**Mathematical Proofs and Concepts****1. Proof that for all integers $ n $, $ n^2 - n + 3 $ is odd.**To prove that $ n^2 - n + 3 $ is odd for all integers $ n $, we need to consider the expression modulo 2. - **Case 1: $ n $ is even** Let $ n = 2k $ for some integer $ k $. Then $ n^2 = (2k)^2 = 4k^2 $, and $ n^2 - n + 3 = 4k^2 - 2k + 3 $. Modulo 2, this simplifies to: \[ 4k^2 \equiv 0 \pmod{2}, \quad 2k \equiv 0 \pmod{2}, \quad 3 \equiv 1 \pmod{2} \] \[ n^2 - n + 3 \equiv 0 - 0 + 1 \equiv 1 \pmod{2} \] Thus, $ n^2 - n + 3 $ is odd.- **Case 2: $ n $ is odd** Let $ n = 2k + 1 $ for some integer $ k $. Then $ n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 $, and $ n^2 - n + 3 = 4k^2 + 4k + 1 - (2k + 1) + 3 $. Modulo 2, this simplifies to: \[ 4k^2 \equiv 0 \pmod{2}, \quad 4k \equiv 0 \pmod{2}, \quad 1 - 1 + 3 \equiv 3 \equiv 1 \pmod{2} \] \[ n^2 - n + 3 \equiv 0 + 0 + 1 \equiv 1 \pmod{2} \] Thus, $ n^2 - n + 3 $ is odd.In both cases, \(

Name: **Mathematical Proofs and Concepts****1. Proof that for all integers
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Description: **Mathematical Proofs and Concepts****1. Proof that for all integers $ n $, $ n^2 - n + 3 $ is odd.**To prove that $ n^2 - n + 3 $ is odd for all integers $ n $, we need to consider the expression modulo 2. - **Case 1: $ n $ is even** L

To prove: For all integers n, n2-n+3 is odd. Assume that n is an integer.

Answered: Prove that for all integers n, n² – n +…

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Prove that for all integers n, n² – n + 3 is odd.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Logic is the analysis of general patterns of thoughts without regard to any specific sense of context.

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**Mathematical Proofs and Concepts**

**1. Proof that for all integers $ n $, $ n^2 - n + 3 $ is odd.**

To prove that $ n^2 - n + 3 $ is odd for all integers $ n $, we need to consider the expression modulo 2.

- **Case 1: $ n $ is even**
Let $ n = 2k $ for some integer $ k $.
Then $ n^2 = (2k)^2 = 4k^2 $, and $ n^2 - n + 3 = 4k^2 - 2k + 3 $.
Modulo 2, this simplifies to:
\[ 4k^2 \equiv 0 \pmod{2}, \quad 2k \equiv 0 \pmod{2}, \quad 3 \equiv 1 \pmod{2} \]
\[ n^2 - n + 3 \equiv 0 - 0 + 1 \equiv 1 \pmod{2} \]
Thus, $ n^2 - n + 3 $ is odd.

- **Case 2: $ n $ is odd**
Let $ n = 2k + 1 $ for some integer $ k $.
Then $ n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 $, and $ n^2 - n + 3 = 4k^2 + 4k + 1 - (2k + 1) + 3 $.
Modulo 2, this simplifies to:
\[ 4k^2 \equiv 0 \pmod{2}, \quad 4k \equiv 0 \pmod{2}, \quad 1 - 1 + 3 \equiv 3 \equiv 1 \pmod{2} \]
\[ n^2 - n + 3 \equiv 0 + 0 + 1 \equiv 1 \pmod{2} \]
Thus, $ n^2 - n + 3 $ is odd.

In both cases, \(

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