If P and Q are two relatively prime number such that `p+q =0 and p lt q .` How fair are possible for (p,q)

Answer

To solve the problem, we need to find pairs of relatively prime numbers \( P \) and \( Q \) such that:

1. \( P + Q = 0 \)
2. \( P < Q \)

### Step-by-Step Solution:

**Step 1: Understand the condition \( P + Q = 0 \)**
From the equation \( P + Q = 0 \), we can express \( Q \) in terms of \( P \):
\[ Q = -P \]

**Hint:** Remember that for two numbers to sum to zero, one must be the negative of the other.

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**Step 2: Substitute \( Q \) in the inequality \( P < Q \)**
Now, substituting \( Q = -P \) into the inequality \( P < Q \):
\[ P < -P \]

**Hint:** This inequality means that \( P \) must be less than its own negative.

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**Step 3: Analyze the inequality \( P < -P \)**
To solve \( P < -P \), we can add \( P \) to both sides:
\[ P + P < 0 \implies 2P < 0 \]
Dividing both sides by 2 gives:
\[ P < 0 \]

**Hint:** This tells us that \( P \) must be a negative number.

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**Step 4: Determine \( Q \) based on \( P < 0 \)**
Since \( P < 0 \), and we have \( Q = -P \), it follows that:
\[ Q > 0 \]

**Hint:** This means \( Q \) is a positive number.

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**Step 5: Check the relative primality of \( P \) and \( Q \)**
For \( P \) and \( Q \) to be relatively prime, their greatest common divisor (GCD) must be 1. However, since \( P \) is negative and \( Q \) is positive, they can only be relatively prime if they have no common factors other than 1.

**Step 6: Conclusion on possible pairs \( (P, Q) \)**
Let’s consider a few examples:
- If \( P = -1 \), then \( Q = 1 \) (relatively prime).
- If \( P = -2 \), then \( Q = 2 \) (not relatively prime).
- If \( P = -3 \), then \( Q = 3 \) (not relatively prime).

The only pair that satisfies both conditions of being relatively prime and the sum being zero is \( (P, Q) = (-1, 1) \).

### Final Answer:
The only possible pair \( (P, Q) \) is:
\[ (P, Q) = (-1, 1) \]

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