명제의 증명과 절대부등식 연습문제 — Sundstrom, Mathematical Reasoning · 3.1절 · 2d번

Sundstrom, Mathematical Reasoning · 3.1절 · 2d번영어 원문

For each of the following, use a counterexample to prove the statement is false.

For each integer $a$ , if 4 divides $\left( a^2 - 1 \right)$ , then 4 divides $(a - 1)$ .

정답·풀이 보기 (영어 원문)

The integer $a = 3$ is a counterexample since $a^2 - 1 = 8$ and $a - 1 = 2$ . Since 4 divides 8 and 4 does not divide 2, this is an example where the hypothesis of the conditional statement is true and the conclusion is false.

한 번 더 — 비슷한 문제Sundstrom, Mathematical Reasoning · 3.1절영어 원문

Construct a know-show table for the following proposition: For all integers $a$ , $b$ , and $c$ with $a \ne 0$ , if $a$ divides $b$ and $a$ divides $c$ , then $a$ divides $(b + c)$ .

다 풀었으면 정답 확인

Conjecture: For all integers $a$ , $b$ , and $c$ with $a \ne 0$ , if $a$ divides $b$ and $a$ divides $c$ , then $a$ divides $b + c$ .

A Know-show table for a proof of the conjecture in Part (3).

| Step | Know | Reason | | --- | --- | --- | | $P$ | $a \mid b$ and $a \mid c$ | Hypothesis | | $P1$ | $\left( {\exists s \in \mathbb{Z}} \right) \left( {b = a \cdot s} \right)$ $\left( {\exists t \in \mathbb{Z}} \right)\left( {c = a \cdot t} \right)$ | Definition of “divides” | | $P2$ | $b + c = as + at$ | Substituting for $b$ and $c$ | | $P3$ | $b + c = a\left( {s + t} \right)$ | Distributive property | | $Q1$ | $s + t$ is an integer | $\mathbb{Z}$ is closed under addition | | $Q$ | $\left. {a } \right\| \left( {b + c} \right)$ | Definition of “divides” | | Step | Show | Reason |

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