Problem 9
Question
Prove each directly. The square of an odd integer is odd.
Step-by-Step Solution
Verified Answer
The square of an odd integer is odd. We can prove this by assuming an odd integer has the form \(k = 2n + 1\), where \(n\) is an integer. Then, \(k^2 = (2n + 1)^2\), and expanding this expression results in \(k^2 = 2(2n^2 + 2n) + 1\). Since \(2n^2 + 2n\) is an integer, the final expression fits the definition of an odd integer, confirming that the square of an odd integer is odd.
1Step 1: Define an odd integer
An integer can be defined as odd if it can be expressed in the form \(2n + 1\), where \(n\) is an integer. Let \(k\) represent an odd integer, then \(k = 2n + 1\) for some integer \(n\).
2Step 2: Calculate the square of the odd integer
Now, find the square of the odd integer \(k\). We will represent this as \(k^2\) and substitute the expression for an odd integer into the equation:
\[k^2 = (2n + 1)^2\]
3Step 3: Expand the expression
Expand the expression using the distributive property:
\begin{align}
k^2 &= (2n + 1)(2n + 1) \\
&= (2n)(2n) + (2n)(1) + (1)(2n) + (1)(1)
\end{align}
4Step 4: Simplify the expression
Continue simplifying the expression:
\begin{align}
k^2 &= 4n^2 + 2n + 2n + 1 \\
&= 4n^2 + 4n + 1 \\
&= 2(2n^2 + 2n) + 1
\end{align}
5Step 5: Prove the square is odd
In the final expression of \(k^2\), we can see that it has the form \(2(2n^2 + 2n) + 1\), which fits the definition of an odd integer, where \(2n^2 + 2n\) is an integer. Therefore, we can conclude that the square of an odd integer is odd.
Key Concepts
Odd IntegerInteger PropertiesDistributive PropertyMathematical Induction
Odd Integer
An odd integer is a number that cannot be divided evenly by 2. In simpler terms, when you divide an odd number by 2, you always have a remainder of 1. Odd integers can be written with a general formula, which is \(2n + 1\), where \(n\) is an integer. This formula helps us identify and work with odd numbers in various mathematical scenarios, including proofs and algebraic manipulations. For example, numbers like 1, 3, 5, and 7 are all odd integers because they can be expressed in the form \(2n + 1\), with \(n\) being different values of integers. Recognizing this form is essential because it lays the groundwork for proving other properties, like showing that the square of an odd integer remains odd.
Integer Properties
Understanding the basic properties of integers is crucial when dealing with mathematical proofs. Integers include all positive and negative whole numbers, as well as zero. Some of the key properties to remember include:
- Addition and Subtraction: The sum or difference of two integers is always an integer.
- Multiplication: The product of two integers is also an integer.
- Odd and Even: Every integer is either odd or even. Even integers can be expressed in the form \(2n\), while odd integers take the form \(2n + 1\).
Distributive Property
The distributive property is a fundamental principle in algebra that allows us to simplify expressions and solve equations. It's expressed as \(a(b+c) = ab + ac\). This means you multiply each term inside a bracket by the term outside. The distributive property is especially useful in expanding and simplifying expressions:
- It helps distribute multiplication across addition or subtraction inside parentheses.
- This property allows us to break down complex expressions into simpler parts.
- First apply \((2n + 1)\) to each term inside another \((2n + 1)\).
- This results in \((2n)(2n) + (2n)(1) + (1)(2n) + (1)(1)\).
Mathematical Induction
Mathematical induction is a powerful method of proof used in mathematics to establish that a given statement is true for all natural numbers. While not directly applied in the provided exercise, it's a vital concept for comprehensive proofs.
- Basic Principle: Establish that the statement is true for the first natural number (often 1).
- Inductive Step: Assume the statement is true for an arbitrary natural number \(k\), and then prove it for \(k+1\).
Other exercises in this chapter
Problem 9
Verify each, where \(f\) denotes a contradiction. (See Table \(1.14 . )\) $$ \sim(p \rightarrow q)=p \wedge \neg q $$
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Let \(P(x) : x^{2} > x, Q(x) : x^{2}=x,\) and the UD \(=\) set of integers. Determine the truth value of each proposition. $$(\exists x)|\mathrm{P}(x) \wedge \m
View solution Problem 9
Verify each, where \(f\) denotes a contradiction. $$\sim(p \rightarrow q) \equiv p \wedge \sim q$$
View solution