문제 은행 · 기초 수학 · 공통수학2 · 집합과 명제

명제와 조건

명제의 참·거짓과 조건의 진리집합, 역·이·대우를 다룬다

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

144문제 · 6 / 8쪽

101.Sundstrom, Mathematical Reasoning · 2.3절 · 4e번영어 원문

Use the roster method to specify the truth set for each of the following open sentences. The universal set for each open sentence is the set of integers Z\mathbb{Z}.

nn is an even integer that is greater than 10.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절 · 2b번
# 명제와 조건계산자동 분류
102.Sundstrom, Mathematical Reasoning · 2.4절영어 원문

For each of the following statements

  • (1) Write the statement in the form of an English sentence that does not use the symbols for quantifiers. - (2) Write the negation of the statement in a symbolic form that does not use the negation symbol. - (3) Write the negation of the statement in the form of an English sentence that does not use the symbols for quantifiers.

(∀a∈R)(a+0=a)\left( \forall a \in \mathbb{R}\right) \left( a + 0 = a \right).

정답·풀이 보기 (영어 원문)
  • For each real number aa, a+0=aa + 0 = a. - (∃a∈R)(a+0≠a)(\exists a \in \mathbb{R})( a + 0 \ne a). - There exists a real number aa such that a+0≠aa + 0 \ne a.
한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절 · 2f번
# 명제와 조건계산자동 분류
103.Sundstrom, Mathematical Reasoning · 2.4절영어 원문

For each of the following statements

  • (1) Write the statement in the form of an English sentence that does not use the symbols for quantifiers. - (2) Write the negation of the statement in a symbolic form that does not use the negation symbol. - (3) Write the negation of the statement in the form of an English sentence that does not use the symbols for quantifiers.

(∀x∈R)[sin⁡(2x)=2(sin⁡x)(cos⁡x)]\left( \forall x \in \mathbb{R} \right) \left[ \sin ( {2x} ) = 2 ( {\sin x} ) ( {\cos x} ) \right].

정답·풀이 보기 (영어 원문)
  • For each real number xx, sin⁡(2x)=2(sin⁡x)(cos⁡x)\sin ( {2x} ) = 2( {\sin x} )( {\cos x} ). - (∃x∈R)(sin⁡(2x)≠2(sin⁡x)(cos⁡x))(\exists x \in \mathbb{R})( \sin ( {2x} ) \ne 2( {\sin x} )( {\cos x} )). - There exists a real number xx such that sin⁡(2x)≠2(sin⁡x)(cos⁡x)\sin \left( {2x} \right) \ne 2\left( {\sin x} \right)\left( {\cos x} \right).
한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절 · 4e번
# 명제와 조건계산자동 분류
104.Sundstrom, Mathematical Reasoning · 2.4절영어 원문

For each of the following statements

  • (1) Write the statement in the form of an English sentence that does not use the symbols for quantifiers. - (2) Write the negation of the statement in a symbolic form that does not use the negation symbol. - (3) Write the negation of the statement in the form of an English sentence that does not use the symbols for quantifiers.

(∀x∈R)(tan⁡2x+1=sec⁡2x)\left( \forall x \in \mathbb{R} \right) \left( \tan ^2 x + 1 = \sec ^2 x \right).

정답·풀이 보기 (영어 원문)
  • For each real number xx, tan⁡2x+1=sec⁡2x\tan ^2 x + 1 = \sec ^2 x. - (∃x∈R)(tan⁡2x+1≠sec⁡2x)\left( \exists x \in \mathbb{R}\right) \left( \tan ^2 x + 1 \ne \sec ^2 x \right). - There exists a real number xx such that tan⁡2x+1≠sec⁡2x\tan ^2 x + 1 \ne \sec ^2 x.
한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절
# 명제와 조건계산자동 분류
105.Sundstrom, Mathematical Reasoning · 2.4절영어 원문

For each of the following statements

  • (1) Write the statement in the form of an English sentence that does not use the symbols for quantifiers. - (2) Write the negation of the statement in a symbolic form that does not use the negation symbol. - (3) Write the negation of the statement in the form of an English sentence that does not use the symbols for quantifiers.

(∃x∈Q)(x2−3x−7=0)\left( \exists x \in \mathbb{Q} \right) \left( x^2 - 3x - 7 = 0 \right).

