MY-ASSIGNMENTEXPERT™可以为您提供my.uq.edu.au MATH1061 Discrete Mathematics离散数学课程的代写代考和辅导服务！
这是昆士兰大学离散数学课程的代写成功案例。
MATH1061课程简介
Course Description: Propositional & predicate logic, valid arguments, methods of proof. Elementary set theory. Elementary graph theory. Relations & functions. Induction & recursive definitions. Counting methods pigeonhole, inclusion/exclusion. Introductory probability. Binary operations, groups, fields. Applications of finite fields. Elementary number theory.
There is no particular assumed background, apart from a level of mathematical sophistication roughly equivalent to completion of Queensland Mathematical Methods (formerly Maths B) at high school or MATH1040. Concurrent enrolment in MATH1061 and MATH1040 is appropriate for students who do not have Mathematical Methods from high school but who have a strong mathematical background, however this will require you to do extra work in MATH1061.
Prerequisites
This course provides an introduction to discrete mathematics. It is likely to be useful for students who are planning on studying more mathematics, those intending to teach, and also for those enrolled in computer science, engineering, science and information technology.
Course Changes in Response to Previous Student Feedback
This is an active learning course. Learning resources have been updated and extended.
MATH1061 Discrete Mathematics HELP（EXAM HELP， ONLINE TUTOR）
Denote $0=|\emptyset|$ and $1=|{\emptyset}|$. Show that for every cardinal number $p$ the following holds:
a) $p+0=p \cdot 1=p$
b) $p^0=0^0=1$
c) $p^1=p$
(recall that equalities between cardinal numbers are represented by some bijections, their summation corresponds to a disjoint union, their product to Cartesian product of sets, and powers are represented by sets of mappings)
Show that for any cardinal numbers $p, q$ and $r$ we have:
a) $p \leq q \Rightarrow p r \leq q r$,
b) $p \leq q \Rightarrow p^r \leq q^r$,
c) if $r \neq 0$, then $p \leq q \Rightarrow r^p \leq r^q$. What happens when $r=0$ ?
Do the statements from problem 2 remain true when we replace inequalities by strict inequalities?
The arithmetics of infinite cardinal numbers is slightly different than expected: For example $\aleph_0+\aleph_0=\aleph_0$, because $\mathbb{N}_0 \cup-\mathbb{N}=\mathbb{Z}$. Similarly, $\aleph_0 \cdot \aleph_0=\aleph_0$, because $|\mathbb{N}||\mathbb{Z}|=|\mathbb{Q}|$.
Denote $\aleph_0=|\mathbb{N}|$ and $c=|\mathbb{R}|=2^{\aleph_0}$. By using inequalities from problem 2 and relations from the lecture show that for any integer $n \in \mathbb{N}, n \geq 2$ holds:
a) $c \leq n c \leq \aleph_0 c \leq c c \leq c^n \leq c^{\aleph_0}=\left(2^{\aleph_0}\right)^{\aleph_0}=c$,
b) $2^{\aleph_0}=n^{\aleph_0}=\aleph_0^{\aleph_0}=c^{\aleph_0}=c$,
c) $2^c=n^c=\aleph_0^c=c^c$
Show that the set of all irrational numbers is uncountable.
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