Administrative info Trouble logging into accounts? Card key access Change office hour room Lecture notes will be on course website PA1 out! Review A proposition is a statement that must be either true or false. EX: 2+2 = 4, 2+2 = 3 Logical operators Unary ¬: negation Binary ∧: and (conjunction) ∨ : or (disjunction) ⊕: xor (exclusive or) =>: implies <=>: iff Truth table Illustrates logical operators P Q P∧Q P∨Q P⊕Q P=>Q P<=>Q T T T T F T T T F F T T F F F T F T T T F F F F F F T T Implications 1+1=3 => 1+1=2 1+1=3 => 1+1=4 Converse 1+1=2 => 1+1=3 1+1=4 => 1+1=3 Contrapositive 1+1≠2 => 1+1≠3 1+1≠3 => 1+1≠4 A predicate is a proposition that depends on variable(s) P(x) = "x > 0" Q(x, y) = "x > y" Quantifiers Universe over which quantifier acts N, Z Make sure to mention this early in semester; later, might elide if obvious Forall ∀x∈N . x > 0 Exists ∃x∈N . x > 0 ====== More quantifiers EX: "the square of a number is always non-negative" ∀x∈N . x^2 >= 0 (true) "there is a number that is its own square" (∃x∈N)(x = x^2) "doubling a number always makes it larger" ∀x∈N . 2x > x (false; x=0) ∀x∈N . 2x >= x (true.) but: latter proposition is false in integers! (x<0) highlights need to include set over which you're quantifying Domain of statement can be further restricted using implications "The square of all natural numbers greater than 5 is greater than 25." ∀x∈N . x > 5 => x^2 > 25 Is predicate true for x = 2? Why? Is predicate true for x = 6? Why? Can combine quantifiers (∀x∈Z)(∀y∈Z)(x^2 - y^2 = (x-y)(x+y)) Equivalent to extra parens around everything after the first quantifier. True or false? ∀x∈N ∃y∈N . x = 2*y (false in integers; true in reals) ∀x∈N ∃y∈N . y = x^2 (true) (note: y is allowed to depend on x; think of this as a game where you pick x, then I pick y, and I win if y=2x; the proposition is true if I have a winning strategy that lets me win no matter what you do) In English: "the square of every natural number is a natural number" ∃x∈N ∃y∈N . y = 2*x (true) ∀x∈N ∀y∈N . y = 2*x (false) Order of quantifiers matters! Not commutative, ∃y∈N ∀x∈N . y = x^2 (false) In English: "there is a natural number that is the square of every natural number", i.e the square of every natural number is the same! De Morgan's law Can move negation inside quantifier, flips quantifier. ¬∃x∈N . ∀y∈N . y <= x ==> ∀x∈N . ¬∀y∈N . y <= x ==> ∀x∈N . ∃y∈N . ¬(y <= x) ==> ∀x∈N . ∃y∈N . y > x EX: (Lewis Carroll, author of "Alice in Wonderland") Universe of Discourse = 'Creatures' "All bears are fierce" "Some bears do not drink Peet's coffee" "Some fierce creatures do not drink Peet's coffee" Let B(x) = "x is a bear" Let F(x) = "x is fierce" Let P(x) = "x drinks Peet's coffee" Express below using ∀, ∃: "All bears are fierce" ∀x∈Creatures . B(x) => F(x) "Some bears do not drink Peet's coffee" why not ∃x∈Creatures . B(x) => ¬P(x)? ∃x∈Creatures . B(x) ∧ ¬P(x) "Some fierce creatures do not drink Peet's coffee" ∃x∈Creatures . F(x) ∧ ¬P(x) ====== Proofs Now that we know how to make a statement that has a definite truth value, we want to be able to prove what the truth value actually is. I'll show you some proofs. The important point is not these particular proofs, but rather to use these as examples to learn what is a valid proof and how to recognize a correct proof. DEF: "axiom" is a proposition that we assume is true without proof EX: Peano axioms for natural numbers DEF: "theorem" is a proposition that we can prove to be true DEF: "conjecture" is a proposition that we think is true but don't know how to prove How to prove implications X => Y? Assume X, prove that Y follows. Thm: P => (P ∨ Q) Proof: By enumeration. Draw a truth table; this holds in all cases. QED. Alternatively: Thm: P => (P ∨ Q) Pf: Assume P. Then P ∨ Q (using the rule that if we know X, we can deduce X ∨ Y). Therefore P => (P ∨ Q). QED. Thm: P => (P ∨ Q) Pf: There are only two cases: 1) P is true. In this case, P ∨ Q is true, and "anything => true", so X => (P ∨ Q) is true for all X, and in particular, P => (P ∨ Q). 