Problem 5
Question
Read the following proof that the sum of two odd numbers is even. Discuss the rules of inference used. Proof: Let \(x\) and \(y\) be odd numbers. Then \(x=2 k+1\) and \(y=2 j+1\) for some integers \(j\) and \(k\). By algebra, $$ x+y=2 k+1+2 j+1=2(k+j+1) $$ Note that \(k+j+1\) is an integer because \(k\) and \(j\) are integers. Hence \(x+y\) is even. Q.E.D.
Step-by-Step Solution
Verified Answer
The sum of two odd numbers is even because their sum can be expressed as 2 times an integer, which is always even.
1Step 1: Define Odd Numbers
Let’s define two odd numbers, say x and y. By definition, an odd number can be expressed in the form 2k+1, where k is an integer. Therefore, let:\( x = 2k + 1 \) and \( y = 2j + 1 \)where j and k are some integers.
2Step 2: Add the Two Odd Numbers
Add the expressions for x and y:\( x + y = (2k + 1) + (2j + 1) \)
3Step 3: Simplify the Sum
Simplify the addition:\( x + y = 2k + 1 + 2j + 1 = 2k + 2j + 2 = 2(k + j + 1) \)
4Step 4: Identify the Result as an Integer
Since k and j are integers, their sum (k + j + 1) is also an integer. Let's call this integer m:\( m = k + j + 1 \)
5Step 5: Conclude the Sum is Even
Since x + y = 2m, and 2 times any integer (m) is always even, we have:\( x + y = 2m \)Therefore, the sum of two odd numbers is even.
Key Concepts
rules of inferenceodd and even numbersalgebraic manipulation
rules of inference
In mathematics, rules of inference help us draw conclusions based on given premises. They are like the logical steps that guide our thought process to reach a conclusion. To understand the proof that the sum of two odd numbers is even, let's look at the key rules of inference applied:
- Definition of odd numbers: We start with the definition. Any odd number can be written as 2k + 1, where k is an integer. This is a premise that we accept as true.
- Addition of numbers: When we add two expressions of the form (2k + 1), we use the rule that allows us to combine like terms. This simplifies our expressions step-by-step.
- Simplification: We utilize algebraic manipulation (another concept explained below) to combine terms and simplify the expression further.
- Conclusion from arithmetic properties: Finally, since integers are closed under addition, the sum of integers is also an integer. Multiplying an integer by 2 will always yield an even number. Hence, we conclude the sum is even using basic properties of numbers.
odd and even numbers
Understanding odd and even numbers is crucial to grasping the proof.
- Odd Numbers: An odd number is any integer not divisible by 2. Mathematically, you can express it as 2k + 1, where k is an integer. For example, 3 (2 * 1 + 1), 5 (2 * 2 + 1), and 7 (2 * 3 + 1) are all odd numbers.
- Even Numbers: An even number is any integer that is divisible by 2. You can write it as 2k, where k is an integer. Examples include 2 (2 * 1), 4 (2 * 2), and 6 (2 * 3).
algebraic manipulation
Another vital concept is algebraic manipulation, which involves rearranging and simplifying expressions to reach a solution.
- Combining Like Terms: In our proof, we combine similar terms to make the equation simpler. For example:
\[ x + y = 2k + 1 + 2j + 1 = 2(k + j + 1) \]shows how terms are grouped and simplified. - Distribution: This rule allows us to distribute a number across terms within parentheses. This is crucial when we write 2(k + j + 1) to show the sum is essentially multiplied by 2.
- Arithmetic Properties: Understanding integer properties helps establish the conclusion. For example, recognizing that multiplying any integer by 2 will always yield an even number ensures the final step of the proof is logical and sound.
Other exercises in this chapter
Problem 4
Write two-column proofs that verify each of the following logical equivalences. \(\neg(A \vee \neg B) \vee(\neg A \wedge \neg B) \cong \neg A\)
View solution Problem 4
Determine a sentence using the and connector \((\wedge)\) that gives the negation of \(A \Longrightarrow B\).
View solution Problem 5
Rewrite the sentence "Fix the toilet or I won't pay the rent!" as a conditional.
View solution Problem 6
Sometimes in constructing a proof we find it necessary to "weaken" an inequality. For example, we might have already deduced that \(x
View solution