Proving Irrationality: A Contradiction Approach

The Logic of Proof by Contradiction

At its heart, proof by contradiction is an indirect way of proving a statement is true. Instead of building a direct argument, you start by playing devil's advocate: you assume the statement you want to prove is false. Then, you follow the logical consequences of that assumption. If this path leads you to a nonsensical result or a direct contradiction of a known fact, you have proven that your initial assumption must have been wrong. If the assumption is wrong, the original statement must be true.

The process follows these general steps:

1. Assume the opposite: Begin by assuming that the statement you wish to prove is false.
2. Deduce the consequences: Use logical steps, definitions, and known theorems to explore what must be true if your assumption holds.
3. Find the contradiction: Show that these consequences lead to a logical impossibility. This is the "reductio ad absurdum" (reduction to absurdity) step.
4. Conclude: Since your assumption led to a contradiction, it must be false. Therefore, the original statement must be true.

To get a formal handle on this method, let's turn to a resource that outlines these steps clearly.

Essential Tools for the Proof

Before we can apply this method to prove $\sqrt{2}$ is irrational, we need to sharpen two specific tools. These are the critical components that make the proof work.

Tool 1: The "Simplest Form" of Rational Numbers (Co-primes)

We know a rational number can be written as a fraction $p/q$. A crucial detail is that any rational number can be reduced to its simplest form. For example, $8/10$ is rational, but it can be simplified to $4/5$. The fraction $4/5$ cannot be simplified further.

When a fraction is in its simplest form, the numerator and denominator are said to be co-prime. This means they share no common factors other than 1. This "co-prime" condition is the key that we will eventually use to create our contradiction.

Tool 2: A Key Theorem on Prime Divisibility

The second tool is a theorem that flows directly from the Fundamental Theorem of Arithmetic (which we studied earlier):

Theorem: If $p$ is a prime number and $p$ divides $a^2$ (where $a$ is a positive integer), then $p$ must also divide $a$.

Why is this true? Think about the prime factorization of $a$. Let's say $a=p_1\times p_2\times \cdots \times p_k$. Then the prime factorization of $a^2$ is simply $a\times a=(p_1\times p_2\times \cdots \times p_k)\times (p_1\times p_2\times \cdots \times p_k)=p_1^2\times p_2^2\times \cdots \times p_k^2$.
For a prime number $p$ to divide $a^2$, it must be one of the primes in this factorization (i.e., one of $p_1,p_2,\dots ,p_k$). If $p$ is one of those primes, it is by definition also a prime factor of $a$.

For example, if we know $a^2$ is divisible by 7 (a prime), we can be certain that $a$ itself must be divisible by 7.

The following video provides an excellent, intuitive foundation for both the concept of co-primes and the properties of even/odd numbers, which is the specific case of our theorem where the prime $p=2$.

The Main Event: Proving $\sqrt{2}$ is Irrational

Now we have all the pieces. Let's assemble them into a formal proof by contradiction.

Statement: $\sqrt{2}$ is irrational.

Proof:

1. Assume the opposite: Let's assume that $\sqrt{2}$ is rational.

2. Deduce the consequences:

   • By the definition of a rational number, we can write $\sqrt{2}=\frac{p}{q}$, where $p$ and $q$ are integers, $q\ne 0$, and crucially, $p$ and $q$ are co-prime.
   • Squaring both sides gives: $2=\frac{p^2}{q^2}$.
   • Rearranging this gives: $p^2=2q^2$.
   • This equation tells us that $p^2$ is an even number (since it is 2 multiplied by some integer $q^2$).
   • Now, we use our key theorem (Tool 2): If $p^2$ is divisible by the prime 2, then $p$ must also be divisible by 2. So, $p$ is an even number.
   • Since $p$ is even, we can write it as $p=2k$ for some integer $k$.
   • Let's substitute this back into our equation $p^2=2q^2$:
     $(2k{)}^2=2q^2$
     $4k^2=2q^2$
     $2k^2=q^2$
   • This new equation tells us that $q^2$ is also an even number.
   • Applying our theorem again: If $q^2$ is even, then $q$ must also be even.

3. Find the contradiction:

   • From our deduction, we have concluded that $p$ is even and $q$ is also even.
   • If both $p$ and $q$ are even, they share a common factor of 2.
   • This contradicts our initial assumption that $p$ and $q$ are co-prime (having no common factors other than 1).

4. Conclude:

   • The only way to resolve this contradiction is to admit that our initial assumption was wrong.
   • Therefore, the assumption that "$\sqrt{2}$ is rational" is false.
   • This proves that $\sqrt{2}$ is irrational.

To see this exact proof laid out in text, you can review the example in the Testbook.com article.

Generalizing the Proof: $\sqrt{3}$ and $\sqrt{5}$

The beautiful thing about this proof structure is that it's not unique to $\sqrt{2}$. The exact same logical flow can be used to prove that $\sqrt{p}$ is irrational for any prime number $p$. You simply replace the number 2 with 3, 5, or any other prime.

Let's watch a video that demonstrates the proof for $\sqrt{3}$. This will help solidify the pattern in your mind and give you a model for how you might explain it to your daughter.

Check Your Understanding

Now it's your turn. Without writing out the full sentences, can you outline the key logical steps to prove that $\sqrt{5}$ is irrational? Think about the equations and the conclusions you would draw at each stage.

Conclusion

Today, we ventured into the world of mathematical proof and mastered a powerful technique. Proof by contradiction is a fundamental tool for establishing truths that are difficult to show directly. By applying it, we have rigorously demonstrated why numbers like $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{5}$ belong to the set of irrational numbers.

Key Takeaways:

• Proof by Contradiction is a logical method where you assume the opposite of a statement and show that this assumption leads to a contradiction.
• The irrationality proofs for $\sqrt{2}$, $\sqrt{3}$, etc., hinge on two key ideas:
  1. Every rational number can be written in a simplest form where the numerator and denominator are co-prime.
  2. If a prime number $p$ divides $a^2$, then it must also divide $a$.
• The contradiction arises when we show that both the numerator ($p$) and the denominator ($q$) must share a prime factor, violating the co-prime assumption.

This lesson concludes the more abstract and proof-based part of our module on Real Numbers. In our next and final lesson for this module, we will return to the properties of rational numbers to answer a practical question: how can we tell from a fraction's denominator whether its decimal form will be terminating or non-terminating?