A6 - Algebraic proof

Algebra can settle a claim about every integer without checking the integers one by one. You will represent whole families of numbers, prove parity and divisibility statements, and spot arguments that leave out cases or use invalid operations.

Representing integer families (Higher)

Can the squares of two consecutive integers ever differ by an even number? For $4$ and $5$ , the difference is $25-16=9$ ; for $10$ and $11$ , it is $121-100=21$ . Both are odd. To settle every case, we need expressions that include every possible pair.

You already know

An identity holds for every permitted value. Expanding $(n+1)^2$ gives $n^2+2n+1$ , including the middle term.

Consecutive integers are $n$ , $n+1$ , $n+2$ , where $n$ is an integer. Adding one moves to the next integer, even from a negative starting point. The list $n$ , $2n$ , $3n$ does not generally describe consecutive integers.

Even, odd and multiple

An even integer has the form $2n$ , and an odd integer has the form $2n+1$ , where $n$ is an integer. A multiple of a fixed integer $k$ has the form $kn$ , with integer $n$ . For example, $0=2(0)$ is even; $1$ is not even.

Consecutive even integers differ by two. Consecutive odd integers also differ by two. Build that gap into the expressions.

Worked example 1: consecutive even integers

Question. Represent three consecutive even integers. Show how your expressions include $-6$ , $-4$ , $-2$ .

Plan. Start with twice an integer and add two each time.

\begin{aligned} &2n,\ 2n+2,\ 2n+4 && \text{use gaps of two} \\ 2n&=-6 && \text{match the first integer} \\ n&=-3 && \text{divide by two} \\ 2(-3)+2&=-4 && \text{second integer} \\ 2(-3)+4&=-2 && \text{third integer} \end{aligned}

Why this form? The factor $2$ makes every member even, while the added twos make the members consecutive within the even numbers.

Answer. $2n$ , $2n+2$ , $2n+4$ , with integer $n$ ; $n=-3$ gives the requested triple.

Check. The two gaps are both $2$ , and all three values are divisible by $2$ .

Change only the family from even to odd. The gap stays two; the starting form changes.

Worked example 2: consecutive odd integers

Question. Represent three consecutive odd integers. Show how your expressions include $-5$ , $-3$ , $-1$ .

Plan. Start one above an even integer, then add two each time.

\begin{aligned} &2n+1,\ 2n+3,\ 2n+5 && \text{odd forms, gaps of two} \\ 2n+1&=-5 && \text{match the first integer} \\ 2n&=-6 && \text{subtract one} \\ n&=-3 && \text{divide by two} \\ 2(-3)+3&=-3 && \text{second integer} \\ 2(-3)+5&=-1 && \text{third integer} \end{aligned}

Why subtract one first? The odd form is twice an integer plus one; both operations must be undone to recover its parameter.

Answer. $2n+1$ , $2n+3$ , $2n+5$ , with integer $n$ ; again $n=-3$ gives the triple.

Check. Each value is one more than an even integer, including the negative values.

Consecutive multiples of five are $5n$ , $5(n+1)$ , $5(n+2)$ . If two integers can vary independently, use different parameters: any two odd integers are $2m+1$ and $2n+1$ . They need not be equal or consecutive. The alternative odd form $2n-1$ is equally valid.

Watch out

Wrong: “ $2n$ is even for any real $n$ .” At $n=\frac12$ , it equals $1$ . Right: state that $n$ is an integer. The condition makes the representation work.

Keep the odd family, but change the starting triple for your turn.

General proofs of parity and divisibility (Higher)

A proof starts with every permitted case and finishes with a form that guarantees the claimed property.

To prove divisibility by $k$ , aim for $kQ$ , where $Q$ is an integer. To prove oddness, aim for $2Q+1$ . The final interpretation connects the algebra to the claim.

These integer forms and proof rules are not on the formula sheet: learn the forms and construct the argument when asked.

Worked example 3: adding two odd integers

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

Plan. Represent the integers independently, add them and factor out two.

Let the odd integers be $2m+1$ and $2n+1$ , where $m,n$ are integers. Then

\begin{aligned} (2m+1)+(2n+1)&=2m+2n+2 && \text{remove the brackets} \\ &=2(m+n+1) && \text{factor out two} \end{aligned}

Why use two letters? “Any two” permits unequal choices; one repeated expression would cover only equal pairs.

Answer. $m+n+1$ is an integer, so the sum is twice an integer and is even.

Check. The allowed pair $-5,9$ gives $4$ , agreeing with the proof. This checks the algebra; the general argument proves the claim.

Change only the operation from addition to multiplication. The same representations now need an odd final form.

Worked example 4: multiplying two odd integers

Question. Prove algebraically that the product of any two odd integers is odd.

Plan. Expand the product, then separate one from an even expression.

For integers $m,n$ ,

\begin{aligned} (2m+1)(2n+1)&=4mn+2m+2n+1 && \text{multiply every term} \\ &=2(2mn+m+n)+1 && \text{separate the final one} \end{aligned}

Why leave the one outside? The target is twice an integer plus one, the definition of an odd integer.

