To explain a rule with algebra, represent the objects in the claim with letters, simplify, and show that the result has the property you claimed, whatever the letters are. The argument then covers every case, not only the ones you checked.
This lesson comes after you have tested a conjecture for counterexamples and before you state the limits of a conclusion. It belongs to International Mathematics investigations and uses skills from algebraic structure.
What do you need before writing the algebra?
You need a way to write the objects in general form. Some common choices are below.
| Object | Algebraic form |
|---|---|
| Any whole number | n |
| Consecutive whole numbers | n, n + 1, n + 2 |
| Even number | 2n |
| Odd number | 2n + 1 or 2n − 1 |
| Consecutive even numbers | 2n, 2n + 2 |
| Consecutive odd numbers | 2n − 1, 2n + 1 |
| Multiple of 5 | 5n |
How to explain a rule, step by step
- State what you are explaining, in words.
- Choose algebraic forms for the objects, and say what n is.
- Write the expression for the quantity in the claim.
- Simplify, expanding or collecting like terms.
- Factorise or rearrange until the property you want is visible.
- Write a concluding sentence saying what it shows and for which values.
Worked example
Claim: the sum of any two consecutive odd numbers is a multiple of 4. First check a few cases: 7 + 9 = 16 and 13 + 15 = 28. Both are multiples of 4, so the claim is worth proving.
Step 1, forms: let n be a positive whole number. The odd numbers 2n − 1 and 2n + 1 differ by 2, so they are consecutive odd numbers.
Step 2, expression: the sum is (2n − 1) + (2n + 1).
Step 3, simplify: 2n − 1 + 2n + 1 = 4n.
Step 4, conclude: 4n is 4 times a whole number, so it is a multiple of 4 for every n. Hence the sum of any two consecutive odd numbers is a multiple of 4.
Link back to the examples: 7 + 9 has 2n − 1 = 7, so n = 4 and 4n = 16. For 13 + 15, n = 7 and 4n = 28. Both match the checks.
The mistake to watch for
A common slip is to write consecutive odd numbers as n and n + 1.
Mistaken setup: n + (n + 1) = 2n + 1.
But n and n + 1 are consecutive whole numbers, so one of them is even. The sum 2n + 1 is odd, not a multiple of 4.
The correction is to check your algebraic forms with a number. If n = 7, then n and n + 1 give 7 and 8, which are not both odd. The forms 2n − 1 and 2n + 1 give 13 and 15 when n = 7, which are both odd and two apart.
A five-second check of the setup saves the whole argument.
Check yourself
1. Show that the sum of any three consecutive whole numbers is a multiple of 3.
Show answer
Let the numbers be n, n + 1, n + 2. Their sum is n + n + 1 + n + 2 = 3n + 3 = 3(n + 1).
This is 3 times a whole number, so it is a multiple of 3 for every n. Check: 10 + 11 + 12 = 33 = 3 × 11.
2. Show that the sum of two consecutive even numbers is never a multiple of 4.
Show answer
Let the numbers be 2n and 2n + 2. Their sum is 4n + 2.
4n is a multiple of 4, so 4n + 2 is 2 more than a multiple of 4, and is never a multiple of 4. Check: 6 + 8 = 14 = 4 × 3 + 2.
3. Matchsticks make a row of n squares. The first square needs 4 sticks and each extra square needs 3. Show that the total is 3n + 1.
Show answer
There are n − 1 extra squares, so the total is 4 + 3(n − 1). Expand: 4 + 3n − 3 = 3n + 1.
Check: n = 4 gives 3 × 4 + 1 = 13, which matches the table of 4, 7, 10, 13.
Where this leads next
Once a rule is explained, the last step is to say how far it applies: state limits of a pattern-based conclusion. The investigations practice set then mixes every stage. The non-calculator working trainer is useful for testing your algebraic result with a few numbers.
Many students can do the algebra once they see the setup. A teacher can help you practise choosing that setup in online one-to-one Mathematics tuition.