# Vlorbik's Diner

## lazy idle little schemer

Posted by vlorbik on September 24, 2010

i’m back at work slinging the math.
i’ve posted a few details in MEZB.

1. ### Anonymoussaid

Mathematics 366: Discrete Mathematical Structures 1
Autumn Quarter 2010

Owen Thomas MW 152
Phone: 292-0244 (Office)
704-4531 (Cel)
Office Hours: TR 4:00-5:20
(and by appointment)

Catalog Number: 12901
Course ID: 115508

TR 5:30–6:48 PM
University Hall 051

Text: Discrete Mathematics with Applications, Susanna S. Epp
(4th Edition; Brooks/Cole 2011 ISBN 0495391328)
Sections Covered: 2.1–2.4… 3.1–3.4… 4.1–4.6… 5.1–5.4… 6.1–6.3… 8.1… 7.1–7.3

Catalogue Description:
Mathematical formalization and reasoning, logic and Boolean algebra; sets, functions, relations, recursive definitions, and mathematical induction; and elementary counting techniques.

Prerequisite: Mathematics 132 or 152.xx
Follow-up Course: Math 566

2. ### vlorbiksaid

http://www.math.ohio-state.edu/~carlson/09sp366.html
carlson’s spring 09 notes

3. ### vlorbiksaid

\smallskip
\centerline{Math 366}
\centerline{Exam 2: November 4, 2010.}
\vskip .5in
\parindent=0

{\bf 1.}
Compute the sum: $\sum_{i=1}^3 i^3$.
\vfil
{\bf 2.}
Prove by induction: $(\forall n \in {\Bbb Z}^+) 6|(7^n -1)$.

(“For every integer $n$ greater than or equal to 1,
six divides $7^n -1$”.)
\vfil\vfil\vfil\vfil
\vfil\eject

{\bf 3.}
Prove that whenever $a$ {\bf mod }$6=3$
and $b$ {\bf mod} $6=2$, it is also true
that $ab$ {\bf mod} $6=0$.

(Remark: this shows that the “Zero
Product Law” $(a \not= 0 \wedge b\not=0) \rightarrow (ab\not=0)$ is false in
certain number systems.)
\vfil
{\bf 4.}
Prove that
$(\forall x, y \in {\Bbb Q}) xy \in {\Bbb Q}\, .$

(“The product of any two rational numbers is
a rational number”.)
\vfil\eject
{\bf 5.}
Prove that there is an odd integer $k$
such that $k$ {\bf mod} $7 = 4$.