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CPSC-121 Indirect proofs_ random stuff_ and number representation

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					     CPSC-121: Indirect proofs, random
      stuff, and number representation
             Week 5 Tutorial outline
                                 Matt Hoffman


1     Proof by contradiction
For a proof by contradiction we want to prove
                       (A ⇒ B) ≡ (¬A ∨ B) ≡ ¬(A ∧ ¬B)
and will assume
                                     A ∧ ¬B
to arrive at some logical contradiction.


2     Examples
    1. Show that “there is no largest integer.”

       Proof. Assume that there is a largest integer, let’s call it x. This means
       that
                                   ∀y ∈ Z : x ≥ y.
       Our number x is an integer, so we can always add one to get another
       integer, say z = x + 1. This means that
                                 x ≥ x + 1 =⇒ 0 ≥ 1.
       This is a contradiction, so our assumption must be wrong, and there is no
       largest integer.

    2. Show that “an integer is even if and only if its square is even.”

       Proof. We have to prove this two ways. Showing that n is even implies
       that n2 is even is easy:
                        n ∈ Z is even =⇒ ∃x ∈ Z : n = 2x
                                       =⇒ n2 = (2x)2 = 2(2x2 )
                                       =⇒ n2 is even


                                           1
   Next we have to prove that n2 even implies n is even. Given that n2 is
   even let’s assume that n is odd. We know that a product is even only if
   at least one of its subparts is even—but this contradicts our assumption
   that n is odd. Therefore our assumption is false, and n must be even.

3. Show that if a, b, c are odd integers, then ax2 + bx + c = 0 has no solution
   in the set of rational numbers.

   Proof. Assume there is a rational solution x, which can be written
                                                 p
                            ∃p, q ∈ N : x =        , q = 0,
                                                 q
                              p
   and here we assume that    q   is in its lowest form. This can be substituted
   in to obtain
                                        2
                                    p          p
                             a              + b + c = 0,
                                    q          q
                                  ap2 + bpq + cq 2 = 0.

   Since we assumed that p was in its lowest form, then p and q cannot both
                           q
   be even. If p is even and q odd, then the cq 2 term is odd, similarly if q
   is even and p is odd then the ap2 term is odd—meaning the polynomial
   is odd. If both p and q are odd, the polynomial will be the sum of 3 odd
   numbers, and thus odd.
   Therefore the polynomial is odd, and since 0 is even, the polynomial can-
   not be equal to 0. This contradiction means our assumption is wrong, and
   thus there is no solution to the polynomial in the rational numbers.

4. Given an eight-by-eight checkerboard.
    (a) Given a domino shape (ie one square black, one square white) and
        the checkerboard with opposite corners removed, prove that the re-
        maining squares cannot be covered exactly by dominoes.
   (b) Given a T-shape, where the center square can be either black or
       white, and a checkerboard with two squares removed from opposite
       corners—prove that the remaining squares cannot be covered exactly
       by the T-shapes.




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3     Incorrect “proof ”
Let x and y be integers, and suppose that x = y.

     x = y =⇒ x2 = xy                               multiply by x
                  2    2           2
            =⇒ x − y = xy − y                       subtract y 2
            =⇒ (x + y)(x − y) = y(x − y)            factor
            =⇒ x + y = y                            divide by (x − y) = 0. error!


    For x = y = 1, this implies that 2 = 1. What’s wrong with the proof?


4     Number representation
    1. Base 10. Numbers in base 10 can be written as

                           21710 = 2(102 ) + 1(101 ) + 7(100 ).

    2. Base 2. If we want to represent things on a computer, we have to use
       bits (transistors)—switches that can be on or off.
      For example
         • 10112 = 1(23 ) + 0(22 ) + 1(21 ) + 1(20 ) = 8 + 2 + 1 = 11.
         • How many numbers can be represented with one bit? 2 (either on or
           off).
         • Adding one more bit doubles the amount, you have all the previous
           bits, either with the last bit as a 0, or a 1.
         • n bits gives 2n distinct numbers.
         • Adding two numbers is just like adding two numbers in base-10.

                                               00110
                                              +10011
                                               11001

    3. Negatives. Negative numbers are usually represented with a minus-sign,
       but all we’ve got are bits. So we can take the easy way out and just let
       one bit represent negative-or-not.
         • Adding/subtracting is a pain.
         • We’ve got two representations for 0,

                                       10 . . . 0 = 00 . . . 0.

    4. Two’s complement. A better way to represent negative numbers.

                                          3
• Wrap the numbers around the n bits. This looks like:

                     {−2n−1 , . . . , 0, . . . , 2n−1 − 1},

  i.e.
                         {−256, . . . , 0, . . . , 255}.
• Positive numbers are just as before, but negative numbers we repre-
  sent as 2n − x. Two easy ways to find −x:
  (a) Easy: flip the bits of x, add 1.
  (b) Easier: ignore everything starting at the right up to and includ-
      ing the first 1, flip every other bit.
• Finally, addition works just the same as above, and adding negative
  numbers just works without needing to check the sign bit.




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