If both a and b are rational numbers, find the values of a and b in the following equalities: $\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = a + b\sqrt{3}$

Question

Accepted Answer

To solve the problem and verify why Option 3 is the correct answer, we need to simplify the expression $\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ and express it in the form $a + b\sqrt{3}$, where both $a$ and $b$ are rational numbers.

### Step-by-Step Solution:

1. **Rationalize the Denominator:**
   To simplify $\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$, we multiply the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{3} - 1$.

$$
   \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)}
   $$

2. **Simplify the Denominator:**
   The denominator becomes:

$$
   (\sqrt{3} + 1)(\sqrt{3} - 1) = (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2
   $$

3. **Expand the Numerator:**
   The numerator is:

$$
   (\sqrt{3} - 1)^2 = (\sqrt{3})^2 - 2 \cdot \sqrt{3} \cdot 1 + 1^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3}
   $$

4. **Combine and Simplify:**
   Now, the expression becomes:

$$
   \frac{4 - 2\sqrt{3}}{2} = \frac{4}{2} - \frac{2\sqrt{3}}{2} = 2 - \sqrt{3}
   $$

This can be written as:

$$
   2 + (-1)\sqrt{3}
   $$

Here, $a = 2$ and $b = -1$.

### Conclusion:

- **Option 3: $a = 2, b = -1$** is indeed the correct answer because the simplified form of the given expression is $2 - \sqrt{3}$, which matches $a + b\sqrt{3}$ with $a = 2$ and $b = -1$.

### Explanation of Incorrect Options:

- **Option 1: $a = -1, b = 2$** is incorrect because it suggests the expression simplifies to $-1 + 2\sqrt{3}$, which does not match our result.
  
- **Option 2: $a = 1, b = 2$** is incorrect because it suggests the expression simplifies to $1 + 2\sqrt{3}$, which does not match our result.
  
- **Option 4: $a = -2, b = 1$** is incorrect because it suggests the expression simplifies to $-2 + \sqrt{3}$, which does not match our result.

Thus, the correct answer is Option 3, where $a = 2$ and $b = -1$.