Question

Prove that $2\sqrt{2}$ is an irrational number.

Hint:

The product of a non-zero rational number (like 2) and an irrational number (like $\sqrt{2}$) is always irrational.

Answer 1

Step 1: Understanding the Concept:
We use the method of contradiction. We assume the number is rational and then show that this assumption leads to a logical impossibility. We also use the known fact that $\sqrt{2}$ is irrational.
Step 2: Logical Assumption:
Let us assume that $2\sqrt{2}$ is a rational number.
Therefore, it can be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers, $q \neq 0$, and $p, q$ are co-prime. \[ 2\sqrt{2} = \frac{p}{q} \]
Step 3: Rearranging the Equation:
Divide both sides by 2: \[ \sqrt{2} = \frac{p}{2q} \] Since $p, q$ and $2$ are integers, $\frac{p}{2q}$ must be a rational number. This implies that $\sqrt{2}$ is a rational number.
However, this contradicts the fact that $\sqrt{2}$ is irrational.
Step 4: Final Answer:
Our assumption was wrong. Therefore, $2\sqrt{2}$ is an irrational number.