A mathematical proof is a logical argument that establishes the truth of a mathematical statement with absolute certainty. Unlike other forms of reasoning, mathematical proofs provide definitive answers through rigorous logical steps. Proofs have key characteristics: they use logical reasoning, follow a step-by-step progression, provide absolute certainty, and are based on established axioms and definitions. The process transforms a conjecture, which is an unproven statement, into a theorem through the construction of a valid proof. This logical structure follows a pattern: given information leads to reasoning, which leads to a conclusion.
There are four main types of mathematical proofs, each with its own approach and applications. Direct proof shows that P implies Q by following a logical chain from the hypothesis to the conclusion. Proof by contradiction assumes the negation of what we want to prove and shows this leads to a logical impossibility. Proof by contrapositive shows that not Q implies not P, which is logically equivalent to P implies Q. Mathematical induction proves statements about all positive integers using a base case and an inductive step. Each method has specific applications: direct proofs work well for straightforward implications, contradiction is powerful for existence proofs, contrapositive is useful when the negation is easier to work with, and induction is essential for proving formulas involving natural numbers.
Let's work through a complete direct proof example. We want to prove that if n is even, then n squared is even. In a direct proof, we start with the given hypothesis and use logical steps to reach the conclusion. Our proof structure is clear: given that n is even, we want to prove that n squared is even using the direct method. Step 1: Since n is even, by definition, n equals 2k for some integer k. Step 2: Therefore, n squared equals 2k squared, which equals 4k squared, which we can write as 2 times 2k squared. Step 3: Since 2k squared is an integer, n squared equals 2 times 2k squared, which means n squared is even by definition. Conclusion: We have shown that if n is even, then n squared is even. This completes our direct proof, demonstrating how each logical step follows from the previous one.
Now let's explore proof by contradiction through the classic example of proving that square root of 2 is irrational. The strategy is to assume the opposite of what we want to prove and show this leads to a logical impossibility. We assume that square root of 2 is rational, meaning it can be expressed as p over q where p and q are integers in lowest terms with no common factors. Squaring both sides, we get 2 equals p squared over q squared, which gives us 2q squared equals p squared. Since p squared equals 2q squared, p squared is even, which means p must be even. If p is even, we can write p as 2k for some integer k. Substituting this back, we get 2q squared equals 4k squared, which simplifies to q squared equals 2k squared. This means q squared is even, so q must also be even. But now we have a contradiction! Both p and q are even, which means they share a common factor of 2, contradicting our assumption that they were in lowest terms. Therefore, our assumption was false, and square root of 2 must be irrational.
Mathematical induction is a powerful proof technique for statements involving positive integers. Let's prove the sum formula: 1 plus 2 plus 3 plus dot dot dot plus n equals n times n plus 1, all divided by 2. Induction has two essential parts: the base case and the inductive step. Think of it like dominoes falling - the first domino must fall, and each domino must knock down the next one. For the base case, we verify the formula for n equals 1. The left side is just 1, and the right side is 1 times 2, divided by 2, which equals 1. So the base case holds. For the inductive step, we assume the formula is true for some value k, then prove it's true for k plus 1. We assume 1 plus 2 plus dot dot dot plus k equals k times k plus 1, divided by 2. We need to prove that 1 plus 2 plus dot dot dot plus k plus k plus 1 equals k plus 1 times k plus 2, divided by 2. Starting with the left side, we can substitute our assumption to get k times k plus 1, divided by 2, plus k plus 1. Factoring out k plus 1, we get k plus 1 times k plus 2, divided by 2, which is exactly what we wanted to prove. By mathematical induction, the formula holds for all positive integers n.