Step 1: Understand the concept of irrational numbers.
An irrational number is a number that cannot be written in the form of a fraction \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \ne 0\). Numbers like \(\sqrt{2}\), \(\pi\), and \(e\) are examples of irrational numbers. It is already a well-known mathematical fact that \(\sqrt{2}\) is an irrational number.
Step 2: Assume the opposite (proof by contradiction).
To prove that \(2\sqrt{2}\) is irrational, we start by assuming the opposite. Suppose that \(2\sqrt{2}\) is a rational number. This means it can be written in the form of a fraction:
\(2\sqrt{2} = \frac{p}{q}\)
where \(p\) and \(q\) are integers and \(q \ne 0\).
Step 3: Express \(\sqrt{2}\) from the equation.
Dividing both sides of the equation by 2, we get:
\(\sqrt{2} = \frac{p}{2q}\)
Here, \(p\) and \(2q\) are integers, and \(2q \ne 0\). This means \(\sqrt{2}\) can be written as a ratio of two integers.
Step 4: Identify the contradiction.
If \(\sqrt{2} = \frac{p}{2q}\), then \(\sqrt{2}\) would be a rational number. However, it is already proven in mathematics that \(\sqrt{2}\) is irrational. This contradicts our assumption.
Step 5: Conclude the proof.
Since our assumption that \(2\sqrt{2}\) is rational leads to a contradiction, the assumption must be false. Therefore, we conclude that \(2\sqrt{2}\) is an irrational number.