MY-ASSIGNMENTEXPERT™可以为您提供sydney MATH1004 Discrete Mathematics离散数学课程的代写代考和辅导服务！
这是悉尼大学 离散数学课程的代写成功案例。
MATH1004课程简介
This unit provides an introduction to fundamental aspects of discrete mathematics, which deals with ‘things that come in chunks that can be counted’. It focuses on the enumeration of a set of numbers, viz. Catalan numbers. Topics include sets and functions, counting principles, discrete probability, Boolean expressions, mathematical induction, linear recurrence relations, graphs and trees.
Prerequisites
At the completion of this unit, you should be able to:
- LO1. identify combinatorial objects involved in counting problems
- LO2. understand how to construct switching circuits representing Boolean functions
- LO3. factor numbers using sieve methods and use the Euclidean algorithm to compute greatest common divisors
- LO4. solve linear recurrence relations by using generating functions or characteristic equations.
MATH1004 Discrete Mathematics HELP（EXAM HELP， ONLINE TUTOR）
Find the negation of $P \rightarrow Q$.
Solution: $\mathrm{P} \rightarrow \mathrm{Q}$ is equivalently written as $(\neg \mathrm{P} \vee \mathrm{Q})$
So, negation of
$$
\begin{aligned}
\mathrm{P} \rightarrow \mathrm{Q} & \equiv \neg(\neg \mathrm{P} \vee \mathrm{Q}) \
& \equiv \neg(\neg \mathrm{P}) \wedge(\neg \mathrm{Q}),(\text { By De-Morgan’s Law) } \
& \equiv \mathrm{P} \wedge(\neg \mathrm{Q})
\end{aligned}
$$
Hence the negation of $\mathrm{P} \rightarrow \mathrm{Q}$ is $\mathrm{P} \wedge(\neg \mathrm{Q})$.
Construct the truth table for $(P \rightarrow Q) \leftrightarrow(\neg P \vee Q)$.
Solution : The given compound statement is $(\mathrm{P} \rightarrow \mathrm{Q}) \leftrightarrow(\neg \mathrm{P} \vee \mathrm{Q})$ where $\mathrm{P}$ and $\mathrm{Q}$ are two atomic statements.
\begin{tabular}{|c|c|c|c|c|c|}
\hline $\mathbf{P}$ & $\mathbf{Q}$ & $\neg \mathbf{P}$ & $\mathbf{P} \rightarrow \mathbf{Q}$ & $\neg \mathbf{P} \vee \mathbf{Q}$ & $(\mathbf{P} \rightarrow \mathbf{Q}) \leftrightarrow(\neg \mathbf{P} \vee \mathbf{Q})$ \
\hline $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ \
$\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ \
$\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ \
$\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ \
\hline
\end{tabular}
Find the negation of the following statement. “If Cows are Crows then Crows are four legged”.
Solution : Let P: Cows are Crows
Q : Crows are four legged
Given statement: If Cows are Crows then Crows are four legged.
$$
\equiv \mathrm{P} \rightarrow \mathrm{Q}
$$
So, the negation is given as $\mathrm{P} \wedge(\neg \mathrm{Q})$ i.e. Cows are Crows and Crows are not four legged.
Example 5 Find the negation of the following statement. He is rich and unhappy.
Solution : LetP $\equiv \mathrm{He}$ is rich
$\mathrm{Q} \equiv \mathrm{He}$ is unhappy
Given statement: He is rich and unhappy
$$
\begin{gathered}
\equiv \mathrm{P} \wedge \mathrm{Q} \
\mathrm{Q}) \equiv \neg \mathrm{P} \vee \neg \mathrm{Q}
\end{gathered}
$$
$$
\text { By De-Morgan’s law } \neg(\mathrm{P} \wedge \mathrm{Q}) \equiv \neg \mathrm{P} \vee \neg \mathrm{Q}
$$
$\equiv \mathrm{He}$ is neither rich nor unhappy.
Prove by constructing truth table
$$
P \rightarrow(Q \vee R) \equiv(P \rightarrow Q) \vee(P \rightarrow R)
$$
Solution : Our aim to prove $P \rightarrow(Q \vee R) \equiv(P \rightarrow Q) \vee(P \rightarrow R)$
Let $P, Q$ and $R$ be three atomic statements.
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline $\mathbf{P}$ & $\mathbf{Q}$ & $\mathbf{R}$ & $\mathbf{Q} \vee \mathbf{R}$ & $\mathbf{P} \rightarrow(\mathbf{Q} \vee \mathbf{R})$ & $\mathbf{P} \rightarrow \mathbf{Q}$ & $\mathbf{P} \rightarrow \mathbf{R}$ & $\begin{array}{c}(\mathbf{P} \rightarrow \mathbf{Q}) \vee \
(\mathbf{P} \rightarrow \mathbf{R})\end{array}$ \
\hline $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ \
$\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{F}$ \
$\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ \
$\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ \
$\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ \
$\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ \
$\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{T}$ \
$\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ \
\hline
\end{tabular}
From the truth table it is clear that $P \rightarrow(Q \vee R) \equiv(P \rightarrow Q) \vee(P \rightarrow R)$.
MY-ASSIGNMENTEXPERT™可以为您提供sydney MATH1004 Discrete Mathematics离散数学课程的代写代考和辅导服务！
