Day 11: mat385

Today:

• Reminders:
  • Test 1 coming up next week, Monday.
  • You may use one side of an 8.5x11 sheet (inches!).
  • The test will cover through 2.5.
  • Homework for 3.1 is due Monday, October 8th.

• Return Homework 2.4
  • #2
  • #20

• Homework 2.5 (with some 2.4) will not be handed in, but a problem will appear on the exam.

• Here are some old tests (note: you'll see no Prolog, of course, in your test, since we didn't cover 1.5):
  • 2006 Spring Test 1 (2006 Spring Test 1 key)
  • 2005 Spring Test 1 (2005 Spring Test 1 key)
  • 2003 Fall Test 1 (2003 Fall Test 1 key)
  • This was a one-hour exam (course took place 3 days per week): 2002 Fall Test 1 (2002 Fall Test 1 key)
  • 2002 Spring Test 1 (2002 Spring Test 1 key) Note: we did proof of correctness in this course; we didn't this time.
• Overview of Chapters 1 and 2

• Section 2.2, #31: a variety of approaches
  • Show that

    Induction works, but is ugly, and involves another induction or appeal to calculus (or to a previous problem -- see Problem 28!). A direct approach is better and perhaps more fun, but involves a good grasp of the factorial function:

    Note: this step is useful whether we're doing the proof directly or by induction. It would have been much easier to deal with the expression on the right-hand side in the induction proof. We already checked the case for n=4. For

    are distinct elements in the product (n-1)!.

    and since (n-1) is a distinct third element in the product (n-1)!, and since (n-1)>1,

    and so

    QED.

    If you go back and do the induction with I trust that you will find it much easier. Give it a whirl!

  • George Elgin suggested a proof by contraposition (pdf) at the end of class.

  • Ken Cooney liked George's idea of proof (using contradiction, actually: see his TheGeorgeProof.doc). He's using contradiction because he assumes P and Q' to show P => Q.

    Notice that Ken asserts a few things (that quantities are decreasing) that one might suppose need to be proven. In class we arrived at a point where something looked obvious, but we proved it by calculus (using the slope of the derivative). Alternatively, we did another induction proof, got to another point that looked obvious, and then -- out of fatigue!;) -- we simply said "Yeah, I believe that!". One more induction would have gotten us down to the "1+1=2" level....

  • Here's the solution key's proof, and you'll see that they did it by invoking the result of another exercise -- that is, they did it by a pair of inductions.

    1. P(k)	hyp
    2. (k+1)! = (k+1)k!	by definition
    3. (k+1)! > k2(k+1)	by P(k)
    4. (k+1)! > (k+1)(k+1)	By Exercise 28, since k > 3
    5. (k+1)! > (k+1)2	QED

    I recall that John asked about problems 28 and 31, so I regret that we didn't start with 28! That way, the strategy might have come to mind immediately....

• Section 3.1: Sets

• Problems from section 3.1 homework:
  • Russell's paradox:

    "Some sets, such as the set of all teacups, are not members of themselves. Other sets, such as the set of all non-teacups, are members of themselves. Call the set of all sets that are not members of themselves "R." If R is a member of itself, then by definition it must not be a member of itself. Similarly, if R is not a member of itself, then by definition it must be a member of itself. Discovered by Bertrand Russell in 1901, the paradox has prompted much work in logic, set theory and the philosophy and foundations of mathematics." (source)

  • Suggestions?

• Section 5.1: Graphs and their representations