2.1 Direct proofs

2.1 Direct proofs

A direct proof shows that a conditional statement p→q is true. Thus we assume that p is true and use axioms, definitions, and previously proven theorems together with rules of inference to prove that q must be true.

2.1.1 Definition

The integer n is even if even if there exists an integer k such that n=2k, and n is odd if there exits an integer k such that n=2k+1. An integer a is a perfect square if there exists an integer b such that a=b2.

2.1.2 Example

Give a direct proof of the theorem. “If n is an odd integer, then n2 is odd”

Solution

Assume that the hypothesis of this conditional statement “n is an odd integer” is true. By definition, n=2k+1, for some integer k. Squaring both sides gives n²=4k²+4k+1=2(2k²+2)+1

2.1 Direct proofs

Let m be the integer 2k²+2k. Thus we obtain


which is an odd number as asserted.

2.1.3 Example

Give a direct proof of the conditional statement “If m and n are perfect squares then mn is also a perfect square”.

Solution

Assume that m and n are both perfect squares. By definition, m=a²and n=b² for some integers a and b. Using associativity and commutativity of multiplication we get, mn=a² b² = a.a.b.b =(ab).(ab) = (ab)².

If we denote the integer ab by c we get mn = c2 which is a perfect square.

Consequently, we have proved the given statement.