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

Step 1: Understanding the Concept:
We use the method of contradiction. We assume the number is rational and show that this leads to a logical inconsistency with the given fact that \(\sqrt{3}\) is irrational.
Step 2: Detailed Explanation:
Let us assume that \(14 - 2\sqrt{3}\) is a rational number.
If it is rational, it can be represented as \(x\), where \(x\) is rational.
\[ 14 - 2\sqrt{3} = x \]
Rearrange the equation to isolate the irrational term:
\[ 14 - x = 2\sqrt{3} \]
\[ \frac{14 - x}{2} = \sqrt{3} \]
Now, evaluate both sides:
On the Left Hand Side (L.H.S.):
Since \(14\) and \(2\) are rational integers and we assumed \(x\) is rational, the difference and division of rational numbers is also rational.
Thus, \(\frac{14 - x}{2}\) is a rational number.
On the Right Hand Side (R.H.S.):
We are given that \(\sqrt{3}\) is an irrational number.
This leads to a contradiction: \(\text{Rational} = \text{Irrational}\).
Our initial assumption that \(14 - 2\sqrt{3}\) is rational is incorrect.
Therefore, \(14 - 2\sqrt{3}\) is an irrational number.
Step 3: Final Answer:
Hence, \(14 - 2\sqrt{3}\) is proved to be irrational.