Use a direct proof to show that the sum of two odd integers is even

Direct proof is one the easiest method to construct proofs.

To use this method, we must prove that a theorem, stated as a conditional statement p-> q is true.

To this, we assume that p is true, and we apply rules of inference, axioms, definitions and previously proved theorems to prove that q is also true. By doing this, we warrantee that truth values of p true and q false never occur.

Let’s see the solution to this example.

We can restate the conjecture we need to prove as follows:

If m and n are odd integers, then m+n is even.

First, let’s use the definition of and odd number.

If n is odd, then n = 2k +1 for some integer k.

m = 2k +1, by definition of odd number

n = 2t +1, by definition of odd number

By doing this, we assumed that p is true. In other words, that m and n are odd integers. Notice that we express m by using k, and n by using t. We shouldn’t assume that k=t. Although it is a possibility, it doesn’t cover all the cases.

m+n = 2k+1 + 2t + 1

Now we should proof that m+n is an even number.

By definition, an even integer p = 2r for some integer r.

m+n = 2k+1 + 2t + 1 = 2k + 2t + 2 = 2(k+t+1) ∎

Let me summarize the steps used to create this proof:

Write the conjecture as a conditional statement p -> q. Remember a conjecture is a statement that we think is true based on some evidence. Once a conjecture is proved, it is a theorem.
Assume p is true. In this case, we assumed p is true by expressing m and n as even numbers. We did this, by using the even number definition.
We applied definitions to add m+n. The we used basic arithmetic rules to transform m+n.
Using the definition of even number, we proved that m+n is even.

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