p : `sqrt(39)` is irrational
To check the validity of statement p , which of the following method we can use

Options

A. Direct method
B. Contrapositive method
C. Contradiction method
D. By giving a counter example

Answer

To determine the validity of the statement \( p: \sqrt{39} \) is irrational, we can analyze the methods provided in the options. Here’s the step-by-step solution:

### Step 1: Understand the Statement
The statement \( p \) asserts that \( \sqrt{39} \) is an irrational number. To validate this, we need to explore different methods of proof.

### Step 2: Evaluate the Direct Method
The direct method involves proving the statement straightforwardly. However, it is known that proving the irrationality of square roots of non-perfect squares directly is often not feasible.

**Conclusion**: The direct method is not suitable for this case.

### Step 3: Evaluate the Contrapositive Method
The contrapositive method is typically used for conditional statements of the form "If A, then B." Since our statement \( p \) is not a conditional statement but rather a simple assertion, this method is not applicable here.

**Conclusion**: The contrapositive method cannot be used for this statement.

### Step 4: Evaluate the Contradiction Method
The contradiction method involves assuming the negation of the statement and showing that this leads to a contradiction. For our statement \( p \), we assume that \( \sqrt{39} \) is rational. If this assumption leads to a contradiction, we can conclude that \( p \) is true.

1. Assume \( \sqrt{39} \) is rational.
2. Then, it can be expressed as \( \frac{a}{b} \) where \( a \) and \( b \) are integers with no common factors.
3. Squaring both sides gives \( 39 = \frac{a^2}{b^2} \) or \( a^2 = 39b^2 \).
4. This implies that \( a^2 \) is divisible by 39. Since 39 is not a perfect square, \( a \) must also be divisible by 3 (one of the prime factors of 39).
5. Let \( a = 3k \) for some integer \( k \). Then substituting back gives \( (3k)^2 = 39b^2 \) or \( 9k^2 = 39b^2 \).
6. Simplifying gives \( k^2 = \frac{39b^2}{9} \), indicating \( k^2 \) is also divisible by 39, leading to further implications about \( a \) and \( b \).
7. This results in a contradiction since both \( a \) and \( b \) cannot have common factors if \( \sqrt{39} \) is rational.

**Conclusion**: The contradiction method can be used to prove that \( \sqrt{39} \) is irrational.

### Step 5: Evaluate the Counterexample Method
The counterexample method is used to disprove statements that are conditional (if-then statements). Since our statement is not conditional, this method is not applicable.

**Conclusion**: The counterexample method cannot be used here.

### Final Conclusion
The only suitable method to check the validity of the statement \( p: \sqrt{39} \) is irrational is the contradiction method. Therefore, the correct answer is the contradiction method.

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