Answer. $2mn+m+n$ is an integer, so the product is odd.

Check. For $-3$ and $7$ , the product is $-21=2(-11)+1$ , which is odd.

Now return to the opening squares. For consecutive integers $n,n+1$ ,

\begin{aligned} (n+1)^2-n^2&=n^2+2n+1-n^2 && \text{expand the square} \\ &=2n+1 && \text{cancel the squared terms} \end{aligned}

This is odd for every integer $n$ . The difference can never be even. A neat consequence is that consecutive squares are separated by successive odd numbers, not by a fixed gap.

Divisibility uses the same reasoning: $5n^2+10n=5(n^2+2n)$ , so it is divisible by five for integer $n$ . Also, three consecutive integers have sum $n+(n+1)+(n+2)=3(n+1)$ , hence divisible by three.

Some arguments split into cases. The product $n(n+1)$ is even: if $n=2k$ , it is $2k(2k+1)$ ; if $n=2k+1$ , it is $2(2k+1)(k+1)$ . Every integer is even or odd, and both cases give twice an integer. Likewise, an even integer plus an odd integer is $2m+(2n+1)=2(m+n)+1$ , which is odd.

Watch out

Wrong: “ $5+n$ contains a $5$ , so it is divisible by five.” The $5$ is a term, rather than a factor of the whole expression. Right: reach $5Q$ and explain why $Q$ is an integer.

Your turn keeps multiplication but changes the family to even integers.

Assessing the validity of a proof (Higher)

A convincing argument must include every permitted case and use valid operations at every step.

Checking $1+2+3=6$ , $2+3+4=9$ and $3+4+5=12$ supports a conjecture about consecutive integers. It does not establish every case. In contrast, one permitted failure settles a universal claim immediately.

Counterexample

A counterexample is one case allowed by a claim for which the claim is false. For example, $n=1$ disproves “every integer has an even square”, since $1^2=1$ is odd. A non-integer cannot disprove a claim restricted to integers.

Inspect the starting expressions before admiring the algebra. Correct algebra about the wrong family is still an incomplete argument.

Worked example 5: a claim about a sum

Question. A student writes: “Any two integers are $2m$ and $2n$ , with integer $m,n$ . Their sum is $2(m+n)$ , so every integer pair has an even sum.” Explain the error and give a counterexample.

Plan. Check what the representations include, then choose an omitted pair.

\begin{aligned} 2m+2n&=2(m+n) && \text{valid for even integers} \\ 3+4&=7=2(3)+1 && \text{test an omitted pair} \end{aligned}

Why choose one odd and one even? The proposed forms exclude odd integers, so an omitted pair can expose the gap.

Answer. The forms represent only even integers. The integers $3,4$ have odd sum $7$ , disproving the claim.

Check. Both values are integers, so the counterexample satisfies the claim's domain.

Change only the operation from a sum to a product. Choose a counterexample suited to the new claim.

Worked example 6: a claim about a product

Question. A student writes: “Any two integers are $2m$ and $2n$ , with integer $m,n$ . Their product is $4mn$ , so every integer pair has an even product.” Explain the error and give a counterexample.

Plan. Identify the same missing family, then test two odd integers.

\begin{aligned} (2m)(2n)&=4mn && \text{valid for even integers} \\ 3\times5&=15=2(7)+1 && \text{an odd product} \end{aligned}

Why use two odd integers this time? One even factor would make the product even and would not disprove the claim.

Answer. The representation excludes odd integers. The pair $3,5$ gives odd product $15$ , so the claim is false.

Check. The calculation supplies a failure, rather than merely a weakness in the proposed proof.

A true claim can also have a faulty proof. Writing both arbitrary odd integers as $2n+1$ shows only that an integer minus itself is zero. Repair the difference argument with independent parameters: $(2m+1)-(2n+1)=2(m-n)$ . The bracket is an integer, so the difference is even.

An invalid operation is another danger. If $a=b=1$ , the following steps are valid:

\begin{aligned} a^2&=ab && \text{use }a=b \\ a^2-b^2&=ab-b^2 && \text{subtract }b^2 \\ (a-b)(a+b)&=b(a-b) && \text{factorise} \end{aligned}

Cancelling $a-b$ would appear to give $a+b=b$ , or $2=1$ . But $a-b=0$ : that cancellation divides by zero.

Watch out

Wrong: cancel a factor without checking whether it can be zero. Right: division by a factor is valid only when the factor is non-zero. An absurd conclusion is a signal to inspect the steps.

For the last check, use a new pair and explain what it establishes.

Quick Recap

State integer conditions and use independent parameters for arbitrary pairs.
Prove parity or divisibility by reaching the appropriate integer form.
Examples support a conjecture; a general argument proves it, and one counterexample can disprove it.
Check that every operation is valid, especially division by a possible zero.

The opening difference is always $(n+1)^2-n^2=2n+1$ , so consecutive integer squares can never differ by an even number.

Last Accessed

Papers 1, 2 and 3

A6 - Algebraic proof

Representing integer families (Higher)

General proofs of parity and divisibility (Higher)

Assessing the validity of a proof (Higher)

Quick Recap

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