MY-ASSIGNMENTEXPERT™可以为您提供ccbcmd.edu MATH243 Discrete Mathematics离散数学课程的代写代考和辅导服务！
MATH243课程简介
Introduces the theoretical tools for describing algorithmic operations; covers set theory, the number system, the nature of proofs, recursion, formal logic, proof by induction, combinatorics, recurrence relations, graph theory, search problems, finite state automata, grammars, and languages. Course offered every fall, spring and may be offered during additional sessions. Same As: CMSC 243. Credit can be earned for one course only.
PREREQUISITE(S) : MATH 165 with a grade of “C” or better
CREDITS: 4
COST: $122/billable hour + fees (in county)
No class sections are currently available for this course.
Prerequisites
Course Description, Objectives, and Expected Learning Outcomes
This course is focused on how to think and use abstract structures in the context of useful, discrete mathematics relevant to computer science. Topics include: Set theory, logic, discrete structures, induction and recursion, graphs and networks, and techniques of proof.
Course Objectives
- To apply mathematical tools to obtain quantitative information relevant to decision making.
- To promote problem-solving and critical thinking skills.
- To prepare st udents for subsequent work in Computer Science/Computer Engineering.
- To develop logic and proving skills.
- To develop effective communication skills.
- To promote the ability to work as part of a team.
- To help strengthen students’ general academic skills.
MATH243 Discrete Mathematics HELP（EXAM HELP， ONLINE TUTOR）
Let $n \in \mathbb{N}$. Show that the following statements are true
(1) $n^5 \bmod 5=n \bmod 5$.
(2) $n^7 \bmod 7=n \bmod 7$.
We can use Fermat’s Little Theorem to prove both statements. Fermat’s Little Theorem states that if $p$ is a prime number and $a$ is an integer not divisible by $p$, then $a^{p-1} \equiv 1 \pmod{p}$.
For (1), we can use $p=5$ and $a=n$. Since $5$ is prime and $n$ is not divisible by $5$ (i.e., $n \bmod 5 \neq 0$), we have $n^4 \equiv 1 \pmod{5}$ by Fermat’s Little Theorem. Multiplying both sides by $n$ gives $n^5 \equiv n \pmod{5}$, which is equivalent to $n^5 \bmod 5 = n \bmod 5$.
For (2), we can use $p=7$ and $a=n$. Since $7$ is prime and $n$ is not divisible by $7$ (i.e., $n \bmod 7 \neq 0$), we have $n^6 \equiv 1 \pmod{7}$ by Fermat’s Little Theorem. Multiplying both sides by $n$ gives $n^7 \equiv n \pmod{7}$, which is equivalent to $n^7 \bmod 7 = n \bmod 7$.
Therefore, both statements are true.
Prove the following statements. Let $n \in \mathbb{N}, n>1$, and $a \in \mathbb{N}$
(1) If $a \bmod n=b \bmod n$, then $(-a) \bmod n=(-b) \bmod b$.
(2) If $p$ is odd prime, then $(-a)^p=-a^p$.
(1) We have $a \equiv b \pmod{n}$, which means there exists an integer $k$ such that $a=b+kn$. Then $-a=-(b+kn)=(-b)+(-k)n$, which means $-a \equiv -b \pmod{n}$. Therefore, $(-a) \bmod n=(-b) \bmod n$.
(2) We can use the binomial theorem to expand $(-a)^p$: $$(-a)^p=(-1)^p a^p=(-1)\cdot(-1)\cdots(-1)\cdot a^p=(-1)^{p-1}a^p.$$ Since $p$ is odd, $p-1$ is even, so $(-1)^{p-1}=1$. Therefore, $(-a)^p=a^p$.
Alternatively, we can use Fermat’s Little Theorem to prove this statement. Since $p$ is odd and $p$ is prime, we have $a^p \equiv a \pmod{p}$ and $(-1)^p \equiv -1 \pmod{p}$. Therefore, $$(-a)^p \equiv (-1)^p a^p \equiv -a \pmod{p}.$$ This implies $(-a)^p=-a^p$, since both $(-a)^p$ and $-a^p$ are congruent to $-a$ modulo $p$.
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