문제 은행 · 기초 수학 · 공통수학2 · 집합과 명제

집합의 뜻과 포함 관계

집합과 원소, 부분집합, 부분집합의 개수를 다룬다

유형·난이도는 우리말 풀이가 붙은 문제는 선생님이 정했고, 나머지는 자동 분류라 다를 수 있어요. 난이도가 정해지지 않은 문제는 난이도로 거를 때 빠집니다.

117문제 · 5 / 6쪽

81.Sundstrom, Mathematical Reasoning · 5.1절 · 12d번영어 원문

Let AA, BB, and CC be subsets of some universal set UU. For each of the following, draw a general Venn diagram for the three sets and then shade the indicated region.

B∪CB \cup C

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.1절 · 3k번
82.Sundstrom, Mathematical Reasoning · 5.1절 · 12e번영어 원문

Let AA, BB, and CC be subsets of some universal set UU. For each of the following, draw a general Venn diagram for the three sets and then shade the indicated region.

A∩(B∪C)A \cap \left( {B \cup C} \right)

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.2절 · 12a번
83.Sundstrom, Mathematical Reasoning · 5.1절 · 12f번영어 원문

Let AA, BB, and CC be subsets of some universal set UU. For each of the following, draw a general Venn diagram for the three sets and then shade the indicated region.

(A∩B)−C\left( {A \cap B} \right) - C

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.2절
84.Sundstrom, Mathematical Reasoning · 5.1절 · 14번영어 원문

One of the properties of real numbers is the so-called Law of Trichotomy,

which states that if a,b∈Ra, b \in \mathbb{R}, then exactly one of the following is true:

  • a<ba \lt b; - a=ba = b; - a>ba > b.

Is the following proposition concerning sets true or false? Either provide a proof that it is true or a counterexample showing it is false. If AA and BB are subsets of some universal set, then exactly one of the following is true:

  • A⊆BA \subseteq B; - A=BA = B; - B⊆AB \subseteq A.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.3절 · 3c번
# 집합의 뜻과 포함 관계개념 확인자동 분류
85.Sundstrom, Mathematical Reasoning · 5.2절영어 원문

Let A={x∈Z∣x is a multiple of 9 }A = \left\{ x \in \mathbb{Z} \mid x \text{ is a multiple of 9 } \right\} and let B={x∈Z∣x is a multiple of 3 }B = \left\{ x \in \mathbb{Z} \mid x \text{ is a multiple of 3 } \right\}.

Is the set AA a subset of BB? Justify your conclusion.

정답·풀이 보기 (영어 원문)

The set AA is a subset of BB. To prove this, we let x∈Ax \in A. Then there exists an integer mm such that x=9mx = 9m, which can be written as

x=3(3m) ⁣x = 3 \left( 3m \right)\!

. Since 3m∈Z3m \in \mathbb{Z}, the last equation proves that xx is a multiple of 3 and so x∈Bx \in B. Therefore, A⊆BA \subseteq B.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.2절
# 집합의 뜻과 포함 관계증명·논술자동 분류
86.Sundstrom, Mathematical Reasoning · 5.2절영어 원문

Let A={x∈Z∣x is a multiple of 9 }A = \left\{ x \in \mathbb{Z} \mid x \text{ is a multiple of 9 } \right\} and let B={x∈Z∣x is a multiple of 3 }B = \left\{ x \in \mathbb{Z} \mid x \text{ is a multiple of 3 } \right\}.

Is the set AA equal to the set BB? Justify your conclusion.

정답·풀이 보기 (영어 원문)

The set AA is not equal to the set BB. We note that 3∈B3 \in B but 3∉A3 \notin A. Therefore, B⊈AB \not \subseteq A and, hence, A≠BA \ne B.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.1절 · 5a번
# 집합의 뜻과 포함 관계증명·논술자동 분류
87.Sundstrom, 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 AA and BB be subsets of some universal set. Then A−B=A∩BcA - B = A \cap B^c.

정답·풀이 보기 (영어 원문)

Proof. Let AA and BB be subsets of some universal set. We will prove that A−B=A∩BcA - B = A \cap B^c by proving that each set is a subset of the other set. We will first prove that A−B⊆A∩BcA - B \subseteq A \cap B^c. Let x∈A−Bx \in A - B. We then know that x∈Ax \in A and x∉Bx \notin B. However, x∉Bx \notin B implies that x∈Bcx \in B^c. Hence, x∈Ax \in A and x∈Bcx \in B^c, which means that x∈A∩Bcx \in A \cap B^c. This proves that A−B⊆A∩BcA - B \subseteq A \cap B^c.

