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Chapter 3: Problem 25

The difference of any even integer minus any odd integer is odd.

Short Answer

Expert verified
The difference between any even integer (\(2n\)) and any odd integer (\(2m+1\)) is given by: \( (2n) - (2m + 1) = 2(n - m) - 1 \) Since \(2(n - m) - 1\) is in the form \(2k + 1\) (where k is any integer), the difference of any even integer minus any odd integer is odd.

Step by step solution

01

General form of even and odd integers

An even integer can be written in the form \(2n\) where n is any integer. Similarly, an odd integer can be written in the form \(2m + 1\) where m is also any integer.
02

Subtract the odd integer from the even integer

Now, we will find the difference between the even integer "e" and the odd integer "o" as follows: \(e - o = (2n) - (2m + 1)\)
03

Simplify the expression

Now, let's simplify the expression by combining the terms: \(e - o = 2n - 2m - 1\)
04

Factor out "2" from the terms with n and m

To make the expression easier to analyze, we can factor out "2" from the terms with n and m: \(e - o = 2(n - m) - 1\)
05

Prove the expression is odd

Now we can observe that our expression is in the form \(2k + 1\) where k is any integer (in this case, k = n - m). As we saw earlier, this is the general form of an odd integer. Therefore, the difference of any even integer minus any odd integer is indeed odd.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Even and Odd Integers
Understanding the difference between even and odd integers is crucial in mathematics. Let's break them down into simpler terms.
Even integers are numbers that can be divided by 2 without leaving a remainder. This makes them look like 2, 4, 6, 8, and so on. In algebraic terms, we represent an even integer as \(2n\), where \(n\) stands for any integer. Odd integers, on the other hand, are numbers that leave a remainder of 1 when divided by 2. Examples would be 1, 3, 5, 7, etc. We can express an odd integer in algebra as \(2m + 1\), where \(m\) is any integer.
  • Even Integer: Takes the form \(2n\)
  • Odd Integer: Takes the form \(2m + 1\)
Recognizing these forms is the first step in solving problems that involve distinguishing between even and odd numbers.
Integer Arithmetic
Integer arithmetic deals with operations involving whole numbers, both positive and negative, and zero. It is the basis for more complex operations in mathematics.
In the context of our problem, the operation we are focusing on is subtraction. If we subtract an odd integer from an even integer, we can expect certain results based on the arithmetic rules of integers. Let's break it down:
  • Subtraction of Integers: The operation where one integer is subtracted from another.
  • Even - Odd: When subtracting, the difference tends to follow a pattern.
Think of it like this, if you view the subtraction of integers as moving from one number to another on a number line, the direction and distance help determine whether the result will end up being odd or even.
Algebraic Expressions
Algebraic expressions are combinations of symbols and numbers, constructed according to specific rules. Understanding these expressions is vital as they form the basis of algebra and help in solving numerous mathematical problems.
In the problem statement, we began with two algebraic expressions: \(2n\) and \(2m + 1\), representing even and odd integers respectively. By subtracting the expression for the odd integer from the even integer, we arrived at: \[2n - (2m + 1) = 2n - 2m - 1\]This expression simplifies our calculation. We then factored out a 2 from terms with \(n\) and \(m\), leading us to: \[2(n - m) - 1\]
  • Simplification: Helps in identifying patterns by using basic arithmetic rules.
  • Factorization: Often reveals hidden characteristics of numbers.
By simplifying to the form \(2k + 1\), where \(k = n - m\), it's evident that the difference between an even and an odd integer leads to an odd number. This deductive reasoning helps in demonstrating key mathematical proofs.

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Most popular questions from this chapter

a. Use proof by contradiction to show that for any integer \(n\), it is impossible for \(n\) to equal both \(3 q_{1}+r_{1}\) and \(3 q_{2}+r_{2}\), where \(q_{1}, q_{2}, r_{1}\), and \(r_{2}\), are integers, \(0 \leq r_{1}<\) \(3,0 \leq r_{2}<3\), and \(r_{1} \neq r_{2}\). b. Use proof by contradiction, the quotient-remainder theorem, division into cases, and the result of part (a) to prove that for all integers \(n\), if \(n^{2}\) is divisible by 3 then \(n\) is divisible by 3 . c. Prove that \(\sqrt{3}\) is irrational.

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