A mathematical proof is a logical argument that shows a statement is true. It’s like a detective laying out evidence to reach a conclusion, except the evidence consists of established facts and the conclusion is a mathematical truth. Every proof stands on the shoulders of two fundamental concepts: definitions and axioms.

Axioms are the bedrock assumptions we agree on. Definitions give us a shared, precise vocabulary. For example, the definition of an even number is an integer that can be expressed as $2k$, where $k$ is also an integer. With these tools, we use logical reasoning to build a case for a new statement, called a theorem.

The most straightforward way to prove a theorem is with a direct proof. Most theorems take the form of an "if-then" statement: "If P is true, then Q is true." In a direct proof, we assume that P (the hypothesis) is true and use a series of logical steps to show that Q (the conclusion) must follow.

First, we start by assuming the "if" part is true. We have two even integers, x and y.

Next, we use our definition of an even number. Since x and y are even, we can write them in their defined form:

Here, a and b can be any integers. Now, we add them together:

Using a basic rule of algebra, we can factor out the 2:

We know that the sum of two integers (a + b) is just another integer. Let's call it c. So, we can write:

This result, $2c$, perfectly matches our definition of an even number. We've shown that the sum x + y must be even. The proof is complete. We started with the hypothesis and arrived directly at the conclusion.

Sometimes, the direct path is messy or hard to see. In these cases, mathematicians turn to indirect proofs. Instead of tackling the problem head-on, they approach it from a different angle. The two most common types are proof by contrapositive and proof by contradiction.

Proof by Contrapositive

Every "if P, then Q" statement has a logically equivalent partner called the contrapositive: "If not Q, then not P." If we can prove the contrapositive is true, then the original statement must also be true. It’s like looking at a reflection to understand the original object.

A direct proof is tricky. How do we go from knowing about $n^2$ to knowing about $n$? Instead, let's form the contrapositive.

Our original statement is:

• P: $n^2$ is even.
• Q: $n$ is even.

The contrapositive is "If not Q, then not P":

• Not Q: $n$ is not even (meaning $n$ is odd).
• Not P: $n^2$ is not even (meaning $n^2$ is odd).

So, we need to prove: If n is an odd integer, then $n^2$ is an odd integer. This is much easier to handle directly.

We start by assuming n is odd. By definition, an odd number can be written as $2k + 1$ for some integer k. So:

Now, let's square it:

We can factor a 2 out of the first two terms:

Since $2k^2 + 2k$ is just another integer, let's call it m. Our equation becomes:

This expression, $2m+1$, is the definition of an odd number. We have successfully proven that if n is odd, then $n^2$ is odd. Because we proved the contrapositive, we have also proven the original, more difficult statement.

Proof by Contradiction

This method is a powerful tool. To prove a statement is true, you start by assuming it's false. Then, you follow the logical consequences of that assumption until you run into a contradiction—a result that's impossible or violates one of your initial assumptions. This absurdity reveals that your initial assumption (that the statement was false) must be wrong, meaning the statement is actually true.

First, we assume the opposite: that $\sqrt{2}$ is a rational number. By definition, a rational number can be written as a fraction $\frac{a}{b}$ where a and b are integers, b is not zero, and the fraction is in its simplest form (meaning a and b have no common factors other than 1).

Let's square both sides and rearrange the equation:

This equation tells us that $a^2$ is an even number, since it's equal to 2 times another integer ($b^2$). And as we just proved with the contrapositive, if $a^2$ is even, then a must also be even. So we can write a as $2k$ for some integer k.

Now, let's substitute $2k$ for a back into our equation:

We can divide both sides by 2:

This means that $b^2$ is also an even number, which in turn means that b must be an even number. And here is the contradiction. We started by assuming the fraction $\frac{a}{b}$ was in its simplest form. But we just showed that both a and b are even, meaning they share a common factor of 2. This is impossible if the fraction is in simplest form.

Our initial assumption has led us to an absurd result. Therefore, the assumption must be false. The square root of 2 cannot be rational.

Ready to check your understanding?

These methods—direct, contrapositive, and contradiction—are the essential tools for building mathematical arguments. They provide a framework for taking what we know and logically discovering what must be true.