To prove that A∩Bc⊆A−BA \cap B^c \subseteq A - B, we let y∈A∩Bcy \in A \cap B^c. This means that y∈Ay \in A and y∈Bcy \in B^c, and hence, y∈Ay \in A and y∉By \notin B. Therefore, y∈A−By \in A - B and this proves that A∩Bc⊆A−BA \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∩BcA - B = A \cap B^c.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.3절
# 집합의 뜻과 포함 관계증명·논술자동 분류
88.Sundstrom, Mathematical Reasoning · 5.2절 · 1a번영어 원문

Let A={x∈R∣x2<4}A = \left\{ {\left. {x \in \mathbb{R}} \right| x^2 \lt 4} \right\} and let B={x∈R∣x<2}B = \left\{ {x \in \mathbb{R}\left. \right| x \lt 2} \right\}.

Is A⊆BA \subseteq B? Justify your conclusion with a proof or a counterexample.

정답·풀이 보기 (영어 원문)

The set AA is a subset of BB. To prove this, we let x∈Ax \in A. Then −2<x<2- 2 \lt x \lt 2. Since x<2x \lt 2, we conclude that x∈Bx \in B and hence, we have proved that AA is a subset of BB.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.2절 · 1b번
# 집합의 뜻과 포함 관계증명·논술자동 분류
89.Sundstrom, Mathematical Reasoning · 5.2절 · 1b번영어 원문

Let A={x∈R∣x2<4}A = \left\{ {\left. {x \in \mathbb{R}} \right| x^2 \lt 4} \right\} and let B={x∈R∣x<2}B = \left\{ {x \in \mathbb{R}\left. \right| x \lt 2} \right\}.

Is B⊆AB \subseteq A? Justify your conclusion with a proof or a counterexample.

정답·풀이 보기 (영어 원문)

The set BB is not a subset of AA. There are many examples of a real number that is in BB but not in AA. For example, −3-3 is in AA, but −3-3 is not in BB.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.2절 · 7b번
# 집합의 뜻과 포함 관계증명·논술자동 분류
90.Sundstrom, Mathematical Reasoning · 5.2절 · 2a번영어 원문

Let AA, BB, and CC be subsets of a universal set UU.

Draw a Venn diagram with A⊆BA \subseteq B and B⊆CB \subseteq C. Does it appear that A⊆CA \subseteq C?

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.1절 · 3f번
91.Sundstrom, Mathematical Reasoning · 5.2절 · 2b번영어 원문

Let AA, BB, and CC be subsets of a universal set UU.

Prove the following proposition:

If

A \subseteq B

and

B \subseteq C

, then

A \subseteq C

.

Note: This may seem like an obvious result. However, one of the reasons for this exercise is to provide practice at properly writing a proof that one set is a subset of another set. So we should start the proof by assuming that A⊆BA \subseteq B and B⊆CB \subseteq C. Then we should choose an arbitrary element of AA.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.3절 · 5e번
# 집합의 뜻과 포함 관계증명·논술자동 분류
92.Sundstrom, Mathematical Reasoning · 5.2절 · 6a번영어 원문

To prove the following set equalities, it may be necessary to use some of the properties of positive and negative real numbers. For example, it may be necessary to use the facts that:

  • The product of two real numbers is positive if and only if the two real numbers are either both positive or both negative. - The product of two real numbers is negative if and only if one of the two numbers is positive and the other is negative.

For example, if x(x−2)<0x \left( x - 2 \right) \lt 0, then we can conclude that either (1) x<0x \lt 0 and x−2>0x - 2 > 0 or (2) x>0x > 0 and x−2<0x - 2 \lt 0. However, in the first case, we must have x<0x \lt 0 and x>2x > 2, and this is impossible. Therefore, we conclude that x>0x > 0 and x−2<0x - 2 \lt 0, which means that 0<x<20 \lt x \lt 2.

Use the choose-an-element method to prove each of the following:

{x∈R∣x2−3x−10<0}={x∈R∣−2<x<5}\left\{x \in \mathbb{R} \mid x^2 - 3x - 10 \lt 0 \right\} = \left\{ x \in \mathbb{R} \mid -2 \lt x \lt 5 \right\}

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.1절 · 4번
# 집합의 뜻과 포함 관계증명·논술자동 분류
93.Sundstrom, Mathematical Reasoning · 5.2절 · 6b번영어 원문

To prove the following set equalities, it may be necessary to use some of the properties of positive and negative real numbers. For example, it may be necessary to use the facts that:

  • The product of two real numbers is positive if and only if the two real numbers are either both positive or both negative. - The product of two real numbers is negative if and only if one of the two numbers is positive and the other is negative.

