Show That Root 3 Is Irrational

Proving √3 is Irrational: A Comprehensive Guide

The quest to prove the irrationality of √3, like that of √2, is a classic exercise in mathematical reasoning. Understanding this proof not only solidifies your grasp of number theory but also illustrates fundamental proof techniques applicable across various mathematical domains. This comprehensive guide will walk you through multiple approaches to proving √3's irrationality, explaining the underlying logic and highlighting the key concepts involved.

What is an Irrational Number?

Before diving into the proof, let's clarify the definition. An irrational number is a real number that cannot be expressed as a fraction p/q, where p and q are integers, and q is not zero. In simpler terms, it can't be written as a simple ratio of two whole numbers. Irrational numbers have decimal representations that neither terminate nor repeat. Famous examples include π (pi) and e (Euler's number), along with the square roots of non-perfect squares, such as √2, √3, √5, and so on.

Proof 1: Using Proof by Contradiction

This is arguably the most common and elegant method to prove the irrationality of √3. It relies on the principle of contradiction, a powerful proof technique where you assume the opposite of what you want to prove and then demonstrate that this assumption leads to a logical contradiction.

1. Assume √3 is Rational:

We begin by assuming the opposite of what we want to prove. Let's assume √3 is rational. This means it can be expressed as a fraction:

√3 = p/q

where p and q are integers, q ≠ 0, and the fraction p/q is in its simplest form (meaning p and q share no common factors other than 1 – they are coprime).

2. Square Both Sides:

Squaring both sides of the equation, we get:

3 = p²/q²

3. Rearrange the Equation:

Multiplying both sides by q², we obtain:

3q² = p²

This equation tells us that p² is a multiple of 3.

4. Deduction: p is a Multiple of 3

If p² is a multiple of 3, then p itself must also be a multiple of 3. This is because if p were not a multiple of 3, its square would also not be a multiple of 3. (Think about it: the prime factorization of p² will contain only the primes that are in the prime factorization of p, doubled in their exponents. Thus, if 3 is not a prime factor of p, it will not be a prime factor of p².) Therefore, we can write p as:

p = 3k

where k is an integer.

5. Substitute and Simplify:

Substitute p = 3k back into the equation 3q² = p²:

3q² = (3k)²

3q² = 9k²

Dividing both sides by 3, we get:

q² = 3k²

6. Deduction: q is a Multiple of 3

This equation shows that q² is also a multiple of 3, and by the same logic as before, q itself must be a multiple of 3.

7. The Contradiction:

We've now shown that both p and q are multiples of 3. This contradicts our initial assumption that p/q is in its simplest form (coprime). If both p and q are multiples of 3, they share a common factor of 3, which is a contradiction.

8. Conclusion:

Since our initial assumption leads to a contradiction, the assumption must be false. Therefore, √3 cannot be expressed as a fraction p/q, and thus, √3 is irrational.

Proof 2: Using the Fundamental Theorem of Arithmetic

This proof leverages the Fundamental Theorem of Arithmetic, which states that every integer greater than 1 can be uniquely represented as a product of prime numbers (ignoring the order of the factors).

1. Assume √3 is Rational:

As before, we assume √3 = p/q, where p and q are coprime integers.

2. Square and Rearrange:

Squaring both sides and rearranging, we get:

3q² = p²

3. Prime Factorization:

Consider the prime factorizations of p and q. Since 3q² = p², the prime factorization of p² must contain at least one factor of 3 (it actually contains an even number of 3's because it is a square). This means that the prime factorization of p must contain at least one factor of 3. Let's represent this as:

p = 3^m * r

where 'm' is a positive integer and 'r' is an integer not divisible by 3.

4. Substitute and Simplify:

Substituting this into 3q² = p², we get:

3q² = (3^m * r)²

3q² = 3^2m * r²

Dividing by 3:

q² = 3^(2m-1) * r²

5. The Contradiction:

This equation shows that q² also contains at least one factor of 3 (specifically an odd number of 3's since the exponent is 2m-1). Therefore, q must contain at least one factor of 3.

But this contradicts our initial assumption that p and q are coprime (they share no common factors). Both p and q are divisible by 3, meaning they have a common factor of 3.

6. Conclusion:

The contradiction arises from our initial assumption that √3 is rational. Hence, √3 is irrational.

Proof 3: A Variation on the Contradiction Method

This proof is similar to the first one but uses a slightly different line of reasoning.

1. Assume √3 is Rational:

Assume √3 = p/q, where p and q are coprime integers.

2. Square and Rearrange:

3q² = p²

3. Consider Parity:

Now let's consider the parity (evenness or oddness) of p and q. Since p² is a multiple of 3, it must be even (because any number multiplied by an even number is even). If p² is even, p must also be even. We can write p as 2k, where k is an integer.

Substitute into the equation:

3q² = (2k)² = 4k²

q² = (4/3)k²

Now, we see that q² must be even (as it's a multiple of 4/3, which is an even number when multiplied by any number), and therefore q must also be even.

4. The Contradiction:

This implies that both p and q are even and therefore share a common factor of 2. This contradicts our initial assumption that p and q are coprime.

5. Conclusion:

The contradiction means that our initial assumption is false, so √3 is irrational.

Conclusion: The Importance of Proof

These proofs demonstrate the beauty and rigor of mathematical reasoning. Proving the irrationality of √3, while seemingly a simple task, underscores important concepts such as proof by contradiction, the fundamental theorem of arithmetic, and the power of logical deduction. Understanding these proofs enhances your mathematical skills and deepens your appreciation for the elegance and precision of mathematics. Remember, each proof offers a slightly different perspective on the same fundamental truth: √3, like many other square roots of non-perfect squares, is an irrational number. The process of exploring different proof methods allows for a more comprehensive understanding of the concept and enhances mathematical intuition.