문제 은행 · 기초 수학 · 공통수학2 · 집합과 명제

집합과 명제 전체 문제

이 단원의 모든 연습문제입니다. 소단원으로 좁혀 보려면 아래 꼬리표를 누르세요.

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283문제 · 10 / 15쪽

181.Sundstrom, Mathematical Reasoning · 2.4절 · 12b번영어 원문

The following exercises contain definitions or results from more advanced mathematics courses. Even though we may not understand all of the terms involved, it is still possible to recognize the structure of the given statements and write a meaningful negation of that statement.

In abstract algebra, a ring

consists of a nonempty set RR and two operations called addition and multiplication. A nonzero element aa in a ring RR is called a zero divisor

provided that there exists a nonzero element bb in RR such that ab=0ab = 0 or ba=0ba = 0. Carefully explain what it means to say that a nonzero element aa in a ring RR is not a zero divisor.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 2.1절 · 11a번
# 명제와 조건계산자동 분류
182.Sundstrom, Mathematical Reasoning · 3.1절영어 원문

Give at least four different examples of integers aa, bb, and cc with a≠0a \ne 0 such that aa divides bb and aa divides cc.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.1절 · 4a번
183.Sundstrom, Mathematical Reasoning · 3.1절 · 4a번영어 원문

Complete the following.

If xx and yy are integers and xy=1xy=1, explain why x=1x=1 or x=−1x=-1.

정답·풀이 보기 (영어 원문)

If xy=1xy = 1, then xx and yy are both divisors of 11, and the only divisors of 11 are −1-1 and 11.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.1절 · 2d번
184.Sundstrom, Mathematical Reasoning · 3.1절 · 15a번영어 원문

Let rr be a positive real number. The equation for a circle of radius rr whose center is the origin is x2+y2=r2x^2 + y^2 = r^2.

Use implicit differentiation to determine dydx\dfrac{dy}{dx}.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.1절 · 2a번
185.Sundstrom, Mathematical Reasoning · 3.1절 · 15b번영어 원문

Let rr be a positive real number. The equation for a circle of radius rr whose center is the origin is x2+y2=r2x^2 + y^2 = r^2.

Let (a,b)\left(a, b \right) be a point on the circle with a≠0a \ne 0 and b≠0b \ne 0. Determine the slope of the line tangent to the circle at the point (a,b)\left(a, b \right).

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.1절
186.Sundstrom, Mathematical Reasoning · 3.1절 · 19a번영어 원문

This type of exercise will appear frequently in the book. In each case, there is a proposed proof of a proposition. However, the proposition may be true or may be false.

  • If a proposition is false, the proposed proof is, of course, incorrect. In this situation, you are to find the error in the proof and then provide a counterexample showing that the proposition is false. - If a proposition is true, the proposed proof may still be incorrect. In this case, you are to determine why the proof is incorrect and then write a correct proof using the writing guidelines that have been presented in this book. - If a proposition is true and the proof is correct, you are to decide if the proof is well written or not. If it is well written, then you simply must indicate that this is an excellent proof and needs no revision. On the other hand, if the proof is not well written, then you must then revise the proof by writing it according to the guidelines presented in this text.
  • Proposition If mm is an even integer, then (5m+4)\left(5m + 4\right) is an even integer. - Proof We see that 5m+4=10n+4=2(5n+2)5m + 4 = 10n + 4 = 2 \left(5n + 2 \right). Therefore, (5m+4)\left(5m + 4 \right) is an even integer.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.2절 · 3a번
187.Sundstrom, Mathematical Reasoning · 3.1절 · 19b번영어 원문

This type of exercise will appear frequently in the book. In each case, there is a proposed proof of a proposition. However, the proposition may be true or may be false.

  • If a proposition is false, the proposed proof is, of course, incorrect. In this situation, you are to find the error in the proof and then provide a counterexample showing that the proposition is false. - If a proposition is true, the proposed proof may still be incorrect. In this case, you are to determine why the proof is incorrect and then write a correct proof using the writing guidelines that have been presented in this book. - If a proposition is true and the proof is correct, you are to decide if the proof is well written or not. If it is well written, then you simply must indicate that this is an excellent proof and needs no revision. On the other hand, if the proof is not well written, then you must then revise the proof by writing it according to the guidelines presented in this text.
  • Proposition For all real numbers xx and yy, if x≠yx \ne y, x>0x > 0, and y>0y >0, then xy+yx>2\dfrac{x}{y} + \dfrac{y}{x} > 2. - Proof Since xx and yy are positive real numbers, xyxy is positive and we can multiply both sides of the inequality by xyxy to obtain (xy+yx)⋅xy>2⋅xyx2+y2>2xy\begin{aligned}\left( \frac{x}{y} + \frac{y}{x} \right) \cdot xy & > 2 \cdot xy \\ x^2 + y^2 & > 2xy\end{aligned} . By combining all terms on the left side of the inequality, we see that x2−2xy+y2>0x^2 - 2xy + y^2 > 0 and then by factoring the left side, we obtain (x−y)2>0\left( x - y \right)^2 > 0. Since x≠yx \ne y, (x−y)≠0\left(x - y \right) \ne 0 and so (x−y)2>0\left( x - y \right)^2 > 0. This proves that if x≠yx \ne y, x>0x > 0, and y>0y >0, then xy+yx>2\dfrac{x}{y} + \dfrac{y}{x} > 2.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.3절 · 6b번
188.Sundstrom, Mathematical Reasoning · 3.1절 · 19c번영어 원문

This type of exercise will appear frequently in the book. In each case, there is a proposed proof of a proposition. However, the proposition may be true or may be false.

