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

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

In this case, we can restate the conjecture as follows:

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

If we assume that p is true (m and n are evens), then we have

Definition: An integer m is even, iff m=2k for some integer k.

m= 2k for some integer k, by definition of even integer.

n=2t for some integer t, by definition of even integer.

Let’s proof that if p is true, then q is also true (q is the proposition m+n is even).

m+n = 2k+2t = 2(k+t), therefore m+n is even by definition of even integer ∎

As in the previous example, we summarize the steps for this proof as follows:

• Express the conjecture as a conditional statement.
• Assume p is true.
• Use definitions to transform p. In this case the definition of even number.
• Use arithmetic rules (we already know those) to transform q.
• Show that q is true by using definitions.

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