Prove that ` 3 + 2 sqrt(5)` is an irrational number

Answer

To prove that \( 3 + 2\sqrt{5} \) is an irrational number, we can use a proof by contradiction. Here’s a step-by-step solution:

### Step 1: Assume the opposite
Assume that \( 3 + 2\sqrt{5} \) is a rational number. This means that it can be expressed in the form \( \frac{P}{Q} \), where \( P \) and \( Q \) are integers and \( Q \neq 0 \).

**Hint**: Start by assuming the opposite of what you want to prove. This is a common technique in proofs.

### Step 2: Rearrange the equation
From our assumption, we can rearrange the equation:
\[ 2\sqrt{5} = \frac{P}{Q} - 3 \]

**Hint**: Isolate the term involving the square root to analyze its properties.

### Step 3: Express \( \sqrt{5} \)
Next, we can express \( \sqrt{5} \):
\[ \sqrt{5} = \frac{1}{2}\left(\frac{P}{Q} - 3\right) \]

**Hint**: Simplifying the equation helps in understanding the relationship between rational and irrational numbers.

### Step 4: Analyze the right-hand side
The right-hand side \( \frac{1}{2}\left(\frac{P}{Q} - 3\right) \) is a rational number because it is a combination of rational numbers (since \( P \) and \( Q \) are integers).

**Hint**: Remember that the sum or difference of rational numbers is also rational.

### Step 5: Conclude about \( \sqrt{5} \)
Since we have expressed \( \sqrt{5} \) as a rational number, we conclude that \( \sqrt{5} \) is rational.

**Hint**: This is where the contradiction arises. Recall that \( \sqrt{5} \) is known to be an irrational number.

### Step 6: Reach a contradiction
The conclusion that \( \sqrt{5} \) is rational contradicts the known fact that \( \sqrt{5} \) is irrational. Therefore, our initial assumption that \( 3 + 2\sqrt{5} \) is rational must be false.

**Hint**: Always check your assumptions and conclusions to identify contradictions.

### Step 7: Conclusion
Thus, we conclude that \( 3 + 2\sqrt{5} \) is an irrational number.

**Final Statement**: Hence, it is proved that \( 3 + 2\sqrt{5} \) is an irrational number.

---