Algebraic Proof [H]
Algebraic Proof
A proof is a rigorous argument showing that a mathematical statement is always true — not just for specific examples, but for every possible case. In GCSE algebra, proofs typically show that sums, products or differences of numbers with particular properties (even, odd, consecutive, etc.) always have a predictable result. The power of algebra is that a single general expression can stand in for every possible number simultaneously.
Algebraic Representations
Choose the right algebraic form before beginning — this is the most important step in any proof.
| Number type | Algebraic form | Why |
|---|---|---|
| Any integer | \( n \) | No restriction on \(n\) |
| Even number | \( 2n \) | Divisible by 2 by definition |
| Odd number | \( 2n+1 \) | One more than an even number |
| Consecutive integers | \( n, n+1, n+2, ldots \) | Each is one more than the previous |
| Consecutive even numbers | \( 2n, 2n+2, 2n+4 \) | Even numbers separated by 2 |
| Consecutive odd numbers | \( 2n+1, 2n+3, 2n+5 \) | Odd numbers separated by 2 |
| Multiple of \( k \) | \( kn \) | Any integer \(n\) times \(k\) |
Proof Structure
A well-presented proof has three parts:
- Define: State clearly what the general expressions represent (e.g. "Let \( n \) be any integer. Then \( 2n \) and \( 2n+2 \) are consecutive even numbers.").
- Calculate: Expand and simplify the algebraic expression.
- Conclude: State explicitly why the result proves the claim (e.g. "This is of the form \( 2k \) where \( k \) is an integer, so it is even, as required.").
Worked Examples
Prove that the product of two odd numbers is always odd.
Let the two odd numbers be \( 2m+1 \) and \( 2n+1 \), where \( m \) and \( n \) are integers.
\[ (2m+1)(2n+1) = 4mn + 2m + 2n + 1 = 2(2mn+m+n) + 1 \]Since \( 2mn+m+n \) is an integer, this is of the form \( 2k+1 \) — an odd number. Therefore the product of two odd numbers is always odd. ✓
Prove that the sum of any four consecutive integers is divisible by 2 but not always by 4.
Let the four consecutive integers be \( n, n+1, n+2, n+3 \).
\[ n + (n+1) + (n+2) + (n+3) = 4n + 6 = 2(2n+3) \]This is always divisible by 2 (the factor of 2 is explicit). It is divisible by 4 only if \( 2n+3 \) is even — but \( 2n+3 \) is always odd (even + odd = odd). So the sum is never divisible by 4. ✓
Prove that the square of an odd number is always odd.
Let the odd number be \( 2n+1 \).
\[ (2n+1)^2 = 4n^2 + 4n + 1 = 4n(n+1) + 1 \]Since \( n(n+1) \) is the product of consecutive integers, one of them is even, so \( n(n+1) \) is even. Thus \( 4n(n+1) \) is divisible by 8 — and certainly even. Adding 1 gives an odd number. ✓
Alternatively: \( 4n^2 + 4n + 1 = 2(2n^2+2n) + 1 \), which is of the form \( 2k+1 \) — odd. ✓
Key Takeaways
- Algebraic proof requires showing a statement is true for all integers, not just checking examples.
- Use \( 2n \) for any even number, \( 2n+1 \) for any odd number, \( n, n+1, n+2 \) for consecutive integers.
- Never use the same letter for two independent general integers — use \( m \) and \( n \) when there are two different general integers.
- The conclusion must explicitly state why the algebra proves the claim: "this is of the form \( 2k \), so it is even".
- When a result is divisible by \( k \), the algebraic expression should contain a clear factor of \( k \).