m+n)$. Explicitly state "$m$ and $n$ are integers."
Mistake 2 — Not Writing a Conclusion
What students do: Stop after showing $= 2(m+n+1)$ without adding a sentence.Why marks are lost: The final "conclusion" mark is explicitly awarded for a sentence such as "therefore the result is always even." Missing it costs 1 mark on almost every proof question.
How to avoid: Always end with "Therefore [restate the claim]." Make it a habit.
Mistake 3 — Verification Instead of Proof
What students do: Show the result for $n = 1, 2, 3$ and claim "it works every time."Why marks are lost: Numerical verification — no matter how many examples — is not a proof. Examiners award 0 marks for "proof" questions answered by examples only.
How to avoid: Never substitute numbers in a proof. Use variables from the very first line.
Mistake 4 — Expanding Brackets Incorrectly
What students do: Write $(2n+3)^2 = 4n^2 + 9$ (forgetting the middle term).Why marks are lost: All subsequent working is wrong, losing multiple method marks and the accuracy mark.
How to avoid: Always use FOIL or the identity $(a+b)^2 = a^2 + 2ab + b^2$. Check: the middle term is $2 \times 2n \times 3 = 12n$.
Mistake 5 — Choosing the Wrong General Form
What students do: Represent "consecutive odd numbers" as $n$ and $n+2$ without ensuring these are odd.Why marks are lost: $n$ and $n+2$ are consecutive even numbers if $n$ is even. The proof is invalid.
How to avoid: For consecutive odd numbers, always use $2n+1$ and $2n+3$, which are guaranteed to be odd for any integer $n$.
Mistake 6 — Incomplete Factorisation in Divisibility Proofs
What students do: Reach $8n + 8$ and state "this contains an 8 so it's a multiple of 8" without factorising.Why marks are lost: The factorisation $8(n+1)$ is required to formally show that 8 is a factor. The statement "contains an 8" is ambiguous and does not constitute a proof.
How to avoid: Always fully factorise: write $8(n+1)$ and then state "since $n+1$ is an integer, this is a multiple of 8."
✅ Final Checklist
Click each item to mark it complete. Progress is saved automatically.
- I can represent even integers as $2n$
- I can represent odd integers as $2n+1$
- I can write consecutive integers as $n, n+1, n+2$
- I can write consecutive even integers as $2n, 2n+2$
- I can write consecutive odd integers as $2n+1, 2n+3$
- I use different variables for two different even/odd numbers
- I can prove sums/products of integers are even or odd
- I can prove divisibility by factorising out the divisor
- I can complete the square to prove an expression is always positive
- I can prove algebraic identities by manipulating one side only
- I always write a clear conclusion sentence naming the property
- I can construct a counter-example to disprove a false statement
- I expand $(a+b)^2$ correctly as $a^2+2ab+b^2$
- I never substitute specific numbers in a proof question
- I have practised at least 3 full Grade 9 proof questions