Algebraic proof

• "Show that" means show every step to reach the answer you're given. A final line that matches isn't enough without the working.
• "Prove" means show something is true in every case, using algebra. Trying a few numbers doesn't prove it.

To show two expressions are the same:

1. Start with one side.
2. Expand and simplify step by step.
3. Finish with exactly the other side.

Example: show that 3(x + 4) − 2(x − 1) ≡ x + 14.

1. 3(x + 4) − 2(x − 1)
2. = 3x + 12 − 2x + 2
3. = x + 14, as required

Proofs about whole numbers start by writing them in algebra, where n is any integer.

1. Write each number in algebra.
2. Do what the question says to them, and simplify.
3. Factorise to show the result has the property needed.

Example: prove that the sum of three consecutive integers is a multiple of 3.

1. Let the integers be n, n + 1 and n + 2
2. n + (n + 1) + (n + 2) = 3n + 3
3. = 3(n + 1), which is a multiple of 3

To show a statement is false, you only need one example where it doesn't work: a counterexample.

1. Try values, including awkward ones: 1, 2, 0, negatives and fractions.
2. When one fails, write it down and explain why it fails.

Example: Sam says, "The sum of two prime numbers is always even." Show that Sam is wrong.

1. 2 and 3 are both prime
2. 2 + 3 = 5, which is odd
3. So the statement is false

Example: show that "x² is always bigger than x" is false.

1. Try x = ½: x² = ¼, which is smaller than ½
2. So the statement is false

Some proofs need expanding brackets, including squares, before you factorise.

Example: prove that the sum of two consecutive odd numbers is a multiple of 4.

1. Let the odd numbers be 2n + 1 and 2n + 3
2. (2n + 1) + (2n + 3) = 4n + 4
3. = 4(n + 1), which is a multiple of 4

Harder example: prove that (n + 3)² − (n + 1)² is always a multiple of 4.

1. (n + 3)² = n² + 6n + 9
2. (n + 1)² = n² + 2n + 1
3. (n² + 6n + 9) − (n² + 2n + 1) = 4n + 8
4. = 4(n + 2), which is a multiple of 4

Put the second square in brackets before subtracting, so every term in it is subtracted.