명제의 증명과 절대부등식 연습문제 — Sundstrom, Mathematical Reasoning · 3.3절 · 2b번

Sundstrom, Mathematical Reasoning · 3.3절 · 2b번영어 원문

Are the following statements true or false? Justify each conclusion.

For all integers $a$ and $b$ , if $a$ is even and $b$ is odd, then $6$ does not divide $\left( a^2 + b^2 \right)$ .

정답·풀이 보기 (영어 원문)

This statement is true. Use a proof by contradiction. So assume that there exist integers $a$ and $b$ such that $a$ is even, $b$ is odd, and 6 divides $a^2 + b^2$ . So there exist integers $m$ and $n$ such that

$a = 2m \qquad \text{ and } \qquad a^2 + b^2 = 6n$

. Substitute $a = 2m$ into the second equation and use algebra to rewrite in the form $b^2 = 2(3n - 2m^2)$ . This means that $b^2$ is even and hence, that $b$ is even. This is a contradiction.

한 번 더 — 비슷한 문제Sundstrom, Mathematical Reasoning · 3.3절 · 2d번영어 원문

Are the following statements true or false? Justify each conclusion.

For all integers $a$ and $b$ , if $a$ is odd and $b$ is odd, then $4$ divides $\left( a^2 + 3b^2 \right)$ .

다 풀었으면 정답 확인

This statement is true. Use a direct proof. Let $a$ and $b$ be integers and assume they are odd. So there exist integers $m$ and $n$ such that

$a = 2m + 1 \qquad \text{ and } \qquad b = 2n + 1$

. We then see that

$\begin{aligned}a^2 + 3b^2 & = 4m^2 + 4m + 1 + 12n^2 + 12n + 3 \\ & = 4\left( m^2 + m + 3n^2 + 3n + 1 \right)\end{aligned}$

. This shows that 4 divides $a^2 + 3b^2$ .

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