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 complete confidence in their conclusions through rigorous logical steps.
Proofs have three key characteristics: they use logical reasoning, follow a step-by-step progression, and provide absolute certainty. The process transforms a conjecture - an unproven statement - into a theorem through rigorous proof.
Every proof follows a standard structure: we begin with a given statement to prove, establish necessary definitions and assumptions, proceed through logical deduction steps, and conclude with Q.E.D. - which stands for 'quod erat demonstrandum' or 'which was to be demonstrated'.
There are several fundamental proof techniques in mathematics, each with its own logical structure and specific applications. Let's explore the four main types of mathematical proofs.
Direct proof uses straightforward logical steps from hypothesis to conclusion. Proof by contradiction assumes the opposite of what we want to prove and shows this leads to an impossibility. Proof by contrapositive proves an equivalent statement, and mathematical induction uses a base case plus an inductive step.
Each proof type has specific applications. For example, we use direct proof to show that if n is even, then n squared is even. Proof by contradiction is perfect for proving that the square root of 2 is irrational. Mathematical induction works well for formulas involving natural numbers, like proving the sum formula.
Let's work through a complete direct proof example. We want to prove that if n is even, then n squared is even. This is a classic example that demonstrates the logical structure of direct proof.
First, we establish our definition: an integer is even if it can be written as 2k for some integer k. This definition is crucial for our proof, as it gives us a concrete way to work with even numbers algebraically.
Now we proceed step by step. Since n is even, we can write n equals 2k for some integer k. Next, we calculate n squared equals 2k squared, which equals 4k squared, which we can rewrite as 2 times 2k squared. Since 2k squared is an integer, n squared has the form 2 times an integer, making it even by definition. Therefore, we have proven our theorem.
Now let's explore proof by contradiction with a classic example: proving that the 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 written as p over q where p and q are integers with no common factors. This assumption will lead us to a contradiction.
Squaring both sides gives us 2 equals p squared over q squared, which means 2q squared equals p squared. Since p squared equals 2q squared, p squared is even, which means p itself must be even. So we can write p as 2k for some integer k.
Substituting p equals 2k into our equation, we get 2q squared equals 4k squared, which simplifies to q squared equals 2k squared. This means q squared is even, so q is also 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 their greatest common divisor is 1. Therefore, our assumption was false, and square root of 2 must be irrational.
Mathematical induction is a powerful proof technique for statements involving natural numbers. We can understand it through the domino analogy: if the first domino falls and each falling domino causes the next to fall, then all dominoes will fall.
Induction has two essential parts: the base case, where we prove the statement for n equals 1, and the inductive step, where we prove that if the statement is true for some k, then it must also be true for k plus 1.
Let's prove the sum formula. For the base case when n equals 1, we have 1 equals 1 times 2 divided by 2, which equals 1. This checks out. For the inductive step, we assume the formula works for k, then prove it works for k plus 1.
We start with 1 plus 2 plus up to k plus k plus 1. Using our assumption, this equals k times k plus 1 over 2, plus k plus 1. Factoring out k plus 1, we get k plus 1 times k over 2 plus 1, which simplifies to k plus 1 times k plus 2 over 2. This completes our inductive step and proves the formula for all positive integers.