  • If a proposition is false, the proposed proof is, of course, incorrect. In this situation, you are to find the error in the proof and then provide a counterexample showing that the proposition is false. - If a proposition is true, the proposed proof may still be incorrect. In this case, you are to determine why the proof is incorrect and then write a correct proof using the writing guidelines that have been presented in this book. - If a proposition is true and the proof is correct, you are to decide if the proof is well written or not. If it is well written, then you simply must indicate that this is an excellent proof and needs no revision. On the other hand, if the proof is not well written, then you must then revise the proof by writing it according to the guidelines presented in this text.
  • Proposition For all integers aa, bb, and cc, if a∣(bc)a \mid \left( bc \right), then a∣ba \mid b or a∣ca \mid c. - Proof We assume that aa, bb, and cc are integers and that aa divides bcbc. So, there exists an integer kk such that bc=kabc = ka. We now factor kk as k=mnk = mn, where mm and nn are integers. We then see that bc=mnabc = mna . This means that b=mab = ma or c=nac = na and hence, a∣ba \mid b or a∣ca \mid c.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.1절 · 3g번
189.Sundstrom, Mathematical Reasoning · 3.1절 · 19d번영어 원문

This type of exercise will appear frequently in the book. In each case, there is a proposed proof of a proposition. However, the proposition may be true or may be false.

  • If a proposition is false, the proposed proof is, of course, incorrect. In this situation, you are to find the error in the proof and then provide a counterexample showing that the proposition is false. - If a proposition is true, the proposed proof may still be incorrect. In this case, you are to determine why the proof is incorrect and then write a correct proof using the writing guidelines that have been presented in this book. - If a proposition is true and the proof is correct, you are to decide if the proof is well written or not. If it is well written, then you simply must indicate that this is an excellent proof and needs no revision. On the other hand, if the proof is not well written, then you must then revise the proof by writing it according to the guidelines presented in this text.
  • Proposition For all positive integers aa, bb, and cc, (ab)c=a(bc)\left( a^b \right)^c = a^{\left( b^c \right)}. - Proof This proposition is false as is shown by the following counterexample: If we let a=2a = 2, b=3b = 3, and c=2c = 2, then (ab)c=a(bc)(23)2=2(32)82=2964≠512\begin{aligned}\left( a^b \right)^c & = a^{\left( b^c \right)} \\ \left( 2^3 \right)^2 & = 2^{\left( 3^2 \right)} \\ 8^2 & = 2^9 \\ 64 & \ne 512\end{aligned} .

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.3절 · 11a번
190.Sundstrom, Mathematical Reasoning · 3.2절영어 원문

In English, write the contrapositive of, “For all real numbers aa and bb, if a≠0a \ne 0 and b≠0b \ne 0, then ab≠0ab \ne 0.”

정답·풀이 보기 (영어 원문)

For all real numbers aa and bb, if ab=0ab = 0, then a=0 or b=0a = 0\text{ or } b = 0.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.2절 · 8c번
191.Sundstrom, Mathematical Reasoning · 3.2절 · 3a번영어 원문

Complete the following.

Write the contrapositive of the following statement: For all positive real numbers aa and bb, if ab≠a+b2\sqrt {ab} \ne \dfrac{{a + b}}{2}, then a≠ba \ne b.

정답·풀이 보기 (영어 원문)

The contrapositive is: For all positive real numbers aa and bb, if a=ba = b, then ab=a+b2\sqrt{ab} = \dfrac{a + b}{2}.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.2절
192.Sundstrom, Mathematical Reasoning · 3.3절 · 3a번영어 원문

Consider the following statement:

For each positive real number

r

, if

r^2 = 18

, then

r

is irrational.

If you were setting up a proof by contradiction for this statement, what would you assume? Carefully write down all conditions that you would assume.

정답·풀이 보기 (영어 원문)

We would assume that there exists a positive real number rr such that r2=18r^2 = 18 and rr is a rational number.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.3절 · 5a번
193.Sundstrom, Mathematical Reasoning · 3.3절 · 7a번영어 원문

Complete the following.

Give an example that shows that the sum of two irrational numbers can be a rational number.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.2절
194.Sundstrom, Mathematical Reasoning · 3.3절 · 7b번영어 원문

Complete the following.

Now explain why the following proof that (2+5)\left( \sqrt{2} + \sqrt{5} \right) is an irrational number is not a valid proof: Since 2\sqrt{2} and 5\sqrt{5} are both irrational numbers, their sum is an irrational number. Therefore, (2+5)\left( \sqrt{2} + \sqrt{5} \right) is an irrational number.

