2.2 - Constructing algebraic proofs

Use algebra to show that a claim holds for every permitted value. You will represent whole families of numbers, reveal factors and prove identities. You will also see why one failed case can settle a question that a thousand successful checks cannot.

Representing a general claim (Higher)

A calculator gives $4^3-4=60$ and $5^3-5=120$ . Both are divisible by $6$ . Is $n^3-n$ divisible by $6$ for every integer $n$ , including negative integers? We need an argument that covers them all.

You already know

An integer is a whole number, including zero and negative whole numbers. Factorising rewrites a sum as a product without changing its value.

Checking examples suggests a pattern. It does not prove an infinite claim: an untested value could behave differently. A variable lets one calculation represent every allowed case.

Proof

A chain of valid reasoning establishing a claim for every case in its stated domain, the set of allowed values. For an arbitrary integer $n$ , $2n$ represents an even integer and $2n+1$ an odd integer.

The integer condition matters. If $n=\frac14$ , then $2n=\frac12$ , which is not an even integer. Consecutive integers are $n$ , $n+1$ and $n+2$ ; consecutive odd integers differ by $2$ . These representations and proof methods are absent from the Higher formulae sheet: learn to construct them.

Worked example 1: consecutive odd integers

Question. Prove that the sum of two consecutive odd integers is even.

Plan. Represent both numbers using one arbitrary integer, then show their sum is twice an integer.

Let $n$ be any integer.

\begin{aligned} \text{numbers}&=2n+1,\ 2n+3 &&\text{odd, separated by }2\\ \text{sum}&=(2n+1)+(2n+3) &&\text{add the numbers}\\ &=4n+4=2(2n+2) &&\text{expose a factor }2 \end{aligned}

Why factor out $2$ ? Being twice an integer is exactly what “even” means.

Answer. Since $2n+2$ is an integer, the sum is even for every integer $n$ .

Check. Take $n=-2$ : the numbers are $-3$ and $-1$ , with even sum $-4$ . This supports the proof but does not replace it.

Change only the parity of the two consecutive numbers. Consecutive even integers still differ by $2$ .

Worked example 2: consecutive even integers

Question. Prove that the sum of two consecutive even integers is even.

Plan. Use $2n$ and $2n+2$ , then factor their sum.

For any integer $n$ ,

\begin{aligned} \text{sum}&=2n+(2n+2) &&\text{represent every pair}\\ &=4n+2 &&\text{collect terms}\\ &=2(2n+1) &&\text{factor }2 \end{aligned}

Why use one variable? It guarantees the second number is exactly $2$ more than the first.

Answer. The sum is even because $2n+1$ is an integer.

Check. With $n=3$ , $6+8=14$ , agreeing with $2(7)$ .

The representation must encode the relationship as well as the type of number.

Watch out

Wrong: “Consecutive odd integers are $n$ and $n+2$ , for any integer $n$ .” This includes even pairs. Correct: use $2n+1$ and $2n+3$ , with $n$ an integer.

Now change the gap to $1$ , so the pair is consecutive integers of different parity.

Proving divisibility (Higher)

An integer is divisible by a positive integer $k$ if it equals $k$ times an integer. Factorisation often reveals this. For example, $n(n+1)$ is even: if $n$ is even it supplies a factor $2$ ; if $n$ is odd, $n+1$ supplies it.

Three consecutive integers include a multiple of $3$ . Their remainders on division by $3$ run through $0$ , $1$ and $2$ , in some order. These arguments also include negative integers and zero. No divisibility proof rule is supplied on the Higher formulae sheet.

Worked example 3: the opening claim

Question. Prove that $n^3-n$ is divisible by $6$ for every integer $n$ .

Plan. Factor into consecutive integers and identify factors $2$ and $3$ .

\begin{aligned} n^3-n&=n(n^2-1) &&\text{take out }n\\ &=(n-1)n(n+1) &&\text{difference of squares} \end{aligned}

Among these three consecutive integers, at least one is even and at least one is a multiple of $3$ .

Why does that give $6$ ? The factors $2$ and $3$ have no common factor greater than $1$ , so their product divides the expression.

Answer. The product, and hence $n^3-n$ , is divisible by $6$ for every integer $n$ .

Check. At $n=-2$ , $-8+2=-6=6(-1)$ . At $n=0$ , the result is $0=6\times0$ , also a multiple.

Shift the block of three consecutive integers one place to the right. The factor argument stays the same, even though the expanded polynomial looks different.

Worked example 4: a shifted consecutive product

Question. Prove that $n^3+3n^2+2n$ is divisible by $6$ for every integer $n$ .

Plan. Factor the polynomial to reveal three consecutive integers.

\begin{aligned} n^3+3n^2+2n&=n(n^2+3n+2) &&\text{common factor }n\\ &=n(n+1)(n+2) &&\text{factor the quadratic} \end{aligned}

These three consecutive integers contain an even factor and a factor divisible by $3$ .

Why factor instead of checking values? The consecutive structure proves both divisibility statements for every integer input.

Answer. Since $2$ and $3$ are coprime, the product is divisible by $6$ .

Check. At $n=2$ , $8+12+4=24=2\times3\times4$ , a multiple of $6$ .

Combining factors needs care when they overlap.

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Number

Algebra

Shape, space and measure

Handling data

2.2 - Constructing algebraic proofs

Representing a general claim (Higher)

Proving divisibility (Higher)

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