Answered: 5. Proven and n² always have the same…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

#5. Thanks.

$**Problem 5**: Prove $ n $ and $ n^2 $ always have the same parity. That is, $ n $ is even if and only if $ n^2 $ is even. ### Explanation Parity refers to whether a number is even or odd. The goal is to prove that a number $ n $ and its square $ n^2 $ share the same parity. This means: 1. If $ n $ is even, then $ n^2 $ is even. 2. If $ n $ is odd, then $ n^2 $ is odd. **Proof:** 1. **Case 1: $ n $ is even.** - If $ n $ is even, there exists an integer $ k $ such that $ n = 2k $. - Then $ n^2 = (2k)^2 = 4k^2 = 2(2k^2) $, which is clearly even because it is divisible by 2. 2. **Case 2: $ n $ is odd.** - If $ n $ is odd, there exists an integer $ k $ such that $ n = 2k + 1 $. - Then $ n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 $. - Notice that $ 4k^2 + 4k $ is divisible by 2, so it is even. Thus, $ 4k^2 + 4k + 1 $ is odd because it is even plus one. Therefore, we have shown that $ n $ and $ n^2 $ always have the same parity.$

Transcribed Image Text:**Problem 5**: Prove $ n $ and $ n^2 $ always have the same parity. That is, $ n $ is even if and only if $ n^2 $ is even. ### Explanation Parity refers to whether a number is even or odd. The goal is to prove that a number $ n $ and its square $ n^2 $ share the same parity. This means: 1. If $ n $ is even, then $ n^2 $ is even. 2. If $ n $ is odd, then $ n^2 $ is odd. **Proof:** 1. **Case 1: $ n $ is even.** - If $ n $ is even, there exists an integer $ k $ such that $ n = 2k $. - Then $ n^2 = (2k)^2 = 4k^2 = 2(2k^2) $, which is clearly even because it is divisible by 2. 2. **Case 2: $ n $ is odd.** - If $ n $ is odd, there exists an integer $ k $ such that $ n = 2k + 1 $. - Then $ n^2 = (2k + 1)^2 = 4k^2 + 4k + 1 $. - Notice that $ 4k^2 + 4k $ is divisible by 2, so it is even. Thus, $ 4k^2 + 4k + 1 $ is odd because it is even plus one. Therefore, we have shown that $ n $ and $ n^2 $ always have the same parity.

Expert Solution