정답·풀이 보기 (영어 원문)
  • There exists a rational number xx such that x2−3x−7=0x^2 - 3x - 7 = 0. - (∀x∈Q)(x2−3x−7≠0)(\forall x \in \mathbb{Q}) \left( x^2 - 3x - 7 \ne 0 \right). - For each rational number xx, x2−3x−7≠0x^2 - 3x - 7 \ne 0.
한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절
# 명제와 조건계산자동 분류
106.Sundstrom, Mathematical Reasoning · 2.4절영어 원문

For each of the following statements

  • (1) Write the statement in the form of an English sentence that does not use the symbols for quantifiers. - (2) Write the negation of the statement in a symbolic form that does not use the negation symbol. - (3) Write the negation of the statement in the form of an English sentence that does not use the symbols for quantifiers.

(∃x∈R)(x2+1=0)\left( \exists x \in \mathbb{R} \right) \left( x^2 + 1 = 0 \right).

정답·풀이 보기 (영어 원문)
  • There exists a real number xx such that x2+1=0x^2 + 1 = 0. - (∀x∈R)( x2+1≠0)(\forall x \in \mathbb{R}) \left( \text{ } x^2 + 1 \ne 0 \right). - For each real number xx, x2+1≠0x^2 + 1 \ne 0.
한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절 · 2b번
# 명제와 조건계산자동 분류
107.Sundstrom, Mathematical Reasoning · 2.4절영어 원문

Use counterexamples to explain why each of the following statements is false.

For each integer nn, (n2+n+1)\left( n^2 + n + 1 \right) is a prime number.

정답·풀이 보기 (영어 원문)

A counterexample is n=4n = 4 since 42+4+1=214^2 + 4 + 1 = 21, and 21 is not prime.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.3절
# 명제와 조건계산자동 분류
108.Sundstrom, Mathematical Reasoning · 2.4절영어 원문

Use counterexamples to explain why each of the following statements is false.

For each real number xx, if xx is positive, then 2x2>x2x^2 > x.

정답·풀이 보기 (영어 원문)

A counterexample is x=14x = \dfrac{1}{4} since 14\dfrac{1}{4} is positive and 2(14)2=182 \left( \dfrac{1}{4} \right)^2 = \dfrac{1}{8} and 18≤14\dfrac{1}{8} \leq \dfrac{1}{4}.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절
# 명제와 조건계산자동 분류
109.Sundstrom, Mathematical Reasoning · 2.4절영어 원문

An integer nn is a multiple of 3 provided that there exists an integer kk such that n=3kn = 3k.

Write this definition in symbolic form using quantifiers by completing the following:

An integer

n

is a multiple of 3 provided that

.

정답·풀이 보기 (영어 원문)

An integer nn is a multiple of 3 provided that (∃k∈Z)(n=3k)\left( \exists k \in \mathbb{Z} \right) \left( n = 3k \right).

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.1절 · 7번
# 명제와 조건계산자동 분류
110.Sundstrom, Mathematical Reasoning · 2.4절영어 원문

An integer nn is a multiple of 3 provided that there exists an integer kk such that n=3kn = 3k.

Give several examples of integers (including negative integers) that are multiples of 3.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절 · 4e번
# 명제와 조건계산자동 분류
111.Sundstrom, Mathematical Reasoning · 2.4절영어 원문

An integer nn is a multiple of 3 provided that there exists an integer kk such that n=3kn = 3k.

Give several examples of integers (including negative integers) that are not multiples of 3.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 1.1절
# 명제와 조건계산자동 분류
112.Sundstrom, Mathematical Reasoning · 2.4절영어 원문

An integer nn is a multiple of 3 provided that there exists an integer kk such that n=3kn = 3k.

Use the symbolic form of the definition of a multiple of 3 to complete the following sentence: “An integer nn is not a multiple of 3 provided that ….”

정답·풀이 보기 (영어 원문)

An integer nn is not a multiple of 3 provided that (∀k∈Z)(n≠3k)\left( \forall k \in \mathbb{Z} \right) \left( n \ne 3k \right).

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절
# 명제와 조건계산자동 분류
113.Sundstrom, Mathematical Reasoning · 2.4절영어 원문

An integer nn is a multiple of 3 provided that there exists an integer kk such that n=3kn = 3k.

Without using the symbols for quantifiers, complete the following sentence: “An integer nn is not a multiple of 3 provided that ….”

정답·풀이 보기 (영어 원문)

An integer nn is not a multiple of 3 provided that for every integer kk, n≠3kn \ne 3k.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절 · 6e번
# 명제와 조건계산자동 분류
114.Sundstrom, Mathematical Reasoning · 2.4절영어 원문

Write the negation of the statement

(∀x∈Z)(∀y∈Z)(x+y=0)\left( {\forall x \in \mathbb{Z}} \right)\left( {\forall y \in \mathbb{Z}} \right)\left( {x + y = 0} \right)

in symbolic form and as a sentence written in English.

