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.
Related posts:
- Use a direct proof to show that the sum of two odd integers is even
- Use a proof by contradiction to prove that the sum of an irrational number and a rational number is irrational
- Prove or disprove that the product of two irrational numbers is irrational
- Show that if n is an integer and n^3+5 is odd, then n is even using a proof by contraposition and a proof by contradiction
- Prove that if m and n are integers and mn is even, then m is even or n is even
