Question

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

Hint:

In these proofs, always isolate the surd (like \(\sqrt{3}\)) on one side. Remember the properties: Rational \(\pm\) Rational = Rational; Rational / Rational = Rational (if divisor \(\neq 0\)).

Answer 1

We are given that \( \sqrt{3} \) is irrational.

We must prove that \[ 14 - 2\sqrt{3} \] is also irrational.

Step 1: Assume the opposite
Suppose \[ 14 - 2\sqrt{3} \] is a rational number.

Let \[ 14 - 2\sqrt{3} = r \] where \( r \) is rational.

Step 2: Rearrange the equation
\[ -2\sqrt{3} = r - 14 \]
Divide both sides by –2:
\[ \sqrt{3} = \frac{14 - r}{2} \]

Step 3: Analyze the right side
Since \( r \) is rational and 14 is rational,
\( 14 - r \) is rational.

Dividing a rational number by 2 also gives a rational number.

Therefore, \[ \frac{14 - r}{2} \] is rational.

This implies: \[ \sqrt{3} \text{ is rational} \] which contradicts the given fact that \( \sqrt{3} \) is irrational.

Step 4: Conclusion
Our assumption was wrong.
Therefore, \[ \boxed{14 - 2\sqrt{3} \text{ is irrational}} \]