If a and b are two distinct non-zero real numbers such that `a -b =a/b=1/b-1/a,` then : (a) `a>0` (b) `a<0` (c) `b<0` (d) `b>0`

Options

A. `a gt 0`
B. `a lt 0`
C. `b lt 0`
D. `b gt 0`

Answer

To solve the problem, we start with the given equations involving two distinct non-zero real numbers \( a \) and \( b \):

1. \( a - b = \frac{a}{b} = \frac{1}{b} - \frac{1}{a} \)

### Step 1: Set the equations equal to each other
From the first part of the equation, we can equate \( a - b \) and \( \frac{1}{b} - \frac{1}{a} \):

\[ a - b = \frac{1}{b} - \frac{1}{a} \]

### Step 2: Simplify the right-hand side
To simplify \( \frac{1}{b} - \frac{1}{a} \), we find a common denominator:

\[ \frac{1}{b} - \frac{1}{a} = \frac{a - b}{ab} \]

### Step 3: Set the equations equal
Now we can set the left-hand side equal to the simplified right-hand side:

\[ a - b = \frac{a - b}{ab} \]

### Step 4: Cancel \( a - b \) (since \( a \neq b \))
Since \( a \) and \( b \) are distinct, we can safely divide both sides by \( a - b \):

\[ 1 = \frac{1}{ab} \]

### Step 5: Rearranging gives us the product of \( a \) and \( b \)
This implies:

\[ ab = 1 \]

### Step 6: Substitute \( b \) in terms of \( a \)
From \( ab = 1 \), we can express \( b \) in terms of \( a \):

\[ b = \frac{1}{a} \]

### Step 7: Use the second equation \( \frac{a}{b} = \frac{1}{b} - \frac{1}{a} \)
Now we can substitute \( b = \frac{1}{a} \) into the second equation:

\[ \frac{a}{\frac{1}{a}} = \frac{1}{\frac{1}{a}} - \frac{1}{a} \]

### Step 8: Simplify both sides
The left-hand side simplifies to:

\[ a^2 \]

The right-hand side simplifies to:

\[ a - \frac{1}{a} \]

So we have:

\[ a^2 = a - \frac{1}{a} \]

### Step 9: Multiply through by \( a \) to eliminate the fraction
Multiplying through by \( a \) (noting \( a \neq 0 \)) gives:

\[ a^3 = a^2 - 1 \]

### Step 10: Rearranging gives us a cubic equation
Rearranging this gives:

\[ a^3 - a^2 + 1 = 0 \]

### Step 11: Analyze the signs of \( a \) and \( b \)
Since we have \( ab = 1 \), both \( a \) and \( b \) must have the same sign (either both positive or both negative) because their product is positive.

### Step 12: Check the inequality \( a > \frac{1}{a} \)
From the earlier steps, we also established that \( a > \frac{1}{a} \) holds true only for positive \( a \). This implies:

\[ a^2 > 1 \implies a > 1 \text{ or } a < -1 \]

However, since \( ab = 1 \) and both numbers are non-zero, we conclude that both \( a \) and \( b \) must be positive.

### Conclusion
Thus, we conclude that:

- \( a > 0 \)
- \( b > 0 \)

### Final Answer
The correct options are:
(a) \( a > 0 \)
(d) \( b > 0 \)