정답·풀이 보기 (영어 원문)

(∃x∈Z)(∃y∈Z)(x+y≠0)\left( {\exists x \in \mathbb{Z}} \right)\left( {\exists y \in \mathbb{Z}} \right)\left( {x + y \ne 0} \right).

There exist integers xx and yy such that x+y≠0x + y \ne 0.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절 · 3e번
# 명제와 조건계산자동 분류
115.Sundstrom, Mathematical Reasoning · 2.4절 · 1a번영어 원문

For each of the following, write the statement as an English sentence and then explain why the statement is false.

(∃x∈Q)(x2−3x−7=0)\left(\exists x \in \mathbb{Q} \right) \left( x^2 - 3x - 7 = 0 \right).

정답·풀이 보기 (영어 원문)

There exists a rational number xx such that x2−3x−7=0x^2 - 3x - 7 = 0. This statement is false since the solutions of the equation are x=3±372x = \dfrac{3 \pm \sqrt{37}}{2}, which are irrational numbers.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절
# 명제와 조건계산자동 분류
116.Sundstrom, Mathematical Reasoning · 2.4절 · 1b번영어 원문

For each of the following, write the statement as an English sentence and then explain why the statement is false.

(∃x∈R)(x2+1=0)\left( \exists x \in \mathbb{R} \right) \left( x^2 + 1 = 0 \right).

정답·풀이 보기 (영어 원문)

There exists a real number xx such that x2+1=0x^2 + 1 = 0. This statement is false since the only solutions of the equation are ii and −i-i, which are not real numbers.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절 · 6c번
# 명제와 조건계산자동 분류
117.Sundstrom, Mathematical Reasoning · 2.4절 · 1c번영어 원문

For each of the following, write the statement as an English sentence and then explain why the statement is false.

(∃m∈N)(m2<1)\left( \exists m \in \mathbb{N} \right) \left( m^2 \lt 1 \right).

정답·풀이 보기 (영어 원문)

There exists a natural number mm such that m2<1m^2 \lt 1. This statement is false because if mm is a natural number, then m≥1m \geq 1 and hence, m2≥1m^2 \geq 1.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절 · 6b번
# 명제와 조건계산자동 분류
118.Sundstrom, Mathematical Reasoning · 2.4절 · 3a번영어 원문

For each of the following statements

  • Write the statement as an English sentence that does not use the symbols for quantifiers. - Write the negation of the statement in symbolic form in which the negation symbol is not used. - Write a useful negation of the statement in an English sentence that does not use the symbols for quantifiers.

(∃x∈Q)(x>2)\left( {\exists x \in \mathbb{Q}} \right)\left( {x > \sqrt 2 } \right).

정답·풀이 보기 (영어 원문)

There exists a rational number xx such that x>2x > \sqrt{2}. The negation is (∀x∈Q)(x≤2)\left( {\forall x \in \mathbb{Q}} \right) \left( {x \leq \sqrt 2 } \right), which is, For each rational number xx, x≤2x \le \sqrt{2}.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절 · 3c번
# 명제와 조건계산자동 분류
119.Sundstrom, Mathematical Reasoning · 2.4절 · 3b번영어 원문

For each of the following statements

  • Write the statement as an English sentence that does not use the symbols for quantifiers. - Write the negation of the statement in symbolic form in which the negation symbol is not used. - Write a useful negation of the statement in an English sentence that does not use the symbols for quantifiers.

(∀x∈Q)(x2−2≠0)\left( {\forall x \in \mathbb{Q}} \right)\left( {x^2 - 2 \ne 0} \right).

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.1절 · 11c번
# 명제와 조건계산자동 분류
120.Sundstrom, Mathematical Reasoning · 2.4절 · 3c번영어 원문

For each of the following statements

  • Write the statement as an English sentence that does not use the symbols for quantifiers. - Write the negation of the statement in symbolic form in which the negation symbol is not used. - Write a useful negation of the statement in an English sentence that does not use the symbols for quantifiers.

(∀x∈Z)(x is even or x is odd )\left( {\forall x \in \mathbb{Z}} \right)\left( {x\text{ is even or } x\text{ is odd } } \right).

정답·풀이 보기 (영어 원문)

For each integer xx, xx is even or xx is odd. The negation is (∃x∈Z)(x is odd and x is even )\left( \exists x \in \mathbb{Z} \right) \left( x \text{ is odd and \(x\) is even } \right), which is, There exists an integer xx such that xx is odd and xx is even.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절 · 3h번
# 명제와 조건계산자동 분류