집합의 뜻과 포함 관계 연습문제 — Sundstrom, Mathematical Reasoning · 2.3절 · 1d번

Sundstrom, Mathematical Reasoning · 2.3절 · 1d번영어 원문

Use the roster method to specify the elements in each of the following sets and then write a sentence in English describing the set.

$\left\{ {x \in \mathbb{N}} \mid x^2 \lt 25 \right\}$

정답·풀이 보기 (영어 원문)

The set of all natural numbers whose square is less than 25, which is $\left\{ {1,2,3,4} \right\}$

한 번 더 — 비슷한 문제Sundstrom, Mathematical Reasoning · 5.2절영어 원문

Prove the following proposition. To do so, prove each set is a subset of the other set by using the choose-an-element method.

Let $A$ and $B$ be subsets of some universal set. Then $A - B = A \cap B^c$ .

다 풀었으면 정답 확인

Proof. Let $A$ and $B$ be subsets of some universal set. We will prove that $A - B = A \cap B^c$ by proving that each set is a subset of the other set. We will first prove that $A - B \subseteq A \cap B^c$ . Let $x \in A - B$ . We then know that $x \in A$ and $x \notin B$ . However, $x \notin B$ implies that $x \in B^c$ . Hence, $x \in A$ and $x \in B^c$ , which means that $x \in A \cap B^c$ . This proves that $A - B \subseteq A \cap B^c$ .

To prove that $A \cap B^c \subseteq A - B$ , we let $y \in A \cap B^c$ . This means that $y \in A$ and $y \in B^c$ , and hence, $y \in A$ and $y \notin B$ . Therefore, $y \in A - B$ and this proves that $A \cap B^c \subseteq A - B$ . Since we have proved that each set is a subset of the other set, we have proved that $A - B = A \cap B^c$ .

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