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.

204문제 · 9 / 11쪽

161.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
162.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
163.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
164.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
165.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
166.Sundstrom, Mathematical Reasoning · 2.4절 · 2a번영어 원문

For each of the following, use a counterexample to show that the statement is false. Then write the negation of the statement in English, without using symbols for quantifiers.

(∀m∈Z)(m2 is even )\left( {\forall m \in \mathbb{Z}} \right)\left( {m^2 \text{ is even } } \right).

정답·풀이 보기 (영어 원문)

m=1m = 1 is a counterexample. The negation is: There exists a natural number mm such that m2m^2 is not even or there exists a natural number mm such that m2m^2 is odd.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.3절 · 4a번
# 명제와 조건증명·논술자동 분류
167.Sundstrom, Mathematical Reasoning · 2.4절 · 2b번영어 원문

For each of the following, use a counterexample to show that the statement is false. Then write the negation of the statement in English, without using symbols for quantifiers.

(∀x∈R)(x2>0)\left( {\forall x \in \mathbb{R}} \right)\left( {x^2 > 0} \right).

정답·풀이 보기 (영어 원문)

x=0x = 0 is a counterexample. The negation is: There exists a real number xx such that x2≤0x^2 \leq 0.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.3절
# 명제와 조건증명·논술자동 분류
168.Sundstrom, Mathematical Reasoning · 2.4절 · 2c번영어 원문

For each of the following, use a counterexample to show that the statement is false. Then write the negation of the statement in English, without using symbols for quantifiers.

For each real number xx, x∈R\sqrt{x} \in \mathbb{R}.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 1.1절 · 1k번
# 명제와 조건증명·논술자동 분류
169.Sundstrom, Mathematical Reasoning · 2.4절 · 2d번영어 원문

For each of the following, use a counterexample to show that the statement is false. Then write the negation of the statement in English, without using symbols for quantifiers.

(∀m∈Z)(m3∈Z)\left( \forall m \in \mathbb{Z} \right) \left( \dfrac{m}{3} \in \mathbb{Z} \right).

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.1절 · 11d번
# 명제와 조건증명·논술자동 분류
170.Sundstrom, Mathematical Reasoning · 2.4절 · 2e번영어 원문

For each of the following, use a counterexample to show that the statement is false. Then write the negation of the statement in English, without using symbols for quantifiers.

(∀a∈Z)(a2=a)\left( {\forall a \in \mathbb{Z}} \right)\left( {\sqrt {a^2 } = a} \right).

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.3절 · 4b번
# 명제와 조건증명·논술자동 분류
171.Sundstrom, Mathematical Reasoning · 2.4절 · 2f번영어 원문

For each of the following, use a counterexample to show that the statement is false. Then write the negation of the statement in English, without using 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)

x=π2x = \dfrac{\pi}{2} is a counterexample. The negation is: 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 · 2.3절
# 명제와 조건증명·논술자동 분류
172.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
173.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
174.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
175.Sundstrom, Mathematical Reasoning · 2.4절 · 3d번영어 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)(2<x<3)\left( {\exists x \in \mathbb{Q}} \right)\left( {\sqrt 2 \lt x \lt \sqrt 3 } \right).

Note: The sentence “2<x<3\sqrt 2 \lt x \lt \sqrt 3” is actually a conjunction. It means 2<x\sqrt 2 \lt x and x<3x \lt \sqrt 3.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.1절 · 11a번
# 명제와 조건계산자동 분류
176.Sundstrom, Mathematical Reasoning · 2.4절 · 3e번영어 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)( If x2 is odd, then x is odd )\left( {\forall x \in \mathbb{Z}} \right)\left( {\text{ If } x^2 \text{ is odd, then } x\text{ is odd } } \right).

정답·풀이 보기 (영어 원문)

For each integer xx, if x2x^2 is odd, then xx is odd. The negation is (∃x∈Z)(x2 is odd and x is even )\left( \exists x \in \mathbb{Z} \right) \left( x^2 \text{ is odd and \(x\) is even } \right), which is, There exists an integer xx such that x2x^2 is odd and xx is even.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절 · 3a번
# 명제와 조건계산자동 분류
177.Sundstrom, Mathematical Reasoning · 2.4절 · 3f번영어 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.

(∀n∈N)\left( {\forall n \in \mathbb{N}} \right) [If nn is a perfect square, then (2n−1)\left( {2^n - 1} \right) is not a prime number].

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.3절
# 명제와 조건계산자동 분류
178.Sundstrom, Mathematical Reasoning · 2.4절 · 3g번영어 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.

(∀n∈N)(n2−n+41 is a prime number )\left( {\forall n \in \mathbb{N}} \right)\left( {n^2 - n + 41} \text{ is a prime number } \right).

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.1절 · 11b번
# 명제와 조건계산자동 분류
179.Sundstrom, Mathematical Reasoning · 2.4절 · 3h번영어 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∈R)(cos⁡(2x)=2(cos⁡x))\left( {\exists x \in \mathbb{R}} \right) \left( {\cos ( {2x} ) = 2( {\cos x} )} \right).

정답·풀이 보기 (영어 원문)

There exists a real number xx such that cos⁡(2x)=2(cos⁡x)\cos \left( 2x \right) = 2 \left( \cos x \right). The negation is (∀x∈R)(cos⁡(2x)≠2(cos⁡x))\left( {\forall x \in \mathbb{R}} \right) \left( \cos \left( 2x \right) \ne 2 \left( \cos x \right) \right), which is, For each real number xx, cos⁡(2x)≠2(cos⁡x)\cos \left( 2x \right) \ne 2 \left( \cos x \right).

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.4절 · 1b번
# 명제와 조건계산자동 분류
180.Sundstrom, Mathematical Reasoning · 2.4절 · 4a번영어 원문

Write each of the following statements as an English sentence that does not use the symbols for quantifiers.

(∃m∈Z)(∃n∈Z)(m>n)\left( {\exists m} \in \mathbb{Z} \right)\left( {\exists n} \in \mathbb{Z} \right)\left( {m > n} \right)

정답·풀이 보기 (영어 원문)
# 명제와 조건계산자동 분류