If a and b are two distinct non-zero real numbers such that `a -b =a/b=1/b-1/a,` then :

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:

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

Let's denote \( k = a - b \). Then we can rewrite the equations as:

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

### Step 1: Equate \( a - b \) and \( \frac{1}{b} - \frac{1}{a} \)

From the first and third equations, we have:

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

### Step 2: Find a common denominator for the right-hand side

The right-hand side can be rewritten as:

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

### Step 3: Substitute into the equation

Now we can substitute this back into our equation:

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

### Step 4: Cancel \( a - b \) (assuming \( a \neq b \))

Since \( a \) and \( b \) are distinct (i.e., \( a - b \neq 0 \)), we can divide both sides by \( a - b \):

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

### Step 5: Rearranging gives us a relationship between \( a \) and \( b \)

This implies:

\[ ab = 1 \]

### Step 6: Substitute \( b \) in terms of \( a \)

From \( ab = 1 \), we can express \( b \) as:

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

### Step 7: Substitute \( b \) back into the equation \( a - b = \frac{a}{b} \)

Now, substituting \( b = \frac{1}{a} \) into \( a - b = \frac{a}{b} \):

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

### Step 8: Rearranging the equation

This gives us:

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

Multiplying through by \( a \) (since \( a \neq 0 \)):

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

### Step 9: Rearranging to form a polynomial equation

Rearranging gives:

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

### Step 10: Analyzing the polynomial

This cubic equation can be analyzed for real roots. However, we can also check the conditions for \( a \) and \( b \) being distinct non-zero real numbers.

### Step 11: Conclusion about the signs of \( a \) and \( b \)

Since \( ab = 1 \) and both \( a \) and \( b \) are non-zero, if one is positive, the other must also be positive. Therefore:

\[ a > 0 \quad \text{and} \quad b > 0 \]