Formulae, identities, proof and functions

Maths · Algebra. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A formula links several quantities. Changing the subject means rearranging it so a different letter stands alone on one side. Use inverse operations, doing the same thing to both sides, in the reverse order to how the formula was built.

An identity (≡) is true for every value, so the two sides must match term by term. A proof uses algebra to show something is true for all numbers, not just the ones you tried. Write an even number as 2n and an odd number as 2n + 1.

A function is a rule that turns an input x into an output f(x). The inverse function undoes it. A composite function applies one function and then another.

Rules to remember

Changing the subject: undo each operation with its inverse (+ and −, × and ÷, square and square root), the same on both sides.
If the new subject appears twice: remove fractions, collect its terms on one side, factorise, then divide.
Identity: expand one side and compare the x² terms, the x terms and the constants.
For any integer n: 2n is even, 2n + 1 is odd, n and n + 1 are consecutive. A multiple of k is k × (an integer).
Inverse function f⁻¹(x): write y = f(x), rearrange to make x the subject, then swap the letter y for x.
Composite function: fg(x) = f(g(x)). Do g first, then f. fg(x) and gf(x) are usually different.

Common mistake

Doing a composite function in the wrong order. In fg(x) the function nearest to x, which is g, is done first.

Worked example

Prove that (n + 3)² − (n − 1)² is a multiple of 8 for every integer n.

Expand: (n + 3)² = n² + 6n + 9 and (n − 1)² = n² − 2n + 1.
Subtract: n² + 6n + 9 − n² + 2n − 1 = 8n + 8.
Factorise: 8n + 8 = 8(n + 1).
n + 1 is an integer, so 8(n + 1) is 8 times an integer.

Answer: The expression equals 8(n + 1), so it is always a multiple of 8.

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