6.01a - Proof with algebra

6.01a - Proof with algebra

Algebra can explain patterns for every integer, including ones you could never finish checking. Represent integers precisely, prove divisibility and parity claims, and judge whether a chain of reasoning really establishes its conclusion.

Higher only.

Representing general integers

Subtract the square of an integer from the square of the next integer: 42−32=74^2-3^2=7, 52−42=95^2-4^2=9, 62−52=116^2-5^2=11. Must the result always be odd? Before proving it, we need symbols that represent every allowed pair, rather than just those three pairs.

An integer can be negative, zero or positive. Let nn be an integer. Its next integer is n+1n+1, so three consecutive integers can be represented by n,n+1,n+2n,n+1,n+2. Consecutive means there are no integers missing between them.

Even and odd integers

An even integer is a multiple of 2, so it can be written as 2n2n, where nn is an integer. An odd integer is one more than an even integer, so it can be written as 2n+12n+1. These forms include negative numbers and zero where appropriate: zero is even, and −3=2(−2)+1-3=2(-2)+1 is odd.

Consecutive even integers differ by 2, so use 2n,2n+2,2n+42n,2n+2,2n+4. Consecutive odd integers also differ by 2: 2n+1,2n+3,2n+52n+1,2n+3,2n+5. The integer condition is essential; if n=0.5n=0.5, then 2n2n would be 1 and would not represent an even integer.

Worked example 1: three consecutive integers

Question. Represent three consecutive integers and simplify their sum.

Plan. Start at an unspecified integer nn and increase by one each time.

Integers: n, n+1, n+2Sum: n+(n+1)+(n+2)=3n+3collect like terms\begin{aligned} \text{Integers: }&n,\ n+1,\ n+2 \\ \text{Sum: }&n+(n+1)+(n+2) \\ &=3n+3 &&\text{collect like terms} \end{aligned}

Answer. The three integers are n,n+1,n+2n,n+1,n+2; their sum is 3n+33n+3, for integer nn.

Check. With n=−4n=-4, the numbers are −4,−3,−2-4,-3,-2. Their sum is −9-9, matching 3(−4)+3=−93(-4)+3=-9.

Now restrict the numbers to consecutive even integers. Only that condition changes, but the step must become 2.

Worked example 2: three consecutive even integers

Question. Represent three consecutive even integers and simplify their sum.

Plan. Start with 2n2n, for integer nn, and increase by two each time.

Integers: 2n, 2n+2, 2n+4Sum: 2n+(2n+2)+(2n+4)=6n+6collect like terms\begin{aligned} \text{Integers: }&2n,\ 2n+2,\ 2n+4 \\ \text{Sum: }&2n+(2n+2)+(2n+4) \\ &=6n+6 &&\text{collect like terms} \end{aligned}

Answer. The integers are 2n,2n+2,2n+42n,2n+2,2n+4; their sum is 6n+66n+6.

Check. With n=−2n=-2, the numbers are −4,−2,0-4,-2,0, whose sum is −6-6. The expression gives 6(−2)+6=−66(-2)+6=-6. No even integer has been skipped.

Use different letters for unrelated integers: two arbitrary odd integers are 2a+12a+1 and 2b+12b+1, for integers a,ba,b. Writing 2n+12n+1 twice would force them to be equal; writing 2n+12n+1 and 2n+32n+3 would force them to be consecutive odd integers. Choose a representation that matches exactly the given condition.

Watch out

n,n+2,n+4n,n+2,n+4 are not necessarily even: n=1n=1 gives three odd integers. Use 2n,2n+2,2n+42n,2n+2,2n+4, with nn an integer, to guarantee evenness.

Change the parity while keeping the spacing correct.

Proving divisibility and parity

A divisibility proof aims for a multiple, with an integer multiplying it.

To show an expression is divisible by 4, rewrite it as 4×an integer4\times\text{an integer}. Merely showing a factor of 2 would prove evenness, but not divisibility by 4. To show oddness, reach 2×an integer+12\times\text{an integer}+1. This final interpretation connects the algebra to the claim.

Worked example 3: the difference between neighbouring squares

Question. Prove that the square of an integer minus the square of the preceding integer is always odd.

Plan. Represent the consecutive integers by nn and n+1n+1, then subtract the earlier integer's square from the later integer's square.

(n+1)2−n2=n2+2n+1−n2expand the square=2n+1cancel n2\begin{aligned} (n+1)^2-n^2&=n^2+2n+1-n^2 &&\text{expand the square} \\ &=2n+1 &&\text{cancel }n^2 \end{aligned}

Answer. Since nn is an integer, 2n+12n+1 is odd. Thus the difference is odd for every pair of consecutive integers, including negative ones.

Check. For the pair −3,−2-3,-2, the difference in the stated order is (−2)2−(−3)2=−5(-2)^2-(-3)^2=-5, which is odd. The result can be negative; oddness does not mean positivity.

Change only the gap between the integers, from 1 to 2. The same expansion now exposes a factor of 4.

Worked example 4: squares of integers two apart

Question. Prove that (n+2)2−n2(n+2)^2-n^2 is divisible by 4 for every integer nn.

Plan. Expand the changed square and factorise the difference.

(n+2)2−n2=n2+4n+4−n2expand=4n+4collect terms=4(n+1)take out factor 4\begin{aligned} (n+2)^2-n^2&=n^2+4n+4-n^2 &&\text{expand} \\ &=4n+4 &&\text{collect terms} \\ &=4(n+1) &&\text{take out factor 4} \end{aligned}

Answer. n+1n+1 is an integer, so 4(n+1)4(n+1) is a multiple of 4. This proves the claim for every integer nn.

Check. At n=−3n=-3, the difference is (−1)2−(−3)2=−8=4(−2)(-1)^2-(-3)^2=-8=4(-2). Negative multiples and zero are divisible by 4 as well.

The later integer need not have the larger square. The first example's negative check shows why a proof must handle signs without silently assuming the integers are positive.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account