Problem 68 - Algebra and Trigonometry

Question

Simplify the compound fractional expression. $$ \frac{1+\frac{1}{c-1}}{1-\frac{1}{c-1}} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{c}{c-2} \).

Step 1: Understand the Given Expression

The problem asks us to simplify the compound fractional expression \( \frac{1+\frac{1}{c-1}}{1-\frac{1}{c-1}} \). This expression contains fractions in both the numerator and the denominator.

Step 2: Combine the Fractions in the Numerator and Denominator

We rewrite the numerator and denominator to have a common denominator. In the numerator, we have \(1 + \frac{1}{c-1} = \frac{c-1}{c-1} + \frac{1}{c-1} = \frac{c-1+1}{c-1} = \frac{c}{c-1}\). In the denominator, we have \(1 - \frac{1}{c-1} = \frac{c-1}{c-1} - \frac{1}{c-1} = \frac{c-1-1}{c-1} = \frac{c-2}{c-1}\).

Step 3: Simplify by Division of Fractions

Divide the fraction in the numerator by the fraction in the denominator. This means we multiply the numerator by the reciprocal of the denominator: \[\frac{\frac{c}{c-1}}{\frac{c-2}{c-1}} = \frac{c}{c-1} \times \frac{c-1}{c-2}.\]

Step 4: Cancel the Common Terms

Notice that \((c-1)\) appears in both the numerator and denominator of the multiplication, so they cancel out: \[ \frac{c}{c-2}.\]

Step 5: Final Simplified Expression

The simplified form of the original expression is \( \frac{c}{c-2} \).

Key Concepts

Simplifying Compound Fractional Expressions

Simplifying a compound fractional expression involves a few easy steps that can make these seemingly complex expressions much more manageable.
The first task is to understand that these expressions are just like regular fractions, but they contain fractions within the numerator or denominator, or both.
Let's break down the original problem:

• The expression given is \( \frac{1+\frac{1}{c-1}}{1-\frac{1}{c-1}} \).
• Notice that both the numerator and denominator contain fractions themselves.

To simplify, the main trick is to first deal with these internal fractions. You achieve this by finding a common denominator for the parts within the main numerator and the main denominator separately.
In our problem, this means converting \(1\) in the numerator to \(\frac{c-1}{c-1}\) and similarly for the denominator. Once they share a denominator, you can combine them.
Simplifying at this stage transforms the complex fraction into simpler term fractions like this: \(\frac{c}{c-1}\) for the numerator and \(\frac{c-2}{c-1}\) for the denominator.
Now, we are set for the next step in simplification.

Understanding Division of Fractions

Dividing fractions can seem challenging at first, but there's a simple rule that makes it easy: multiplying by the reciprocal.
When faced with a division of fractions, you actually just flip the second fraction (known as the reciprocal) and multiply.
In our given example, once we have reduced it to \(\frac{\frac{c}{c-1}}{\frac{c-2}{c-1}}\), you can apply this rule.

• Take the numerator: \( \frac{c}{c-1} \).
• Take the reciprocal of the denominator: \( \frac{c-1}{c-2} \).
• Multiply them together: \( \frac{c}{c-1} \times \frac{c-1}{c-2} \).

Multiplying these brings us to a simpler expression.
The division of fractions transforms a division problem into multiplication, which is easier to handle and often leads directly to simplification.

The Role of Reciprocals in Simplifying Fractional Expressions

The concept of a reciprocal is crucial in simplifying fractional expressions, especially in division.
A reciprocal of a fraction is created by flipping its numerator and denominator. For example, the reciprocal of \( \frac{a}{b} \) is \( \frac{b}{a} \).
In division of fractions, you don't actually "divide" in the traditional sense. Instead, you multiply the first fraction by the reciprocal of the second. This not only simplifies the arithmetic but often helps reduce other terms as well.

• In the problem we are discussing, the denominator's reciprocal was \( \frac{c-1}{c-2} \).
• By using the reciprocal for multiplication, common terms such as \((c-1)\) cancel out.

This role of the reciprocal streamlines many fraction problems down to simpler and more understandable forms.
In the end, understanding and applying reciprocals allows you to turn complicated expressions into neatly simplified results, allowing for deeper comprehension of fractional arithmetic.