Week 3 - Friday

 What did we talk about last time?  Proving universal statements  Disproving existential statements  Rational numbers  Proof formatting

 An anthropologist studying on the Island of Knights and Knaves is told that an astrologer and a sorcerer are waiting in a tower  When he goes up into the tower, he sees two men in conical hats  One hat is blue and the other is green  The anthropologist cannot determine which man is which by sight, but he needs to find the sorcerer  He asks, "Is the sorcerer a Knight?"  The man in the blue hat answers, and the anthropologist is able to deduce which one is which  Which one is the sorcerer?

 After determining that the man in the green hat is the sorcerer, the sorcerer asks a question that has a definite yes or no answer  Nevertheless, the anthropologist, a naturally honest man, could not answer the question, even though he knew the answer  What's the question?

 A real number is rational if and only if it can be expressed at the quotient of two integers with a nonzero denominator  Or, more formally, r is rational   a, b  Z  r = a/b and b  0

 The reciprocal of any rational number is a rational number

 Math moves forward not by people proving things purely from definitions but also by using existing theorems  "Standing on the shoulders of giants"  Given the following: 1. The sum, product, and difference of any two even integers is even 2. The sum and difference of any two odd integers are even 3. The product of any two odd integers is odd 4. The product of any even integer and any odd integer is even 5. The sum of any odd integer and any even integer is odd  Using these theorems, prove that, if a is any even integer and b is any odd integer, then (a 2 + b 2 + 1)/2 is an integer

 If n and d are integers, then n is divisible by d if and only if n = dk for some integer k  Or, more formally:  For n, d  Z,  n is divisible by d   k  Z  n = dk  We also say:  n is a multiple of d  d is a factor of n  d is a divisor of n  d divides n  We use the notation d | n to mean "d divides n"

 Prove that for all integers a, b, and c, if a | b and b | c, then a | c  Steps:  Rewrite the theorem in formal notation  Write Proof:  State your premises  Justify every line you infer from the premises  Write QED after you have demonstrated the conclusion

 For all integers a and b, if a | b and b | a, then a = b  How could we change this statement so that it is true?  Then, how could we prove it?

 For any integer n > 1, there exist a positive integer k, distinct prime numbers p 1, p 2, …, p k, and positive integers e 1, e 2, …, e k such that  And any other expression of n as a product of prime numbers is identical to this except, perhaps, for the order in which the factors are written

 Let m be an integer such that  8∙7 ∙6 ∙5 ∙4 ∙3 ∙2 ∙m = 17∙16 ∙15 ∙14 ∙13 ∙12 ∙11 ∙10  Does 17 | m?  Leave aside for the moment that we could actually compute m

 If you have a premise consisting of clauses that are ANDed together, you can split them up  Each clause can be used in your proof  What if clauses are ORed together?  You don't know for sure that they're all true  In this situation, you use a proof by cases  Assume each of the individual possibilities is true separately  If the proof works out in all possible cases, it still holds

 For a direct proof using cases, follow the same format that you normally would  When you reach your cases, number them clearly  Show that you have proved the conclusion for each case  Finally, after your cases, state that, since you have shown the conclusion is true for all possible cases, the conclusion must be true in general

 For any integer n and any positive integer d, there exist unique integers q and r such that  n = dq + r and 0 ≤ r < d  This is a fancy way of saying that you can divide an integer by another integer and get a unique quotient and remainder  We will use div to mean integer division (exactly like / in Java )  We will use mod to mean integer mod (exactly like % in Java)  What are q and r when n = 54 and d = 4?

 As another way of looking at our earlier definition of even and odd, we can apply the quotient-remainder theorem with the divisor 2  Thus, for any integer n  n = 2q + r and 0 ≤ r < 2  But, the only possible values of r are 0 and 1  So, for any integer n, exactly one of the following cases must hold:  n = 2q + 0  n = 2q + 1  We call even or oddness parity

 Prove that, given any two consecutive integers, one is even and the other is odd  Hint Divide into two cases:  The smaller of the two integers is even  The smaller of the two integers is odd

 Theorem: for all integers n, 3n 2 + n + 14 is even  How could we prove this using cases?  Be careful with formatting

 For any real number x, the floor of x, written  x , is defined as follows:   x  = the unique integer n such that n ≤ x < n + 1  For any real number x, the ceiling of x, written  x , is defined as follows:   x  = the unique integer n such that n – 1 < x ≤ n

 Prove or disprove:   x, y  R,  x + y  =  x  +  y   Prove or disprove:   x  R,  m  Z  x + m  =  x  + m

 Give the floor for each of the following values  25/4    Now, give the ceiling for each of the same values  If there are 4 quarts in a gallon, how many gallon jugs do you need to transport 17 quarts of werewolf blood?  Does this example use floor or ceiling?

 Indirect argument  Proof by contradiction  Irrationality of the square root of 2  Infinite number of primes

 Turn in Assignment 2 by midnight tonight!  Keep reading Chapter 4