Is √3 a Rational Number?

To determine whether sqrt{3} is a rational number, we must first understand the definitions and properties of rational and irrational numbers. A rational number can be expressed as the quotient of two integers p and q, where q neq 0. Conversely, an irrational number cannot be expressed as such a quotient.

Proof by Contradiction

Let us prove that sqrt{3} is irrational using a proof by contradiction. We begin by assuming the opposite: that sqrt{3} is rational.

Step 1: Assumption

Let sqrt{3} be rational. This means we can express it as ab, where a and b are integers and b neq 0.

Step 2: Squaring Both Sides

Squaring both sides of the equation, we get:

sqrt{3}2ab2

which simplifies to:

a2b23

Multiplying both sides by b2, we obtain:

a23b2

Step 3: Prime Factor Analysis

Consider the prime factorization of a2 and b2. According to the Fundamental Theorem of Arithmetic, every positive integer has a unique prime factorization. Squaring a number ensures that all prime factors have even powers.

From a23b2, we know that a2 has a factor of 3. This implies that a itself must be divisible by 3. Let a 3k for some integer k.

Substituting a 3k back into the equation, we get:

3k23b2

Simplifying, we find:

k2b2

This implies that b2 is also divisible by 3. Therefore, b must be divisible by 3.

Step 4: Contradiction

Since both a and b are divisible by 3, their greatest common divisor is at least 3. This contradicts our initial assumption that the fraction ab is in its simplest form. Therefore, our assumption that sqrt{3} is rational must be false.

Conclusion

Thus, sqrt{3} is an irrational number.

Furthermore, the number sqrt{3}5 is also irrational because any non-zero rational number multiplied with an irrational number results in an irrational number.

Rational numbers can be expressed as simple fractions, while irrational numbers have non-repeating, non-terminating decimal representations.