Prove The Square Root Of 5 Is Irrational

Proving the Square Root of 5 is Irrational: A Step-by-Step Guide

This article provides a clear and concise proof that the square root of 5 (√5) is an irrational number. Understanding this proof requires a grasp of basic number theory concepts, but we'll break it down step-by-step, making it accessible to a wider audience. This proof utilizes the method of contradiction, a common and powerful technique in mathematics.

What are Rational and Irrational Numbers?

Before diving into the proof, let's quickly 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 goes on forever without repeating. Famous examples include π (pi) and e (Euler's number). We aim to demonstrate that √5 falls into the latter category.

Proof by Contradiction: The Strategy

We'll use proof by contradiction. This method assumes the opposite of what we want to prove and then shows that this assumption leads to a logical contradiction. If the assumption leads to a contradiction, it must be false, thus proving the original statement true.

Step 1: The Assumption

Let's assume, for the sake of contradiction, that √5 is rational. This means it can be expressed as a fraction:

√5 = 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).

Step 2: Manipulation and Simplification

Square both sides of the equation:

5 = p²/q²

Rearrange the equation:

5q² = p²

This equation tells us that p² is a multiple of 5. Since 5 is a prime number, this implies that p itself must also be a multiple of 5. We can express this as:

p = 5k (where k is an integer)

Substitute this back into the equation 5q² = p²:

5q² = (5k)²

5q² = 25k²

Divide both sides by 5:

q² = 5k²

This equation now tells us that q² is also a multiple of 5, and therefore, q must also be a multiple of 5.

Step 3: The Contradiction

We've now shown that both p and q are multiples of 5. This directly contradicts our initial assumption that the fraction p/q is in its simplest form (i.e., that p and q share no common factors). We've reached a logical contradiction.

Step 4: The Conclusion

Since our initial assumption (that √5 is rational) leads to a contradiction, the assumption must be false. Therefore, the square root of 5 is irrational.

In Summary:

This proof demonstrates the irrationality of √5 using the elegant method of contradiction. By showing that the assumption of rationality leads to a self-contradictory conclusion, we definitively prove that √5 is, indeed, an irrational number. This proof can be adapted to prove the irrationality of the square roots of other non-perfect squares.