Algebraic proof: even, odd and consecutive numbers
What this lesson covers
- An example shows a statement can be true; only algebra shows it is always true.
- Even numbers can be written 2n, odd numbers 2n + 1, and consecutive numbers n, n + 1, n + 2.
Examples are not proof
- Three consecutive numbers, three times: a multiple of 3 each time.
- But there are infinitely many cases nobody has tried.
- Algebra checks every case at once.
| Three in a row | Sum | Multiple of 3? |
|---|---|---|
| 4 + 5 + 6 | 15 | yes, 3 × 5 |
| 10 + 11 + 12 | 33 | yes, 3 × 11 |
| 99 + 100 + 101 | 300 | yes, 3 × 100 |
| Every other case? | never checked | examples cannot say |
Writing any number with a letter
- n stands for any integer at all.
- An even number is 2n. An odd number is 2n + 1.
- Consecutive numbers go n, n + 1, n + 2.
| Kind of number | Written as | Why |
|---|---|---|
| Even | 2n | two times an integer |
| Odd | 2n + 1 | one more than an even |
| Consecutive | n, n + 1, n + 2 | each one more than the last |
| Multiple of k | kn | k times an integer |
| Consecutive even | 2n, 2n + 2 | even numbers go up in 2s |
Two numbers, two letters
- Using 2n + 1 twice means the same odd number twice.
- Two different odd numbers are 2m + 1 and 2n + 1.
- Say that m and n are integers, once, at the start.
| Writing | Means | Covers |
|---|---|---|
| 2n + 1 and 2n + 1 | one odd number, twice | only 7 + 7, 9 + 9 |
| 2m + 1 and 2n + 1 | any two odd numbers | every case |
| m and n | both integers | say so at the start |
Proof: three consecutive integers
- Let the numbers be n, n + 1 and n + 2.
- Add them and collect: 3n + 3.
- Take out the common factor: 3(n + 1).
| Step | Working | Reason |
|---|---|---|
| Define | n, n + 1, n + 2 | n is any integer |
| Add | n + (n + 1) + (n + 2) | the sum in the claim |
| Collect | 3n + 3 | like terms |
| Factorise | 3(n + 1) | take out the 3 |
Writing the last line
- Say what the algebra shows.
- Say why that settles it: name the integer.
- Then restate the claim, in words.
| Part | Wording |
|---|---|
| The algebra | The sum is 3(n + 1). |
| Why it settles it | n + 1 is an integer, so this is a multiple of 3. |
| The claim again | Three consecutive integers add to a multiple of 3. |
| Too vague | So it works. |
Proof: odd plus odd is even
- Two odd numbers: 2m + 1 and 2n + 1.
- The sum collects to 2m + 2n + 2.
- Factorise out 2: two times an integer is even.
| Step | Working | Reason |
|---|---|---|
| Define | 2m + 1 and 2n + 1 | m, n integers |
| Add | 2m + 2n + 2 | collect like terms |
| Factorise | 2(m + n + 1) | take out the 2 |
| Conclude | m + n + 1 is an integer | so the sum is even |
Proof: odd times odd is odd
- Expand the two brackets: four terms.
- Three of them have a factor of 2. The 1 does not.
- Two times an integer, plus one, is odd.
| Step | Working | Reason |
|---|---|---|
| Define | 2m + 1 and 2n + 1 | m, n integers |
| Expand | 4mn + 2m + 2n + 1 | every term by every |
| Group evens | 2(2mn + m + n) + 1 | take 2 out of three |
| Conclude | 2 × integer, plus 1 | so the product is odd |
Proof: consecutive squares
- The larger square take the smaller: (n + 1)² − n².
- Expand, and the n² terms cancel, leaving 2n + 1.
- n + (n + 1) is also 2n + 1. They are equal.
| Step | Working | Note |
|---|---|---|
| Define | n and n + 1 | n is any integer |
| Difference | (n + 1)² − n² | larger square first |
| Expand, n² cancels | 2n + 1 | n² − n² = 0 |
| Their sum | n + (n + 1) | also 2n + 1 |
| Check: 7 and 8 | 64 − 49 = 15 | 7 + 8 = 15 |
A number trick
- Think of a number, add 4, multiply by 3, subtract 6.
