Problem 20 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-26: Prove that the statement is true for every positive integer \(n\). 2 is a factor of \(n^{2}+n\).

Step-by-Step Solution

Answer

For any positive integer \( n \), \( n^2+n=n(n+1) \) is even since one of \( n \) or \( n+1 \) is even.

Step 1: Understand the Problem

The problem asks us to show that for every positive integer \( n \), the expression \( n^2 + n \) is divisible by 2. This means \( n^2 + n \) is an even number because even numbers are exactly those integers that are divisible by 2.

Step 2: Factor the Expression

Notice that the expression \( n^2 + n \) can be factored as \( n(n + 1) \). This expresses the original polynomial as a product of two consecutive integers \( n \) and \( n + 1 \).

Step 3: Use Properties of Consecutive Integers

Understand that a pair of consecutive integers \( n \) and \( n + 1 \) will always include one even number. This is because if \( n \) is even, \( n + 1 \) is odd, and vice versa. Therefore, their product \( n(n + 1) \) is always even.

Step 4: Conclude the Proof

Since \( n(n + 1) \) is always even, it follows that \( n^2 + n \) is always divisible by 2. Hence, for every positive integer \( n \), 2 is a factor of \( n^2 + n \). This completes the proof that the statement is true for every positive integer \( n \).

Key Concepts

Even Numbers

In mathematics, even numbers are numbers that can be divided by 2 with no remainder. This concept is central to the proof at hand because we need to show that the expression \( n^2 + n \) yields an even number. By definition, an even number can be expressed in the form \( 2k \), where \( k \) is an integer. This means any even number, when divided by 2, results in a whole number.

Understanding even numbers helps us see why divisibility by 2 is important in number theory. If \( n^2 + n \) is even, it means that 2 evenly divides this expression, leaving no remainder. Thus, proving \( n^2 + n \) is even confirms it meets the criteria for having 2 as a factor, which is what the exercise problem requires us to demonstrate.

Consecutive Integers

Consecutive integers are integers that come one after another. For any integer \( n \), the next consecutive integer is \( n+1 \). The key point about consecutive integers is that one will always be even while the other is odd. This characteristic is crucial in proving our expression is divisible by 2.

Consider the expression \( n^2 + n \) which we can rewrite as \( n(n + 1) \). This is the product of two consecutive integers. Because one of these integers \( n \) or \( n+1 \) must be even, it follows that their product \( n(n + 1) \) is also even. This feature of consecutive integers simplifies many problems in number theory, especially when demonstrating properties involving divisibility and parity (oddness or evenness).

• Consecutive integers always contain an even number.
• The product of any two consecutive integers is even.

Divisibility

Divisibility is a fundamental concept in mathematics that deals with the ability of one number to be divided by another without leaving a remainder. In our exercise, we are dealing with divisibility by 2, which means that we need to show \( n^2 + n \) can be divided by 2 evenly.

If a number is divisible by 2, it is even, as noted earlier. The reason divisibility is so important in this context is because it allows us to confirm certain properties about numbers and expressions. For instance, proving \( n^2 + n \) is divisible by 2 tells us that no matter what positive integer \( n \) we start with, the resulting expression will always meet this divisibility standard. Recognizing how factors work and how to factor expressions helps in identifying and proving such properties of numbers.

• Divisibility by a number means the quotient is an integer with no remainder.
• Divisibility rules simplify complex arithmetic reasoning.
• Mathematical proofs often rely on demonstrating divisibility.