Note: You may even assume that we have proven that 5\sqrt{5} is an irrational number. (We have not proven this.)

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.1절 · 1a번
195.Sundstrom, Mathematical Reasoning · 3.3절 · 15b번영어 원문

Three natural numbers aa, bb, and cc with a<b<ca \lt b \lt c are called a Pythagorean triple

provided that a2+b2=c2a^2 + b^2 = c^2. For example, the numbers 3, 4, and 5 form a Pythagorean triple, and the numbers 5, 12, and 13 form a Pythagorean triple.

Determine two other Pythagorean triples. That is, find integers a,ba, b, and cc such that a2+b2=c2a^2 + b^2 = c^2.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.1절 · 3b번
196.Sundstrom, Mathematical Reasoning · 3.3절 · 16a번영어 원문

Consider the following proposition: There are no integers aa and bb such that b2=4a+2b^2 = 4a + 2.

Rewrite this statement in an equivalent form using a universal quantifier by completing the following:

For all integers

a

and

b

,

.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 3.3절 · 6b번
197.Sundstrom, Mathematical Reasoning · 5.1절영어 원문

Let the universal set be U={1,2,3,4,5,6}U = \left\{ {1,2,3,4,5,6} \right\}, and let

A={1,2,4} ⁣,B={1,2,3,5} ⁣,C={x∈U∣x2≤2} ⁣A = \left\{ {1,2,4} \right\}\!, B = \left\{ {1,2,3,5} \right\}\!, C = \left\{ {\left. {x \in U} \right|x^2 \leq 2} \right\}\!

.

In each of the following, fill in the blank with one or more of the symbols ⊂,⊆,=,≠,∈, or ∉\subset, \\ \subseteq, =, \ne, \in ,\text{ or } \notin so that the resulting statement is true. For each blank, include all symbols that result in a true statement. If none of these symbols makes a true statement, write nothing in the blank.

AA_____BB

정답·풀이 보기 (영어 원문)

⊄,⊈≠\not \subset, \not \subseteq \ne

한 번 더 — 비슷한 문제 풀기Sundstrom, Mathematical Reasoning · 5.1절
198.Sundstrom, Mathematical Reasoning · 5.1절영어 원문

Let the universal set be U={1,2,3,4,5,6}U = \left\{ {1,2,3,4,5,6} \right\}, and let

A={1,2,4} ⁣,B={1,2,3,5} ⁣,C={x∈U∣x2≤2} ⁣A = \left\{ {1,2,4} \right\}\!, B = \left\{ {1,2,3,5} \right\}\!, C = \left\{ {\left. {x \in U} \right|x^2 \leq 2} \right\}\!

.

In each of the following, fill in the blank with one or more of the symbols ⊂,⊆,=,≠,∈, or ∉\subset, \\ \subseteq, =, \ne, \in ,\text{ or } \notin so that the resulting statement is true. For each blank, include all symbols that result in a true statement. If none of these symbols makes a true statement, write nothing in the blank.

5_____BB

정답·풀이 보기 (영어 원문)
199.Sundstrom, Mathematical Reasoning · 5.1절영어 원문

Let the universal set be U={1,2,3,4,5,6}U = \left\{ {1,2,3,4,5,6} \right\}, and let

A={1,2,4} ⁣,B={1,2,3,5} ⁣,C={x∈U∣x2≤2} ⁣A = \left\{ {1,2,4} \right\}\!, B = \left\{ {1,2,3,5} \right\}\!, C = \left\{ {\left. {x \in U} \right|x^2 \leq 2} \right\}\!

.

In each of the following, fill in the blank with one or more of the symbols ⊂,⊆,=,≠,∈, or ∉\subset, \\ \subseteq, =, \ne, \in ,\text{ or } \notin so that the resulting statement is true. For each blank, include all symbols that result in a true statement. If none of these symbols makes a true statement, write nothing in the blank.

AA_____CC

정답·풀이 보기 (영어 원문)
200.Sundstrom, Mathematical Reasoning · 5.1절영어 원문

Let the universal set be U={1,2,3,4,5,6}U = \left\{ {1,2,3,4,5,6} \right\}, and let

A={1,2,4} ⁣,B={1,2,3,5} ⁣,C={x∈U∣x2≤2} ⁣A = \left\{ {1,2,4} \right\}\!, B = \left\{ {1,2,3,5} \right\}\!, C = \left\{ {\left. {x \in U} \right|x^2 \leq 2} \right\}\!

.

In each of the following, fill in the blank with one or more of the symbols ⊂,⊆,=,≠,∈, or ∉\subset, \\ \subseteq, =, \ne, \in ,\text{ or } \notin so that the resulting statement is true. For each blank, include all symbols that result in a true statement. If none of these symbols makes a true statement, write nothing in the blank.

{1,2}\left\{ {1, 2} \right\}_____AA

정답·풀이 보기 (영어 원문)