### Algebraic Proofs

These worked example mainly focus on the proof involving even and odd numbers that often appear in GCSE/iGCSE examination papers.

In this chapter, an *even* number is denoted as *2n* and an *odd* number as *2n + 1*, where *n* is an integer.

**E.g.1**

Show that the sum of squares of two consecutive odd numbers is a multiple of 4 added to 2.

Let the numbers be (2n + 1) and (2n + 3).

(2n + 1)^{2} + (2n + 3)^{2} = 4n^{2} + 4n + 1 + 4n^{2} + 12n + 9

8n^{2} + 16n + 10 = 4[n^{2} + 4n + 2] + 2

**E.g.2**

Show that the sum of squares of two consecutive even numbers is a multiple of 4.

Let the numbers be (2n) and (2n + 2).

(2n)^{2} + (2n + 2)^{2} = 4n^{2} + 4n^{2} + 8n + 4

8n^{2} + 8n + 4 = 4[2n^{2} + 2n + 1]

**E.g.3**

Show that the difference of squares of two consecutive numbers is always an odd number.

Let the numbers be x and (x + 1).

(x + 1)^{2} - x^{2} = x^{2} + 2x + 1 - x^{2}

= 2x + 1, an odd number regardless of x.

**E.g.4**

Show that the difference of squares of two consecutive odd numbers is a multiple of 8.

Let the numbers be (2n + 1) and (2n + 3).

(2n + 3)^{2} - (2n + 1)^{2} = 4n^{2} + 12n + 9 - 4n^{2} - 4n - 1

8n^{2} + 8n + 8 = 8[n^{2} + n + 1]

**E.g.5**

Show that the expression, x^{2} - 6x + 11, is always positive regardless of x.

By completing the square, x^{2} - 6x + 11 = (x - 3)^{2} + 2

Since, being a square, (x - 3)^{2} is always positive, so is (x - 3)^{2} + 2.

**E.g.6**

Show that the sum of four consecutive integers is always even.

Let the numbers be x, x + 1, x + 2 and x + 3

The sum = x + x + 1 + x + 2 + x + 3 = 4x + 6 = 2(2x + 3), an even number.