Prove that `5-3/7 sqrt(3)` is an irrational number.

Answer

To prove that \( 5 - \frac{3}{7} \sqrt{3} \) is an irrational number, we will use the method of contradiction.

### Step-by-Step Solution:

1. **Assume the given expression is rational:**
Suppose \( 5 - \frac{3}{7} \sqrt{3} \) is a rational number.

2. **Express the assumption mathematically:**
If \( 5 - \frac{3}{7} \sqrt{3} \) is rational, then it can be written as \( \frac{a}{b} \) where \( a \) and \( b \) are integers and \( b \neq 0 \).

\[ 5 - \frac{3}{7} \sqrt{3} = \frac{a}{b} \]

3. **Isolate the irrational part:**
Rearrange the equation to isolate the term involving \( \sqrt{3} \):

\[ 5 - \frac{a}{b} = \frac{3}{7} \sqrt{3} \]

4. **Simplify the left-hand side:**
Let \( 5 - \frac{a}{b} = \frac{c}{d} \) where \( \frac{c}{d} \) is a rational number (since the difference of two rational numbers is rational).

\[ \frac{c}{d} = \frac{3}{7} \sqrt{3} \]

5. **Equate the rational and irrational parts:**
Since \( \frac{c}{d} \) is rational and \( \frac{3}{7} \sqrt{3} \) is irrational, this leads to a contradiction because a rational number cannot equal an irrational number.

6. **Conclusion:**
Hence, our initial assumption that \( 5 - \frac{3}{7} \sqrt{3} \) is rational is incorrect. Therefore, \( 5 - \frac{3}{7} \sqrt{3} \) must be irrational.