题目

Problem

Given that $n\in\mathbb{N}$ , use algebra to prove by contradiction that

\text{“if }n^2-4n+5\text{ is even then }n\text{ is odd”}

(4)

题目中文翻译

已知 $n\in\mathbb{N}$ ，用代数中的反证法证明：

\text{“如果 }n^2-4n+5\text{ 是偶数，那么 }n\text{ 是奇数”}

解答

We prove the statement by contradiction.

Assume that

n^2-4n+5

is even, but $n$ is not odd.

Since $n\in\mathbb N$ , this means $n$ is even.

So let

n=2k

where $k$ is an integer.

Then

\begin{aligned} n^2-4n+5 &=(2k)^2-4(2k)+5 \\ &=4k^2-8k+5 \\ &=4(k^2-2k)+5 \end{aligned}

Now $4(k^2-2k)$ is even, and $5$ is odd.

Therefore

4(k^2-2k)+5

is odd.

So $n^2-4n+5$ is odd, contradicting the assumption that $n^2-4n+5$ is even.

Hence the assumption is false.

Therefore, if $n^2-4n+5$ is even, then $n$ is odd.