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.

**Related posts:**

- Use a direct proof to show that the sum of two even integers is even
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- Prove or disprove that the product of two irrational numbers is irrational
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