Handouts For Class (Organized by the day on which we will discuss the material in class)
February 1: No Handout
January 31: Section 7.3
January 30: Sections 7.1-7.2 Read Section 4.1 in the text.
January 27: Section 6.1
January 26: No handout. Exploration
January 25: Section 2.4, Section 3.4
January 24: Section 3.3, Section 5.3,
January 23: Section 3.2, Section 5.1
January 20: Section 2.3, Section 3.1
January 19: Section 1.9, Today's Advertisement, More on Cantor's Theorem (for those who are interested)
January 18: Section 1.8, Section 2.2
January 17: Section 1.7, Section 2.1
January 16: Section 1.5, Section 1.6
January 13: Section 1.3, Section 1.4
January 12: Section 1.2, (I recommend that you print or download copies of these handouts before class.)
January 11: Syllabus, Section 1.1, Room Visiting Problem (I will provide a copy of this document for you in class - no need to print it)
Download The Entire Text (Chapters may be edited, especially later chapters)
Entire Text (Section 1.11 updated on 1/11/23) This document is over 300 pages long! You may want to read this online.
In-Class Explorations
January 19: Hats and Gloves
January 17: Penny (or pen!) Game
January 16: Gerrymandering
January 12: Square Output
January 11: Exploration - Gnomons
Solutions to Selected Homework Problems
6.1.15) We did this two ways. There are C(17,3) ways to choose three people. How many ways have three together? 17. How many have two together and one other? There are 17 pairs, each of which has 13 options for the other choice. Thus C(17,3)-17-17*13 = 442. We could also choose 1 individual (17 choices), then choose the other two using a gap analysis argument. There are 14 people not next to the one we chose. Remove the two we are going to choose, and that leaves 12 people, so 13 gaps. Put one in one gap, the second in another, so there are 13*12 choices for the gaps. But we've introduced an ordering, so the number of ways to choose our 3 people is 17*13*12/(3!)=442.
Section 2.4 problems 1-5, 6-13
Section 2.3 problems 1-2, problems 3a-e, problems 3f-j (m is identical to a proof in the notes!)
Section 2.2 problems 1-2, problems 3a-d, problems 3e-i
Section 2.1 problem 2, Proofs of all 19 Logic Rules
Section 1.8 - all problems - Venn Diagrams
Section 1.3 - all problems collected so far
Problem 4 from Section 1.3: Because \(x\in \mathbb E\), \(x=2n\) where \(n\in \mathbb Z\). Because \(y\in \mathbb E\), \(y=2m\) where \(m\in \mathbb Z\). Then \(xy=(2n)(2m)=2(2nm)\), and because \(2nm\in \mathbb E\) we have proven that \(xy\in \mathbb E\). Commentary: Do you see we we need to write \(xy=2(2nm)\)? Because the definition of \(\mathbb E\) involves writing our number as \(2\) times some integer. The definition tells you what to do!
Section 1.2 - all problems - Mathematica File
Problem from the web: Use Exercise 3 from Section 1.2 to prove that \(f:\mathbb R\rightarrow\mathbb R\) by \(f(x)=15x+8\) is one-to-one. What did you prove in Exercise 3? That any function of the form \(f(x)=ax+b\) is one-to-one as long as \(a\neq 0\). How do we use this exercise? Here's the proof. "Because \(15\neq 0\), we know by Exercise 3 from Section 1.2 that \(f(x)=15x+18\) is one-to-one." That's all? YES! If we've shown that every function of the form \(f(x)=ax+b\) is one-to-one, we can apply that knowledge to any function having the proper form.
Section 1.2, problem 3: You were invited, encouraged, begged to model your proof after the ones we did in class. You could have simply cut the proof of Examples 1.2.5 and 1.2.9 into your notes as well making very few changes. Your proof could have been this short: "If \(a_1\) and \(a_2\) are real numbers and \(f(a_1)=f(a_2)\), then \(a( a_1) + b = a(a_2) + b \Rightarrow a_1=a_2\). Thus \(f\) is one-to-one. To show that \(f\) is onto, we choose \(y\in \mathbb R\) and let \(x=\frac{1}{a}(y-b)\in \mathbb R\). Then \(f(x)=y\), and because \(y\) was arbitrarily chosen, \(f\) is onto." That's all it took!
Section 1.1 - Mathematica File
Section 1.1, problem 7: It's not possible to get all of the books rightside up. If you start with all 7 books upside down and turn them over four at a time, there will always be an odd number of books upside down! There are five possibilities for turning over four books: all were upside down, three were upside down and one was rightside up, two were upside down and two were rightside up, one was upside down and three were rightside up, and all four were rightside up. In all of these cases, the number of upside down books is changed by an even number! I'll post a graph if I get a chance.
January 13 - function iteration
January 12 - define a function - Online Version
January 23 - discrete graphs - Online Version
January 24 - zero divisor graphs - Online Version
January 26 - spirograph
January 30 - voting triangles
February 1 - function iteration (complete file) - attracting orbit1 - attracting orbit 2
January 12 - define a function
January 24 - Multiplication Tables
January 25 - Euclidean Algorithm
January 31 - Apportionment