- Divide by 3, then take away the number you started with.
- Five gives 2. Twenty gives 2. Why?
| Instruction | Start with 5 | Start with 20 |
|---|---|---|
| Think of a number | 5 | 20 |
| Add 4 | 9 | 24 |
| Multiply by 3 | 27 | 72 |
| Subtract 6 | 21 | 66 |
| Divide by 3, take away start | 7, then 2 | 22, then 2 |
Proving the trick
- Call the starting number n.
- Each instruction becomes one line of algebra.
- The n cancels at the end, so the answer is always 2.
| Instruction | Algebra | Simplified |
|---|---|---|
| Think of a number | n | n |
| Add 4, times 3 | 3(n + 4) | 3n + 12 |
| Subtract 6 | 3n + 12 − 6 | 3n + 6 |
| Divide by 3 | (3n + 6) ÷ 3 | n + 2 |
| Take away n | n + 2 − n | 2, for every n |
Disproving: one counterexample
- To show a claim is always true, you need a proof.
- To show it is false, you need one case where it fails.
- Nine is odd and not prime. That one case is enough.
| Question | To show true | To show false |
|---|---|---|
| Always true? | a proof for every case | one case that fails |
| How many examples? | examples are never enough | one is enough |
| Odd means prime? | no proof exists | 9 = 3 × 3 is not |
Is n² + n + 41 always prime?
- n = 1, 2, 3 give 43, 47, 53. All prime.
- It stays prime for every n from 0 to 39.
- At n = 40 it is 1681, which is 41 × 41.
| n | n² + n + 41 | Prime? |
|---|---|---|
| 1 | 43 | yes |
| 2 | 47 | yes |
| 3 | 53 | yes |
| 0 to 39 | forty values | every one prime |
| 40 | 1600 + 40 + 41 = 1681 | no: 41 × 41 |
The shape of every proof
- Define, form, simplify.
- Factorise into the form the claim needs.
- Then conclude in a sentence.
| Step | What you write |
|---|---|
| 1. Define | Let n be an integer. Write each number using it. |
| 2. Form | Write the sum, product or difference in the claim. |
| 3. Simplify | Expand and collect like terms. |
| 4. Target form | Factorise to 2( ), 2( ) + 1 or k( ). |
| 5. Conclude | Name the integer and property, restate the claim. |
What to take away
- Seven lines. The first is why the rest exist.
| Examples | show it can be true, never always |
| Even, odd | 2n and 2n + 1 |
| Consecutive | n, n + 1, n + 2 |
| Two different numbers | two different letters |
| Target form | 2( ) even, 2( ) + 1 odd, k( ) multiple of k |
| One counterexample | is enough to disprove |
| Last line | say why it is true, then restate the claim |
Your turn: four questions
- Pause here and write each one as a full proof.
- Question 1 has two different numbers: two letters.
- Every proof ends with a sentence.
| Question | |
|---|---|
| 1 | Prove that an even number plus an odd number is always odd. |
| 2 | Prove the sum of four consecutive integers is always even. |
| 3 | Find a counterexample: the sum of two primes is always even. |
| 4 | Prove (2n + 3)² − (2n + 1)² is always a multiple of 8. |
Show answer
1. Let the numbers be 2m and 2n + 1, with m and n integers. The sum is 2m + 2n + 1 = 2(m + n) + 1. Since m + n is an integer, this is one more than an even number, so the sum is always odd. 2. n + (n + 1) + (n + 2) + (n + 3) = 4n + 6 = 2(2n + 3). Since 2n + 3 is an integer, the sum is always even. 3. 2 + 3 = 5. Both are prime and 5 is odd, so the claim is false. 4. Expanding gives 4n² + 12n + 9 − 4n² − 4n − 1 = 8n + 8 = 8(n + 1). Since n + 1 is an integer, the difference is always a multiple of 8.
The answers
- 2m + 2n + 1 = 2(m + n) + 1, so odd.
- 4n + 6 = 2(2n + 3), so even.
- 2 + 3 = 5, which is odd. False.
- 8n + 8 = 8(n + 1), a multiple of 8.
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