Problem 11 - Discrete Mathematics with Applications

Question

Prove each directly. The product of any even integer and any odd integer is even.

Step-by-Step Solution

Answer

Let the even integer be represented as \(2n\) and the odd integer be represented as \(2m + 1\), where n and m are integers. Find the product of these two numbers: \((2n)(2m + 1) = 2n(2m) + 2n(1) = 4nm + 2n\). Factor out the common factor of 2: \(2(2mn + n)\). Since the product is a multiple of 2, it is an even number. Thus, we have proven that the product of any even integer and any odd integer is always even.

Step 1: Define even and odd integers

Let's represent the even integer as 2n, where n is an integer. Likewise, represent the odd integer as 2m + 1, where m is also an integer.

Step 2: Find the product of the even and odd integers

We need to find the product of these two numbers. To do this, multiply the even integer (2n) by the odd integer (2m + 1): \( (2n)(2m + 1) \)

Step 3: Distribute the even integer

Distribute the 2n to both terms in the parenthesis: \( 2n(2m) + 2n(1) \)

Step 4: Simplify the expression

Simplify each term in the expression to find the product: \( 4nm + 2n \)

Step 5: Factor out the common factor of 2

Notice that 2 is a common factor in both terms of the expression. We can factor it out to demonstrate that the product is even: \( 2(2mn + n) \)

Step 6: Conclude the proof

Since we have factored out a 2 from the expression, we can see that the product of the even and odd integers is a multiple of 2, making it an even number. Therefore, we have proved that the product of any even integer and any odd integer is always even.

Key Concepts

Even and Odd Numbers

Understanding even and odd numbers is essential to solving many number theory problems.
An **even number** is any integer that can be expressed as twice another integer. In mathematical terms, an even number can be written as \( 2n \), where \( n \) is an integer. Since even numbers can be divided evenly by 2, this characteristic helps identify them.
On the other hand, an **odd number** cannot be divided evenly by 2. Instead, odd numbers are represented as \( 2m + 1 \), where \( m \) is an integer. This representation ensures that when an integer \( m \) is used, the resulting number will not divide evenly by 2, giving a remainder of 1.

• Example of even numbers: 2, 4, 8, 10.
• Example of odd numbers: 1, 3, 5, 7.

Recognizing these properties is crucial when dealing with integer multiplication, as we'll see in the related proofs.

Direct Proof

A direct proof is a straightforward approach to proving mathematical statements. In the context of the provided exercise, we use a direct proof to show that the product of an even integer and an odd integer is always even.
Here's how a direct proof typically works:

• Start with known facts or definitions. In this case, the definitions of even and odd numbers.
• Apply logical and mathematical operations to these facts.
• Reach a conclusion that directly demonstrates the statement we're trying to prove.

For this exercise, starting with the definition of the even integer \( 2n \) and the odd integer \( 2m + 1 \), we can illustrate that their multiplication results in an even number.
By expanding \( (2n)(2m + 1) \) using distribution, we show each step of the process leading to a product with a common factor of 2. Factoring this out validates that the product is indeed even.

Integer Multiplication

Multiplying integers is a fundamental math concept that, when combined with our understanding of even and odd numbers, can provide interesting outcomes as in the exercise.
To multiply integers, one must follow basic arithmetic rules. When dealing with \( (2n)(2m + 1) \), distribute one integer across the terms of the other.

• Distribute: \( 2n(2m + 1) \) becomes \( 2n \cdot 2m + 2n \cdot 1 \)
• Simplify: This results in \( 4nm + 2n \)

Next, identify and factor out any common terms from the expression to simplify.
Recognizing that both terms, \( 4mn \) and \( 2n \), share a common factor of 2, allows for further simplification: \( 2(2mn + n) \), confirming the product is even.
This method effectively shows how integer multiplication confirms our proposition and why examining individual terms in such products is essential.