Question

Prove that \( 2\sqrt{3} \) is an irrational number.

Hint:

This proof strategy works for any number of the form \( k\sqrt{n} \), where \(k\) is a non-zero rational number and \( \sqrt{n} \) is irrational. The key is to isolate the irrational part and show that it equals a rational part, which creates a contradiction.

Answer 1

Step 1: Understanding the Concept:
To prove that a number is irrational, we often use the method of proof by contradiction. We start by assuming the opposite (that the number is rational) and then show that this assumption leads to a logical contradiction.
Step 2: Key Formula or Approach:
1. A rational number can be expressed in the form \( \frac{p}{q} \), where \(p\) and \(q\) are integers, \(q \neq 0\), and \(p\) and \(q\) are co-prime (have no common factors other than 1).
2. We will use the established fact that \( \sqrt{3} \) is an irrational number.
Step 3: Detailed Explanation:
Let us assume, to the contrary, that \( 2\sqrt{3} \) is a rational number.
By definition, we can write it in the form \( \frac{p}{q} \):
\[ 2\sqrt{3} = \frac{p}{q} \] where \(p\) and \(q\) are co-prime integers and \(q \neq 0\).
Now, let's isolate \( \sqrt{3} \) in the equation:
\[ \sqrt{3} = \frac{p}{2q} \] Since \(p\) and \(q\) are integers, \(p\) and \(2q\) are also integers. Since \(q \neq 0\), it follows that \(2q \neq 0\).
Therefore, \( \frac{p}{2q} \) is a rational number.
Our equation shows that \( \sqrt{3} \) is equal to a rational number \( \frac{p}{2q} \). This implies that \( \sqrt{3} \) is a rational number.
However, this contradicts the well-known fact that \( \sqrt{3} \) is an irrational number.
This contradiction has arisen because of our incorrect initial assumption that \( 2\sqrt{3} \) is rational.
Therefore, our assumption must be false.
Hence, \( 2\sqrt{3} \) is an irrational number.
\[ \text{Hence Proved.} \]