Problem 97 - College Algebra

Question

Determine whether each statement is true or false If the statement is false, make the necessary change(s) to produce a true statement. $$\frac{n !}{(n-1) !}-\frac{1}{n-1}$$

Step-by-Step Solution

Answer

The original statement seems to be a misunderstanding. The correct interpretation of the provided expression could be to simplify it, and the simplified form of the given expression is \( n - \frac{1}{n-1} \).

Step 1: Understanding the given expression

The given expression is \( \frac{n !}{(n-1) !} - \frac{1}{n-1} \). Here, \( n! \) denotes the factorial of 'n', which is the product of all positive integers up to 'n'. So, if 'n' is 5, then \( n! \) would be \( 5*4*3*2*1 \). The term \( (n-1)! \) is calculated in the same way but up to \( n-1 \). The expression thus involves division of factorials and direct subtraction operation.

Step 2: Simplifying the first term

Let's simplify the first term \( \frac{n !}{(n-1) !} \). We know that \( n! = n*(n-1)! \). So if we replace \( n! \) by \( n*(n-1)! \) in the numerator, the expression will become \( \frac{n *(n-1) !}{(n-1) !} \). \( (n-1)! \) in the numerator and denominator will cancel out, so the term simplifies to 'n'.

Step 3: Simplifying the expression

Now, substitute 'n' in place of the first term in the original expression. The expression now will be \( n - \frac{1}{n-1} \). This statement can never be true or false as it is an expression, not an equation. Therefore, the original statement seems to be a misunderstanding.

Key Concepts

Simplifying expressions

When dealing with algebraic expressions, the ability to simplify them is crucial for ease of computation and understanding. Simplifying an expression typically involves reducing it to its most straightforward form while maintaining the same value. It often includes operations like factoring, combining like terms, or canceling out common factors.
In the original exercise, we were given the expression \( \frac{n!}{(n-1)!} - \frac{1}{n-1} \). Our first step in simplification targeted the term \( \frac{n!}{(n-1)!} \). Knowing that \( n! = n \times (n-1)! \), we can directly substitute \( n \times (n-1)! \) for \( n! \). The \((n-1)!\) in both numerator and denominator cancels out, simplifying to just \( n \).
This step is crucial in transforming complex expressions into manageable ones, making further calculations much more straightforward. Recognize these patterns and simplification techniques as foundational tools in algebra.

Division of factorials

Factorials are a unique way of representing numbers as products, which simplifies the notation for products of consecutive integers. The division of factorials often occurs in combinations and permutations and can be simplified by canceling repeated terms.
For example, consider the division \( \frac{n!}{(n-1)!} \). Since the factorial \( n! \) is defined as \( n \times (n-1) \times (n-2) \times \ldots \times 1 \), and \( (n-1)! \) is \( (n-1) \times (n-2) \times \ldots \times 1 \), division allows us to cancel out all terms from \((n-1)!\) in both the numerator and the denominator. Hence, \( \frac{n!}{(n-1)!} = n \).
Understanding this concept helps in solving problems that involve permutations and combinations, as well as simplifying more complex mathematical expressions involving factorials. Division of factorials is essentially about identifying and eliminating repetitive multiplication.

Subtraction of terms

Subtraction in algebra involves finding the difference between two expressions or terms. It's important to ensure compatibility among the terms involved to simplify or solve expressions effectively.
After simplifying \( \frac{n!}{(n-1)!} \) to \( n \) in the original problem, the expression morphed into \( n - \frac{1}{n-1} \). Subtraction isn't inherently complex; however, it often requires a common understanding of the structure of both terms involved, particularly when fractions are involved.
In this case, further simplification or interpretation isn't straightforward because the expression is not an equation set to zero. Instead, the result remains an expression reflecting the difference between \( n \) and \( \frac{1}{n-1} \). Recognizing when subtraction alone does not resolve the expression's complexity is an important skill in algebra, providing insight into when additional simplification or context is necessary.