Proving a statement is true requires defining general cases. Proof falls into two categories: arithmetic proofs and algebraic proofs.

Arithmetic proofs rely on two facts, both of which may be taken as true:

i) 2*n*, where *n* is a positive integer, is always even (2*n* + 1 is always odd);

ii) the sum of an odd and an even number is odd (addition is **commutative**).

An arithmetic proof will require algebra to support the **general case**.

Algebraic proofs will rely on building an expression from given facts, or rearranging expressions to prove identities.

Show that the product of two consecutive odd numbers will always be odd.

Answer:

Let the two consecutive odd numbers be 2*n*+1 and 2*n*+3

(2*n* + 1)(2*n* + 3)

= 4*n*^{2} + 4*n* + 3

= 2(2*n*^{2} + 2) + 3

The expression in brackets is multiplied by 2 and will therefore always be even. The remainder is an odd number. An even number added to an odd number is always odd.

A circle has a radius *x*. A smaller circle, of radius *x* - 2, is removed from the circle. The area of the shape that is left is 16π.

Show that *x* - 5 = 0.

Answer:

The area of the larger circle is π*x*^{2}

The area of the smaller circle is π(*x* - 2)^{2}

The remaining area is π*x*^{2} - π(*x* - 2)^{2}

= π*x*^{2} - π(*x*^{2} - 4*x* + 4)

= π*x*^{2} - π*x*^{2} + 4*πx* - 4π

= 4π(*x* - 1)

The area of the circle is given as 16π, therefore 4π(*x* - 1) = 16π

dividing both sides by 4π:

*x* - 1 = 4, or *x* - 5 = 0.

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