Problem 13 - Precalculus: Mathematical for Calculus

Question

Show that \(n^{2}+n\) is divisible by 2 for all natural numbers \(n\)

Step-by-Step Solution

Answer

Yes, \(n^2+n\) is divisible by 2 for all natural numbers \(n\).

Step 1: Understand the problem

We need to prove that \(n^2+n\) is divisible by 2 for every natural number \(n\). This means that when \(n^2+n\) is divided by 2, there will be no remainder.

Step 2: Express the expression in a factorized form

Recognize that \(n^2+n\) can be rewritten as \(n(n+1)\). This expression represents the product of two consecutive integers, \(n\) and \(n+1\).

Step 3: Analyze the properties of consecutive integers

Recall that in any set of two consecutive integers, one is always even. For example, if \(n\) is even, \(n+1\) is odd, and vice versa.

Step 4: Conclude divisibility by 2

Since one of the integers \(n\) or \(n+1\) is guaranteed to be even, their product \(n(n+1)\) must also be even. Therefore, \(n^2+n\) is divisible by 2.

Key Concepts

Natural Numbers

Natural numbers are the simplest form of numbers that we start to learn when counting. These numbers include all the positive integers starting from 1 and go on infinitely: 1, 2, 3, 4, and so on.
Natural numbers do not include zero or negative numbers. They are important in many areas of mathematics as they are the foundation for more complex number systems.
When solving problems involving natural numbers, like the one in our original exercise, we use their properties to understand divisibility and other mathematical operations. Since natural numbers are used to solve real-world counting problems, they hold a key role in problem-solving and logical thinking.

• Properties: Non-zero, positive integers
• Examples: 1, 2, 3, 4,...
• Basic operations: addition, subtraction, multiplication

When dealing with these numbers, it's essential to understand how their properties apply to larger concepts, like divisibility and even or odd nature, ensuring you have a strong foundation to tackle exercises efficiently.

Consecutive Integers

Consecutive integers are a sequence of numbers where each number is one more than the previous one. For example, if we start with 3, the next consecutive integers would be 4, 5, 6, and so forth.
Understanding consecutive integers is important when dealing with problems of divisibility and parity, as demonstrated in our original solution.
Consider the expression from the exercise: it uses the consecutive integers, \(n\) and \(n+1\). By their nature, consecutive numbers always include one even and one odd number.

• Properties: Always include both an even and an odd number in any two consecutive integers
• Example: The sequence 5, 6, 7 is consecutive; 5 is odd, 6 is even, and 7 is odd.
• Applications: Used in defining sequences, solving algebraic expressions

This property that every pair of consecutive integers contains one even number is why expressions involving consecutive integers, like \(n(n+1)\), guarantee that the product is divisible by 2.

Even and Odd Numbers

Even and odd numbers are two distinct categories used to classify integers based on divisibility by 2. An even number is any integer that can be divided by 2 without leaving a remainder. Conversely, an odd number leaves a remainder of 1 when divided by 2.
These properties are fundamental in many mathematical calculations, including our exercise. When we have a product of consecutive integers like \(n(n+1)\), the nature of even and odd numbers helps us determine divisibility.
For example:

• If \(n\) is even, then \(n+1\) is odd, making the product \(n(n+1)\) even.
• If \(n\) is odd, then \(n+1\) is even, again resulting in an even product.

Understanding these characteristics of even and odd numbers allows us to conclude about the divisibility of expressions like \(n^2+n\) by 2 effectively.
This principle aids in problem-solving, where recognizing the pattern of evenness in mathematical expressions can simplify complex calculations.