If `f(x)=(x)/(x-1)`, then prove that `(f(a))/(f(a+1))=f(a^(2))`

Answer

To prove that \(\frac{f(a)}{f(a+1)} = f(a^2)\) for the function \(f(x) = \frac{x}{x-1}\), we will follow these steps:

### Step 1: Calculate \(f(a)\)

We start by substituting \(a\) into the function \(f(x)\):

\[ f(a) = \frac{a}{a - 1} \]

### Step 2: Calculate \(f(a + 1)\)

Next, we substitute \(a + 1\) into the function:

\[ f(a + 1) = \frac{a + 1}{(a + 1) - 1} = \frac{a + 1}{a} \]

### Step 3: Calculate \(\frac{f(a)}{f(a + 1)}\)

Now we will find the ratio \(\frac{f(a)}{f(a + 1)}\):

\[ \frac{f(a)}{f(a + 1)} = \frac{\frac{a}{a - 1}}{\frac{a + 1}{a}} = \frac{a}{a - 1} \cdot \frac{a}{a + 1} \]

This simplifies to:

\[ \frac{a^2}{(a - 1)(a + 1)} \]

### Step 4: Simplify the expression

We can recognize that \((a - 1)(a + 1) = a^2 - 1\). Thus, we have:

\[ \frac{f(a)}{f(a + 1)} = \frac{a^2}{a^2 - 1} \]

### Step 5: Calculate \(f(a^2)\)

Now we will calculate \(f(a^2)\):

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

### Step 6: Compare the two expressions

Now we can compare \(\frac{f(a)}{f(a + 1)}\) and \(f(a^2)\):

\[ \frac{f(a)}{f(a + 1)} = \frac{a^2}{a^2 - 1} = f(a^2) \]

### Conclusion

Thus, we have shown that:

\[ \frac{f(a)}{f(a + 1)} = f(a^2) \]

This completes the proof.

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