Show That Root 3 Is Irrational

Proving √3 is Irrational: A Step-by-Step Guide

Meta Description: This article provides a clear and concise proof that the square root of 3 is an irrational number, using the method of proof by contradiction. Learn the steps involved and understand the underlying mathematical concepts.

The question of whether √3 is rational or irrational is a classic problem in number theory. Understanding this proof is crucial for grasping fundamental concepts in mathematics. This article will guide you through a rigorous proof, explaining each step clearly. We'll use the widely accepted method of proof by contradiction.

Understanding Rational and Irrational Numbers

Before diving into the proof, let's define our terms. A rational number can be expressed as a fraction p/q, where p and q are integers, and q is not zero. Examples include 1/2, 3/4, and -5/7. An irrational number cannot be expressed as such a fraction; its decimal representation is non-terminating and non-repeating. Examples include π (pi) and e (Euler's number).

Proof by Contradiction: The Strategy

We'll use proof by contradiction, a powerful technique in mathematics. This involves assuming the opposite of what we want to prove and then showing that this assumption leads to a contradiction. If the assumption leads to a contradiction, it must be false, and therefore the original statement must be true.

Proving √3 is Irrational

1. The Assumption: Let's assume, for the sake of contradiction, that √3 is rational. This means we can express it 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 have no common factors other than 1).

2. Squaring Both Sides: Squaring both sides of the equation, we get:

3 = p²/q²

3. Rearranging the Equation: Multiplying both sides by q², we obtain:

3q² = p²

This equation tells us that p² is a multiple of 3. Since 3 is a prime number, this implies that p itself must also be a multiple of 3. We can express this as:

p = 3k (where k is an integer)

4. Substituting and Simplifying: Substituting p = 3k into the equation 3q² = p², we get:

3q² = (3k)²

3q² = 9k²

Dividing both sides by 3, we get:

q² = 3k²

This equation shows that q² is also a multiple of 3, and therefore q must be a multiple of 3.

5. 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 (they have no common factors). The only way this contradiction can be resolved is if our initial assumption—that √3 is rational—is false.

6. The Conclusion: Therefore, √3 must be irrational.

This proof demonstrates the elegance and power of proof by contradiction. By assuming the opposite and arriving at a contradiction, we conclusively prove that the square root of 3 is indeed an irrational number. This fundamental concept is a building block for further exploration in number theory and related fields.