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Chapter 3: Problem 40

The negative of any odd integer is odd.

Short Answer

Expert verified
The given statement is true. If \(x\) is an odd integer represented as \(2n + 1\), then the negation of \(x\), denoted as \(-x\), can be simplified to \(-2n - 1\), which is also an odd integer as it can be represented in the form \(2m + 1\), where \(m = -n\).

Step by step solution

01

Assume an odd integer

Let's assume any odd integer as \(x\). By definition of an odd integer, it can be represented as \(2n + 1\) where n is an integer. So, \(x = 2n + 1\).
02

Negate the odd integer

Now, we need to find the negative of \(x\), which we can represent as \(-x\). The negation of \(x\) can be found by multiplying it by \(-1\). \(-x = -1(2n + 1)\)
03

Simplify the expression

Let's simplify the expression above: \(-x = -2n - 1\)
04

Check for oddness

The resulting expression can be rewritten in the form of an odd integer if it can be represented as \(2m + 1\), where m is an integer. We can rewrite the expression: \(-x = 2(-n) - 1\). Now, let m = -n. Since n is an integer, m is also an integer. Hence, the negation of any odd integer is also an odd integer, as \(-x = 2m + 1\) (or specifically, \(-x = 2(-n) - 1\)). This proves the statement: "The negative of any odd integer is odd."

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Key Concepts

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

Integer Negation
Integers include both positive numbers, negative numbers, and zero. Integer negation involves changing the sign of an integer. Therefore, to negate an integer means to multiply it by -1. This operation flips its position on the number line.
  • If the original integer is positive, the result is negative.
  • If the original integer is negative, the result is positive.

For odd integers, such as those expressed in the form of 2n + 1 (where I is an integer), negation follows the same principle. For example, multiplying the odd integer 2n + 1 by -1 gives us -2n - 1. This result transforms the positive odd number into a negative odd number, highlighting the inseparable nature of property retention during integer negation.
Proof by Structure
Proof by structure involves logically demonstrating that a mathematical property or statement is universally true, by relying on the inherent characteristics or construction of the objects involved.
In proving that the negative of an odd integer is also odd, we utilize the structured definition of odd integers. We start with an integer expressed in its standard odd form 2n + 1 and show through structured steps that its negation -2n - 1 retains this format.
  • Step 1: Assume the odd integer form.
  • Step 2: Apply negation.
  • Step 3: Simplify and observe structural retention.
  • Step 4: Re-identify as an odd integer form 2m + 1 where I is an integer.
This technique does not rely on individual examples, but on the structural properties of the expressions involved, ensuring a comprehensive proof applicable to all relevant cases.
Integer Representation
Understanding integer representation is crucial for identifying the properties of numbers. Integers can be characterized in various forms, but one of the simplest and most intuitive representations is based on even and odd categories. For odd integers, the standard representation is 2n + 1, indicating they fall one unit above an even integer.
  • Standard Representation: Any integer can be expressed as n (for even) or 2n + 1 (for odd).
  • Even vs Odd: Even integers are divisible by 2 ( 2n), while odd integers are defined by their remainder being 1 when divided by 2.
  • Re-identifying: Negative odd integers maintain the form -2n 鈥� 1, being a unit below a negative even number.

When discussing operations like negation, having a clear representation aids in visualizing and understanding how the properties of integers are preserved.

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Most popular questions from this chapter

If \(n\) is an integer and \(n>1\), then \(n !\) is the product of \(n\) and every other positive integer that is less than \(n\). For example, \(5 !=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\). a. Write \(6 !\) in standard factored form. b. Write \(20 !\) in standard factored form. c. Without computing the value of \((20 !)^{2}\) determine how many zeros are at the end of this number when it is written in decimal form. Justify your answer.

Evaluate the expressionsa. 28 div 5 b. 28 mod 5

Let \(N=2 \cdot 3 \cdot 5 \cdot 7+1\). What remainder is obtained when \(N\) is divided by 2 ? 3 ? 5 ? 7 ? Is \(N\) prime? Justify your answer.

The product of any four consecutive integers is divisible by 8 .

Write an algorithm that accepts the numerator and denominator of a fraction as input and produces as output the numerator and denominator of that fraction written in lowest terms. (The algorithm may call upon the Euclidean algorithm as needed.)
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