2) P is false. In this case, "false => anything", so P => anything is true, and in particular, P => (P ∨ Q). In either case, P => (P ∨ Q) is true. These are the only two cases possible. QED. Notice how this is the same proof, just much more elaborate. How to prove existential statement? Give an example. (Sometimes called "proof by example.") Thm: ∃x∈N . x = x^2 Pf: 0 = 0^2 = 0 How to prove universal statement? Much harder; need to prove for every instance. In general, consider arbitrary instance x, prove it for x without making any assumptions about x, conclude that your proof works for every x. Theorem: For every n∈N, n^3 - n is divisible by 3. Proof: Consider any n∈N. n^3 - n = (n-1)n(n+1). One of these is divisible by three, since they are consecutive. Three divides the RHS, so it must divide the LHS. So 3 divides n^3. This is true for any n we choose; our choice of n was completely arbitrary, so it must be true for all n. Theorem: For every n∈N, n^3 - n is divisible by 3. Proof: Consider any n∈N. There are three cases: case 0: n=3k for some k∈N: n^3 - n = (3k)^3 - 3k = 27k^3 - 3k, which is divisible by 3. case 1: n=3k+1 for some k∈N: n^3 - n = (3k+1)^3 - (3k+1) = 27k^3 + 27k^2 + 6k, which is divisible by 3. case 2: n=3k+2 for some k∈N: n^3 - n = (3k+2)^3 - (3k+2) = 27k^3 + 54k^2 + 33k + 6, which is divisible by 3. These 3 cases exhaust all the possibilities, and in every case, n^3 - n is divisible by 3. So therefore n^3 - n is divisible by 3 for any n we choose; our choice of n was completely arbitrary, so it must be true for all n. Question for break Can you prove the following: Theorem: For every n∈N, n is even <=> n^2 is even. Try splitting into two. Theorem: For all n∈N, n is even => n^2 is even. Proof: Assume n is even. Then n=2k for some k∈N. Plugging in, n^2 = (2k)^2. So n^2 = 4k^2 = 2 * (2k), which is twice an even number. This means that n^2 is even. Q: Why is this OK? Why can we just assume that n is even? Proof by contrapositive Theorem: For all n∈N, if n^2 is even, then n is even How would you go about proving this? We can assume that n^2 = 2k for some k, but then what? What can we say about sqrt(2k)? Instead, let's look at the contrapositive of the statement: if n is odd, then n^2 is odd This seems more promising, since we don't have to deal with sqrts. Theorem: For all n∈N, if n^2 is even, then n is even Proof: Let n be arbitrary. We prove the contrapositive that if n is odd, then n^2 is odd. Assume that n is odd. Then n = 2k+1 for some k∈N. Then n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2l + 1, where l = 2k^2+2k. Thus, n^2 is odd. QED. Having proven both the last two theorems ∀n∈N . n even => n^2 even ∀n∈N . n^2 even => n even we can conclude that ∀n∈N . n even <=> n^2 even Proof by contradiction Assume opposite of what we are trying to prove. Show that it leads to an impossible situation. So our assumption must have been false. Theorem: x^5 - x + 1 = 0 has no solution in the rationals. Lemma: If x^5 - x + 1 = 0 and x=a/b for a,b∈Z, then both a and b are even. Proof of lemma: We have (a/b)^5 - a/b + 1 = 0. Multiplying by b^5, we get a^5 - ab^4 + b^5 = 0; Now consider possibilities for parity of a and b: Case 1: a odd, b odd Then we get odd - odd + odd = even. Impossible. Case 2: a even, b odd Then we get even - even + odd = even. Impossible. Case 3: a odd, b even Then we get odd - even + even = even. Impossible. Case 4: a even, b even Then we get even - even + even = even. Possible. Since case 4 is the only one that is possible, it must be that a and b are both even. Proof of theorem: Suppose (for a contradiction) that there exists some rational number x such that x^5 - x + 1 = 0. Then we can write x = a/b, with a∈Z, b∈N, b ≠ 0. There may be many such a and b, so let's choose the a and b that make b minimal, so that a and b have no common factors (i.e. a/b is a reduced fraction). By the lemma, it must be that a and b are even. But then a/b is not a reduced fraction, since 2 is a common factor of both a and b! (Alternative: But then let a'=a/2, b'=b/2, then x=a'/b'. So b is not minimal!) This is a contradiction, so our initial assumption must have been wrong.