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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

The square of any integer has the form \(4 k\) or \(4 k+1\) for some integer \(k\).

Observe that $$ \begin{aligned} 7524 &=7 \cdot 1000+5 \cdot 100+2 \cdot 10+4 \\ &=7(999+1)+5(99+1)+2(9+1)+4 \\ &=(7 \cdot 999+7)+(5 \cdot 99+5)+(2 \cdot 9+2)+4 \\ &=(7 \cdot 999+5 \cdot 99+2 \cdot 9)+(7+5+2+4) \\ &=(7 \cdot 111 \cdot 9+5 \cdot 11 \cdot 9+2 \cdot 9)+(7+5+2+4) \\ &=(7 \cdot 111+5 \cdot 11+2) \cdot 9+(7+5+2+4) \\ &=(\text { an integer divisible by } 9) \end{aligned} $$ \(\begin{aligned} 7524 &=7 \cdot 1000+5 \cdot 100+2 \cdot 10+4 \\\ &=7(999+1)+5(99+1)+2(9+1)+4 \\ &=(7.999+7)+(5 \cdot 99+5)+(2 \cdot 9+2)+4 \\\ &=(7 \cdot 999+5 \cdot 99+2 \cdot 9)+(7+5+2+4) \\ &=(7 \cdot 111 \cdot 9+5 \cdot 11 \cdot 9+2 \cdot 9)+(7+5+2+4) \\ &=(7 \cdot 111+5 \cdot 11+2) \cdot 9+(7+5+2+4) \\ &=(\text { an integer divisible by } 9) \\ &+(\text { the sum of the digits of } 7524) \end{aligned}\) Since the sum of the digits of 7524 is divisible by 9,7524 can be written as a sum of two integers each of which is divisible by 9 . It follows from exercise 15 that 7524 is divisible by \(9 .\) Generalize the argument given in this example to any nonnegative integer \(n\). In other words, prove that for any nonnegative integer \(n\), if the sum of the digits of \(n\) is divisible by 9 , then \(n\) is divisible by 9 ,

Given any integer \(n\), if \(n>3\), could \(n, n+2\), and \(n+4\) all be prime? Prove or give a counterexample.

Fill in the blanks in the following proof by contraposition that for all integers \(n\), if \(5 X n^{2}\) then \(5 X n\). Proof (by contraposition): [The contrapositive is: For all integers \(n\), if \(5 \mid n\) then \(5\left\lfloor n^{2}\right.\).] Suppose \(n\) is any integer such that \(\frac{(\mathrm{a})}{-}\) [We must show that (b) ] By definition of divisibility, \(n=\) (c) for some integer \(k\). By substitution, \(n^{2}=\frac{(\mathrm{d})}{-5\left(5 k^{2}\right), \text { But } 5 k^{2} \text { is an integer because it }}\) is a product of integers. Hence \(n^{2}=5 \cdot\) (an integer), and so (e) \([\) as was to be shown \(]\).

For all nonnegative real numbers \(a\) and \(b, \sqrt{a b}=\sqrt{a} \sqrt{b}\). (Note that if \(x\) is a nonnegative real number, then there is a unique nonnegative real number \(y\), denoted \(\sqrt{x}\), such that \(\left.y^{2}=x_{0}\right)\)
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