rove that the sum of an even integer and an odd integer is odd.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Direct Proof: Proving the Sum of an Even Integer and an Odd Integer is Odd**
To prove that the sum of an even integer and an odd integer is odd, let's define the problem using basic algebraic expressions.
1. **Definition of Even and Odd Integers:**
- An even integer can be expressed as \(2n\), where \(n\) is an integer.
- An odd integer can be expressed as \(2m + 1\), where \(m\) is an integer.
2. **Sum of Even and Odd Integers:**
- Let the even integer be \(2n\).
- Let the odd integer be \(2m + 1\).
3. **Calculating the Sum:**
\[
(2n) + (2m + 1) = 2n + 2m + 1
\]
4. **Rearranging the Expression:**
\[
2n + 2m + 1 = 2(n + m) + 1
\]
5. **Analysis of the Result:**
- The expression \(2(n + m)\) is even, as it is a multiple of 2.
- Adding 1 to an even number results in an odd number.
6. **Conclusion:**
- Thus, the sum \(2(n + m) + 1\) is odd.
By this direct proof, we have demonstrated that the sum of an even integer and an odd integer is indeed always odd.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F13912691-4f2a-48b8-8b2a-dac872326113%2Fec966141-06ed-4122-b698-55fb0fa34085%2Fsgxcf5e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Direct Proof: Proving the Sum of an Even Integer and an Odd Integer is Odd**
To prove that the sum of an even integer and an odd integer is odd, let's define the problem using basic algebraic expressions.
1. **Definition of Even and Odd Integers:**
- An even integer can be expressed as \(2n\), where \(n\) is an integer.
- An odd integer can be expressed as \(2m + 1\), where \(m\) is an integer.
2. **Sum of Even and Odd Integers:**
- Let the even integer be \(2n\).
- Let the odd integer be \(2m + 1\).
3. **Calculating the Sum:**
\[
(2n) + (2m + 1) = 2n + 2m + 1
\]
4. **Rearranging the Expression:**
\[
2n + 2m + 1 = 2(n + m) + 1
\]
5. **Analysis of the Result:**
- The expression \(2(n + m)\) is even, as it is a multiple of 2.
- Adding 1 to an even number results in an odd number.
6. **Conclusion:**
- Thus, the sum \(2(n + m) + 1\) is odd.
By this direct proof, we have demonstrated that the sum of an even integer and an odd integer is indeed always odd.
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