Question

Prove that $14 - 2\sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is irrational.

Hint:

For any combination of rational and irrational numbers like $a - b\sqrt{c}$, rearranging to isolate the radical $\sqrt{c}$ on one side will automatically establish the proof by contradiction!

Answer 1

Step 1: Understanding the Question:
We need to prove that the expression $14 - 2\sqrt{3}$ is an irrational number.
We are given the fact that $\sqrt{3}$ is an irrational number.
This proof can be done using the method of contradiction.

Step 2: Key Formula or Approach:
We assume the contrary, i.e., $14 - 2\sqrt{3}$ is a rational number.
A rational number can be expressed in the form $\frac{p}{q}$ (or as a single variable $r$), where $p$ and $q$ are integers and $q \neq 0$.
We will rearrange the terms to isolate $\sqrt{3}$ on one side and show that it leads to a logical contradiction.

Step 3: Detailed Explanation:

• Let us assume on the contrary that $14 - 2\sqrt{3}$ is a rational number $r$:
\[ 14 - 2\sqrt{3} = r \quad (\text{where } r \text{ is rational}) \]

• Rearrange the terms to isolate the radical term $2\sqrt{3}$:
\[ 2\sqrt{3} = 14 - r \]

• Divide both sides by 2 to completely isolate $\sqrt{3}$:
\[ \sqrt{3} = \frac{14 - r}{2} \]

• Analyze the components on the right-hand side:
- Since $r$ is rational, $14 - r$ is also rational (as the difference of two rational numbers is rational).
- Therefore, $\frac{14 - r}{2}$ is also rational (as the division of a rational number by a non-zero rational number is rational).

• Since the right-hand side is rational, the left-hand side $\sqrt{3}$ must also be rational.
This contradicts the given fact that $\sqrt{3}$ is irrational.

• Our initial assumption that $14 - 2\sqrt{3}$ is a rational number is incorrect.


Step 4: Final Answer:
Therefore, $14 - 2\sqrt{3}$ is an irrational number.