Algebraic proof

Key points

•Algebraic proof uses variables to represent numbers and shows that a statement is true for all numbers in a specified set.

•Choose expressions that match the conditions, such as consecutive numbers or even numbers.

•To prove divisibility, simplify the expression and factor out the claimed divisor.

•To prove a result is always odd or even, use the structure of the expression to show its parity.

Proving divisibility by 16

Consider (x+10)^2-(x+2)^2, where x is a positive integer.

Expand both squares:

(x+10)^2=x^2+20x+100

(x+2)^2=x^2+4x+4

Subtract and simplify:

(x^2+20x+100)-(x^2+4x+4)=16x+96

Factor out 16:

16x+96=16(x+6)

Since x is an integer, x+6 is an integer. The expression is therefore 16 multiplied by an integer, so it is divisible by 16.

Expressions for common types of integers

Use variables to represent numbers in ways that preserve their defining properties.

Type of number

Algebraic expression

Even number

2n

Odd number

2n+1

Three consecutive integers

n,\ n+1,\ n+2

Three consecutive even integers

2n,\ 2n+2,\ 2n+4

Three consecutive odd integers

2n+1,\ 2n+3,\ 2n+5

Two even integers, not necessarily consecutive

2n,\ 2m

Two odd integers, not necessarily consecutive

2n+1,\ 2m+1

For consecutive even or odd numbers, the difference between successive numbers is 2, because adding 1 would change parity. When numbers are not specified as consecutive, use separate variables to avoid implying that they are.

Sum of three consecutive even integers

Represent the numbers as 2n, 2n+2, and 2n+4. Their sum is

2n+(2n+2)+(2n+4)=6n+6=6(n+1)

This is 6 multiplied by an integer, so the sum is divisible by 6.

Product of two odd integers

Represent two odd integers as 2n+1 and 2m+1. Their product is

(2n+1)(2m+1)=4mn+2n+2m+1

The terms 4mn, 2n, and 2m are all even. Their sum is even, and adding 1 makes the result odd. Therefore, the product of two odd integers is always odd.

Difference of squares of consecutive integers

Let two consecutive integers be n and n+1. The difference of their squares is

(n+1)^2-n^2=n^2+2n+1-n^2=2n+1

Their sum is also

n+(n+1)=2n+1

Since both expressions equal 2n+1, the difference of the squares equals the sum of the two consecutive integers.