Problem 20 - Algebra for College Students

Question

n^{2}-n \text { is divisible by } 2

Step-by-Step Solution

Answer

The expression \( n^2 - n \) is always divisible by 2 because it involves the product of two consecutive integers.

Step 1: Understand the Problem

We need to prove that the expression \( n^2 - n \) is divisible by 2 for any integer \( n \). When an expression is divisible by 2, it means that it results in an even number.

Step 2: Factorize the Expression

Notice that the expression \( n^2 - n \) can be factorized. It can be written as \( n(n - 1) \). This factorization helps in understanding the behavior of the product regarding divisibility by 2.

Step 3: Analyze Consecutive Integers

The expression \( n(n-1) \) consists of two consecutive integers, \( n \) and \( n-1 \). Recall that every pair of consecutive integers includes one even number since one must be divisible by 2. Therefore, their product \( n(n-1) \) must also be even.

Step 4: Conclude the Divisibility

Since \( n(n-1) \) results in an even number for any integer \( n \), \( n^2 - n \) is always divisible by 2.

Key Concepts

Factorization

Factorization is a crucial concept in algebra. It involves breaking down complex expressions into simpler terms or products that are easier to handle. In the case of the expression \( n^2 - n \), factorization allows us to express it as \( n(n-1) \). This new form is particularly useful because multiplying \( n \) by \( n-1 \) shows the structure of the expression more clearly.

• Factorization can simplify expressions and make solving equations easier.
• It reveals underlying properties, such as divisibility, by showing an expression in terms of its components.

In the original problem, factorization was used to determine that \( n^2 - n \) is divisible by 2. By rewriting the expression as a product of two consecutive integers, we can easily analyze its divisibility.

Consecutive Integers

Consecutive integers are numbers that follow each other in order, without any gap between them. For example, if \( n \) is a number, then \( n \) and \( n-1 \) are consecutive integers. Understanding consecutive integers is essential when analyzing products, like in the factor \( n(n-1) \) derived from \( n^2 - n \).

• One of the defining characteristics of consecutive numbers is that one of them must always be even.
• This property is particularly useful when considering divisibility, such as divisibility by 2.

In our case, each pair of consecutive integers \( n \) and \( n-1 \) guarantees that the product \( n(n-1) \) contains an even number, thereby making the entire product even.

Even Numbers

An even number is any integer divisible by 2 without a remainder. Examples include 2, 4, 6, and 8. To determine if an algebraic expression like \( n^2 - n \) is even, factorization and analysis of the integers involved are useful tools.

• An even number can be written in the form \( 2k \), where \( k \) is an integer.
• Multiplying any integer by an even number results in an even product.

In the expression \( n(n-1) \), because one of the terms is always even due to them being consecutive, the result is an even number. Therefore, \( n^2 - n \) is guaranteed to be divisible by 2 for every integer \( n \).