Prove that 5 - √3 is an irrational number.

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

### Step 1: Assume \( 5 - \sqrt{3} \) is rational
Let us assume, for the sake of contradiction, that \( 5 - \sqrt{3} \) is a rational number. By definition, a rational number can be expressed as the ratio of two integers \( p \) and \( q \), where \( q \neq 0 \) and \( p, q \) are coprime integers (i.e., their greatest common divisor is 1). So, we can write:

\( 5 - \sqrt{3} = \frac{p}{q} \)

### Step 2: Isolate \( \sqrt{3} \)
Rearranging the equation to isolate \( \sqrt{3} \), we get:

\( \sqrt{3} = 5 - \frac{p}{q} \)

Simplify further:

\( \sqrt{3} = \frac{5q - p}{q} \)

### Step 3: Analyze \( \sqrt{3} \)
Here, \( \frac{5q - p}{q} \) is the ratio of two integers because \( 5q - p \) and \( q \) are integers. This implies that \( \sqrt{3} \) is rational.

### Step 4: Contradiction
However, it is a well-known fact that \( \sqrt{3} \) is an irrational number (it cannot be expressed as the ratio of two integers). This contradicts our assumption that \( 5 - \sqrt{3} \) is rational.

### Step 5: Conclusion
Since assuming \( 5 - \sqrt{3} \) to be rational leads to a contradiction, we conclude that \( 5 - \sqrt{3} \) is an irrational number.