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Chapter 3: Problem 28

For all integers \(n\), if \(n\) is odd then \(n^{2}\) is odd.

Short Answer

Expert verified
We are given that \(n\) is odd, which means it can be represented as \(n = 2k + 1\), where \(k\) is an integer. We want to show that \(n^2\) is also odd. By calculating and simplifying, we find that \(n^2 = (2k + 1)^2 = 4k^2 + 4k + 1\). We can rewrite this expression as \(n^2 = 2(2k^2 + 2k) + 1\), which is in the form of an odd integer. Hence, we have proven that if \(n\) is odd, then \(n^2\) is also odd.

Step by step solution

01

1. Understand the definition of an odd integer

An integer is considered odd if it can be written in the form \(2k + 1\) where \(k\) is also an integer. Essentially, an odd integer is one more than an even integer (which is a multiple of 2).
02

2. Represent the integer n as an odd integer

Since we are given that \(n\) is odd, we can represent \(n\) as \(2k + 1\), where \(k\) is an integer.
03

3. Calculate the square of n, represented as an odd integer

Now we need to calculate \(n^2\). Since we represented \(n\) as \(2k + 1\), we can write the expression for \(n^2\) as follows: \[ (2k + 1)^2 \]
04

4. Expand the squared expression

In order to prove that \(n^2\) is odd, we first need to expand the squared expression. We can do this using the distributive property (also known as the FOIL method), which is: \[(a + b)^2 = a^2 + 2ab + b^2 \] Applying the distributive property to our expression, we get: \[(2k + 1)^2 = (2k)^2 + 2(2k)(1) + (1)^2 \]
05

5. Simplify the expanded expression

Simplify the expanded expression by performing the multiplications and combining like terms: \[ (2k)^2 + 2(2k)(1) + (1)^2 = 4k^2 + 4k + 1 \]
06

6. Show that the result is odd

Now, we can show that \(n^2\) is indeed odd by rearranging the expression and factoring out a 2: \( n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 \) We can see that the expression is now in the form of an odd integer (i.e., \(2j + 1\)), where \(j = 2k^2 + 2k\) is an integer since \(k\) is also an integer. Thus, we have proven that if n is odd, then \(n^2\) is also odd.

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Key Concepts

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

Odd and Even Integers
Odd and even integers form the foundation of many topics in discrete mathematics. An integer is either odd or even, which helps in various proofs and solutions.

An odd integer can be defined as any integer that can be expressed in the form \(2k + 1\), where \(k\) is an integer. This means when you divide an odd number by 2, there is always a remainder of 1. For example, 3, 5, and 7 are all odd integers.

On the other hand, even integers are expressed as \(2k\), where \(k\) is an integer. Dividing an even number by 2 leaves no remainder. Examples of even integers include 2, 4, and 6.

Understanding these distinctions is crucial when analyzing and performing operations on numbers, especially in problems and proofs involving algebraic expressions in discrete math.
Mathematical Proof
Mathematical proofs are essential in validating statements or theorems within mathematics, ensuring they hold true for all cases specified.

A proof involves a logical sequence of statements and justifications, leading from hypotheses to conclusion. In our exercise, we are proving that the square of an odd integer is also odd. This requires understanding properties of odd integers, carrying through algebraic manipulations, and logical reasoning to arrive at the result that if \(n\) is an odd integer, \(n^2\) will be in the form \(2j + 1\), where \(j\) is an integer.

Proof techniques like direct proof, as used here, involve starting with the given hypothesis and applying known definitions to derive the required outcome. Such structured argument enhances clarity and communication in mathematics.
Expansion and Simplification
Expansion and simplification are algebraic techniques used to transform expressions, making them easier to work with or prove properties.

In the problem, we focus on expanding \((2k + 1)^2\). This operation uses the distributive property, known as FOIL (First, Outside, Inside, Last), to reformat the expression:
  • First, calculate \( (2k)^2 \).
  • Then, compute \( 2 \cdot (2k) \cdot 1 \).
  • Finally, add \( 1^2 \).

These steps result in \(4k^2 + 4k + 1\), which is simplified further to demonstrate that it takes the form of an odd number, \(2(2k^2 + 2k) + 1\).

Mastering expansion and simplification not only aids in proving properties within exercises but also builds competence in handling complex algebraic expressions.

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Most popular questions from this chapter

Assume that \(m\) and \(n\) are particular integers. \(\begin{array}{ll}\text { a. Is } 6 m+8 n \text { even? } & \text { b. Is } 10 m n+7 \text { odd? }\end{array}\) c. If \(m>n>0\), is \(m^{2}-n^{2}\) composite?

Definition: The least common multiple of two nonzero integers \(a\) and \(b\), denoted \(\operatorname{lcm}(a, b)\), is the positive integer \(c\) such that a. \(a \mid c\) and \(b \mid c\) b. for all integers \(m\), if \(a \mid m\) and \(b \mid m\), then \(c \mid m\). Prove that for all integers \(a\) and \(b, \operatorname{gcd}(a, b) \mid \operatorname{lcm}(a, b)\).

Theorem: For all integers \(k\), if \(k>0\) then \(k^{2}+2 k+1\) is composite. "Proof: Suppose \(k\) is any integer such that \(k>0\). If \(k^{2}+\) \(2 k+1\) is composite, then \(k^{2}+2 k+1=r \cdot s\) for some integers \(r\) and \(s\) such that \(1

Consider the statement "For all real numbers \(r\), if \(r^{2}\) is irrational then \(r\) is irrational." a. Write what you would suppose and what you would need to show to prove this statement by contradiction. b. Write what you would suppose and what you would need to show to prove this statement by contraposition.

The product of any even integer and any integer is even.
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