For example, if x(x−2)<0x \left( x - 2 \right) \lt 0, then we can conclude that either (1) x<0x \lt 0 and x−2>0x - 2 > 0 or (2) x>0x > 0 and x−2<0x - 2 \lt 0. However, in the first case, we must have x<0x \lt 0 and x>2x > 2, and this is impossible. Therefore, we conclude that x>0x > 0 and x−2<0x - 2 \lt 0, which means that 0<x<20 \lt x \lt 2.

Use the choose-an-element method to prove each of the following:

{x∈R∣x2−5x+6<0}={x∈R∣2<x<3}\left\{x \in \mathbb{R} \mid x^2 - 5x + 6 \lt 0 \right\} = \left\{ x \in \mathbb{R} \mid 2 \lt x \lt 3 \right\}

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.1절 · 3j번
# 집합의 뜻과 포함 관계증명·논술자동 분류
94.Sundstrom, Mathematical Reasoning · 5.2절 · 6c번영어 원문

To prove the following set equalities, it may be necessary to use some of the properties of positive and negative real numbers. For example, it may be necessary to use the facts that:

  • The product of two real numbers is positive if and only if the two real numbers are either both positive or both negative. - The product of two real numbers is negative if and only if one of the two numbers is positive and the other is negative.

For example, if x(x−2)<0x \left( x - 2 \right) \lt 0, then we can conclude that either (1) x<0x \lt 0 and x−2>0x - 2 > 0 or (2) x>0x > 0 and x−2<0x - 2 \lt 0. However, in the first case, we must have x<0x \lt 0 and x>2x > 2, and this is impossible. Therefore, we conclude that x>0x > 0 and x−2<0x - 2 \lt 0, which means that 0<x<20 \lt x \lt 2.

Use the choose-an-element method to prove each of the following:

{x∈R∣x2≥4}={x∈R∣x≤−2}∪{x∈R∣x≥2}\left\{ x \in \mathbb{R} \mid x^2 \geq 4 \right\} = \left\{ x \in \mathbb{R} \mid x \leq -2 \right\} \cup \left\{ x \in \mathbb{R} \mid x \geq 2 \right\}

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.1절 · 3d번
# 집합의 뜻과 포함 관계증명·논술자동 분류
95.Sundstrom, Mathematical Reasoning · 5.2절 · 7a번영어 원문

Let AA and BB be subsets of some universal set UU. Prove each of the following:

A∩B⊆AA \cap B \subseteq A

정답·풀이 보기 (영어 원문)

Let x∈A∩Bx \in A \cap B. Then, x∈Ax \in A and x∈Bx \in B. This proves that if x∈A∩Bx \in A \cap B, then x∈Ax \in A, and hence, A∩B⊆AA \cap B \subseteq A.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.1절 · 4번
# 집합의 뜻과 포함 관계증명·논술자동 분류
96.Sundstrom, Mathematical Reasoning · 5.2절 · 7b번영어 원문

Let AA and BB be subsets of some universal set UU. Prove each of the following:

A⊆A∪BA \subseteq A \cup B

정답·풀이 보기 (영어 원문)

Let x∈Ax \in A. Then, the statement “x∈Ax \in A or x∈Bx \in B” is true. Hence, A⊆A∪BA \subseteq A \cup B.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.2절 · 7a번
# 집합의 뜻과 포함 관계증명·논술자동 분류
97.Sundstrom, Mathematical Reasoning · 5.2절 · 7c번영어 원문

Let AA and BB be subsets of some universal set UU. Prove each of the following:

A∩A=AA \cap A = A

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.1절 · 3n번
# 집합의 뜻과 포함 관계증명·논술자동 분류
98.Sundstrom, Mathematical Reasoning · 5.2절 · 7d번영어 원문

Let AA and BB be subsets of some universal set UU. Prove each of the following:

A∪A=AA \cup A = A

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.1절 · 3c번
# 집합의 뜻과 포함 관계증명·논술자동 분류
99.Sundstrom, Mathematical Reasoning · 5.2절 · 7f번영어 원문

Let AA and BB be subsets of some universal set UU. Prove each of the following:

A∪∅=AA \cup \emptyset = A

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.2절 · 1a번
# 집합의 뜻과 포함 관계증명·논술자동 분류
100.Sundstrom, Mathematical Reasoning · 5.2절 · 9번영어 원문

Is the following proposition true or false? Justify your conclusion with a proof or a counterexample.

For all sets

A

and

B

that are subsets of some universal set

U

, the sets

A \cap B

and

A - B

are disjoint.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.2절 · 12a번
# 집합의 뜻과 포함 관계개념 확인자동 분류