SUNY Geneseo Department of Mathematics
Friday, March 23 - Monday, March 26
Math 239 01
Spring 2018
Prof. Doug Baldwin
Sample exam questions and solutions.
How I converted numeric grades to letter grades last time I taught this course. Not an exact guide to how I’ll do it this semester, but it might give you some sense of how you’re doing.
(From this course plus some general comments from calc 3)
Use of Canvas, organization of materials in it, formal proofs from class.
Open-notes exams.
Grading scheme, especially for exams.
Face-to-face grading. I think this is important for giving immediate and timely feedback on your work, so I plan to continue doing it, even though some people listed it as something they’d like me to stop.
Solving problem sets as expected gets 80%, with more points for going “above and beyond.” This is frustrating to some people, but it sends an important message that meeting expectations isn’t perfection, and that you can rise beyond just meeting expectations, so I plan to keep doing it. But realize that given the way numeric grades convert to letter grades, you can also choose to do what’s asked, get 80s on problem sets, and still probably get an A or B for the course, depending on how you do on the exams.
Face-to-face grading. See above
Solving problem sets as expected only gets 80%. See above.
Random names. I’ll use them as a backup mechanism for getting discussions going with wide participation, but will try first to run discussions through volunteers offering to speak.
Late penalties due to lack of time slots before “Grade By” date. Let me know by the “complete by” date if you can’t get an appointment, and I’ll work out some appropriate extension.
Review key points from readings at start of class. I don’t do this explicitly, but try to do it via my choice of in-class exercises and “key points” sections in the notes.
Provide reviews before exams, including good textbook problems to practice on. These are good ideas, and I can use some class time for reviews, as long as they’re based on your questions.
Typeset lectures, equations.
Use whiteboard instead of computer for in-class notes. I can do this if you feel a strong need, but the cost would probably be no more Canvas notes -- recreating a complete set of electronic notes from ephemeral whiteboard ones is much more time-consuming than cleaning up electronic notes started before or during class.
Let students use whiteboard to explain things. Excellent idea.
Put information on Canvas to help you estimate letter grades. Good idea, see earlier note.
Hour exam 2 is Wednesday, March 28.
It will cover material not covered on the first exam, through (at most) simple induction (e.g., quantifiers, proof via the contrapositive, proofs about biconditionals, proof by contradiction, proof in cases, maybe some induction).
The rules and format will otherwise be similar to the first exam, particularly the open-book, open-computer rule.
Define a “fully parenthesized arithmetic expression” to be a string of one of the following forms:
Give some examples or counter-examples of fully parenthesized arithmetic expressions.
Prove that for all integers n ≥ 0, every fully parenthesized arithmetic expression with n operators has n+1 constants.
One idea: think of an infinitely long expression such as (1) + ((2) + ((3) + .... and then do induction on “prefixes” of it. But this runs into a need to define exactly what the prefixes are, and there are plenty of expressions beside this one that the theorem should apply to.
Another idea: do induction on the number of operators in an expression. The inductive step would be based on the second part of the definition, and would somehow use the induction hypothesis on the subexpressions E1 and E2. This is a promising idea, and is the one we’ll use in the formal proof.
Theorem: for all integers n ≥ 0, every fully parenthesized arithmetic expression with n operators has n+1 constants.
Proof: we will use strong induction on n to prove that for all integers n ≥ 0, every fully parenthesized arithmetic expression with n operators has n+1 constants.
For the base case, consider n = 0. The only fully parenthesized arithmetic expressions with 0 operators are constants, which have 1 constant.
For the induction step, we assume that for some k ≥ 0, any fully parenthesized arithmetic expression with m operators, 0 ≤ m ≤ k, has m+1 constants; we then show that any fully parenthesized arithmetic expression with k+1 operators has k+2 constants. Since k+1 is at least 1, a fully parenthesized arithmetic expression with k+1 operators has to be of the form (E1) * (E2) where “*” denotes any one of the operators. Let m1 be the number of operators in E1 and m2 the number in E2. Since expression (E1) * (E2) has k+1 operators and “*” accounts for one of them, we have m1 + m2 = k. Moreover, the number of operators in an expression can’t be negative, so 0 ≤ m1 ≤ k, and similarly 0 ≤ m2 ≤ k. Thus the induction hypothesis applies to both E1 and E2, i.e., E1 has m1+1 constants and E2 has m2+1. The total number of constants in expression (E1) * (E2) is thus
Having established both the base case and the induction step, we have now shown by strong induction that for all integers n ≥ 0, every fully parenthesized arithmetic expression with n operators has n+1 constants. QED.
You can apply mathematical methods (theorems and proofs, etc.) to mathematical notations and ideas.
Reading and using definitions.
Strong Induction; note that it may have more than one base case but doesn’t have to.
What would you like to review or talk about?
Quantifiers: Fundamentally, they introduce variables by associating them with sets of possible values. The two forms are “for all” and “there exists.” For example, “for all integers n...”, “for some (aka there exists a) real number r....”
Modular congruence: Notated as a ≡ b (mod n), e.g., 6 ≡ 2 (mod 4). This can be interpreted in either of two equivalent ways: a = qn + b for some integer q, e.g., 6 = 1×4 + 2, or n divides (b-a), e.g., 4 divides (6-2). Proofs involving congruences will typically translate between them and one of these two forms.
The solution to the case-based proof in the sample questions (sample question 2): The decision to use a case-based proof is based on the fact that the function is defined piecewise, i.e., defined in cases; each case in the proof corresponds to a case in the definition.
Sets and their operations.
Read section 5.1.