Day 8: mat385

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Today:

• Announcements:
  • Reminder: turn off your rectangles.

  • Roll.

  • Your 1.4 homework is returned.

    • Graded 18, 26, 42 (uh oh, nothin' but evens....):

      • 18: Same trick as I mentioned last time: given \[ (\exists x)[R(x) \lor S(x)] \longrightarrow (\exists x)R(x) \lor (\exists x)S(x) \] replace with \[ (\exists x)[R(x) \lor S(x)] \longrightarrow [((\exists x)R(x))' \longrightarrow (\exists x)S(x) \] and use the deduction method.

        There was a fair amount of illegal existential generalization. So here is an example:

        1. $(\exists x)[R(x) \lor S(x)]$ (hyp)
        2. $R(a) \lor S(a)$ (1, ei)
        3. $(\exists x)[R(x)] \lor S(a)$ (2, eg) -- whoops, illegal eg
        That's illegal. From a true wff, P(a), write $(\exists x)P(x)$. So line 2 is true, so we can write $(\exists x)[R(x) \lor S(x)]$ -- but that's not very helpful!

      • 26: Same deal. I saw this illegal move:
        1. $(\forall y)[Q(x,y) \longrightarrow P(x)]$ (hyp)
        2. $Q(x,a) \longrightarrow P(x)$ (1, ui)
        3. $(\exists y)Q(x,y)] \longrightarrow P(x)$ (2, eg) -- whoops, illegal eg
        We can legally say $(\exists y)(Q(x,y)] \longrightarrow P(x)) (2, eg)$

        It's about scope.

      • 42: Wicked! It's really important here to write your hypotheses in a form conducive to making the case; and to write the hypotheses correctly. For example, I saw these:

        • $(\forall x)(\forall y)(F(x)\land C(y) \longrightarrow O(x,y))$ -- which says all farmers own all cows.

        • $(\forall x)(\exists y)(F(x)\land C(y) \longrightarrow O(x,y))$ -- which is true providing there's anything at all in the set which isn't a cow! (There's that existential quantifier with the implication....)

    • Graded 2.1 homework: #4, 61, and 74

      • 4: Show why these are counterexamples.

      • 61: Make the argument: works for $n=1$; for natural numbers $n \ge 2$, \[ n+\frac{1}{n} \ge 2+\frac{1}{n} \ge 2 \]

        Alternative strategy: MacKenzie (and another one or two): Assume $n>0$ \[ (n-1)^2>0 \implies n^2-2n+1>0 \implies n^2+1>2n \implies n+\frac{1}{n}>2 \]

        This is better, because it shows that it's true for a bigger set:

        Why does the argument fail for $n<0$?

      • 74: You didn't seem to struggle with this argument much: just putting the facts together.

• Today:

  • Last time: Recursion.

    We didn't quite finish, however, so we'll look at a few more examples, before moving along.

  • Today we consider Recurrence Relations.