Show that reciprocal of `3+2sqrt(2)` is an irrational number

Answer

To show that the reciprocal of \(3 + 2\sqrt{2}\) is an irrational number, we can follow these steps:

### Step 1: Define the Reciprocal
Let \( x = \frac{1}{3 + 2\sqrt{2}} \).

### Step 2: Assume \( x \) is Rational
Assume for the sake of contradiction that \( x \) is a rational number. This means we can express \( x \) in the form \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \).

### Step 3: Manipulate the Expression
From our assumption, we have:
\[ x = \frac{1}{3 + 2\sqrt{2}} = \frac{p}{q} \]
Cross-multiplying gives:
\[ p = q(3 + 2\sqrt{2}) \]
This can be rewritten as:
\[ p = 3q + 2q\sqrt{2} \]

### Step 4: Isolate the Irrational Part
Rearranging gives:
\[ p - 3q = 2q\sqrt{2} \]
Now, we can isolate \( \sqrt{2} \):
\[ \sqrt{2} = \frac{p - 3q}{2q} \]

### Step 5: Analyze the Right Side
Since \( p \) and \( q \) are integers, \( p - 3q \) is also an integer. Therefore, the right side \( \frac{p - 3q}{2q} \) is a fraction of integers, which means it is rational (as long as \( q \neq 0 \)).

### Step 6: Contradiction
However, we know that \( \sqrt{2} \) is an irrational number. This leads to a contradiction because we cannot have an irrational number equal to a rational number.

### Step 7: Conclusion
Since our assumption that \( x \) is rational led to a contradiction, we conclude that \( x \) must be irrational. Therefore, the reciprocal of \( 3 + 2\sqrt{2} \) is indeed an irrational number.

### Final Statement
Thus, we have shown that:
\[ \frac{1}{3 + 2\sqrt{2}} \text{ is an irrational number.} \]

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