cse 20 discrete math - home | computer science and...
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CSE 20 DISCRETE MATH Winter 2017 http://cseweb.ucsd.edu/classes/wi17/cse20-ab/
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Today's learning goals • Distinguish between a theorem, an axiom, lemma, a corollary, and a conjecture. • Recognize direct proofs • Recognize proofs by contraposition • Recognize proofs by contradiction • Recognize fallacious “proofs” • Evaluate which proof technique(s) is appropriate for a given proposition
• Direct proof • Proofs by contraposition • Proofs by contradiction • Proof by cases • Constructive existence proofs
• Correctly prove statements using appropriate style conventions, guiding text, notation, and terminology
Textbook references: Sections 1.7-1.8 (with reference to 1.6)
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Review: quantified statements Rosen p. 64 #1
In which domain is this statement not true? A. All real numbers in the closed interval [0,1]. B. The set of integers {1,2,3}. C. All real numbers. D. All positive real numbers. E. All positive integers.
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And in the other direction … Rosen p. 66 #23
• "The product of two negative real numbers is positive."
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And in the other direction … Rosen p. 66 #23
• "The difference between a real number and itself is zero."
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And in the other direction … Rosen p. 66 #23
• "A negative real number does not have a square root that is a real number."
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And in the other direction … Rosen p. 66 #23
• "Every positive real number has exactly two square roots."
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For the sake of argument Rosen p. 81
"A proof is a valid argument that establishes the truth of a statement"
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For the sake of argument Rosen p. 69
"A proof is a valid argument that establishes the truth of a statement"
Valid: truth of the conclusion (final statement) is guaranteed by the truth of
the premises (preceding statements)
Argument: sequence of statements ending with a conclusion
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For the sake of argument Rosen p. 81
Axioms: statements
assumed to be true
Rules of inference+definitions
Theorem Proposition
Fact Result Lemma
Corollary
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What rules of inference are allowed? Rosen 1.6
Argument form is valid if no matter which propositions are
substituted in, the conclusion is true if all the premises are true
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Fancy names Rosen 1.6
• Modus ponens • Modus tollens • Hypothetical syllogism … • Universal instantiation • Universal generalization • Existential instantiation • Existential generalization
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Translating to proof strategy Rosen p. 82
• Modus ponens Direct proof To prove that "If P, then Q" - Assume P is true. - Use rules of inference, axioms, definitions to… - … conclude Q is true.
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Translating to proof strategy Rosen p. 83
• Modus tollens Proof by contraposition To prove that "If P, then Q" - Assume Q is false. - Use rules of inference, axioms, definitions to… - … conclude P is also false.
Both modus ponens and modus tollens apply when proving a
conditional statement
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Main connective
What's the main logical structure in the following sentence? A. Conditional (p à q) B. Universal statement C. Conjunction ("AND") D. Inequality (>=) E. What does this question even mean?
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Main connective
Are these statements true?
I. II.
A. Both are true. B. Both are false. C. I is true and II is false. D. II is true and I is false. E. I don't know how to evaluate the truth value of these sentences.
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Universal conditionals
can be translated to
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Universal conditionals Rosen p. 88
To prove this statement is false :
To prove this statement is true :
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Universal conditionals Rosen p. 88, 82
To prove this statement is false :
Find a counterexample. To prove this statement is true :
Select a general element of the domain, use rules of inference (e.g direct proof, proof by contrapositive, etc.) to prove that the conditional statement is true of this element, c conclude that it holds of all members of the domain.
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Universal conditionals
Claim: is false. Proof:
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Universal conditionals
Claim: is false. Proof: Since this is a universal conditional statement, to prove that it is false, it's enough to find one counterexample. That is, we need a specific value of n which is a positive integer and for which
A. n2 +1 > 2n B. n2 +1 < 2n C. n2 +1 <= 2n D. n2 +1 >= 2n E. What's a universal conditional?
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Universal conditionals
Claim: is false. Proof: Since this is a universal conditional statement, to prove that it is false, it's enough to find one counterexample. That is, we need a specific value of n which is a positive integer and for which n2 +1<2n. Consider ….
A. n = 0 B. n = 1 C. n = 4 D. n = 5 E. None of the above
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Universal conditionals
Claim: is false. Proof: Since this is a universal conditional statement, it's enough to find one counterexample. That is, we need a specific value of n which is a positive integer and for which n2 +1<2n. Consider n = 5. Then, for this example, the LHS of the inequality evaluates to 52+1 = 26 and the RHS evaluates to 25=32. Therefore, LHS < RHS, contradicting the statement. Thus, there is (at least) one counterexample to the universal statement and hence the universal statement is false.
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Universal conditionals
Claim: is true. Proof: WTS Let n be a positive integer (a general element of the domain). For a direct proof, assume
A. n ≥ 1 B. n2 + 1 ≥ 2n C. n ≤ 4 D. n2 + 1 < 2n E. None of the above
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Universal conditionals
Claim: is true. Proof: WTS Let n be a positive integer (a general element of the domain). For a direct proof, assume that 1 ≤ n ≤ 4. Goal : WTS that n2 + 1 ≥ 2n
How can we use the hypothesis?
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Universal conditionals
Claim: is true. Proof: WTS Let n be a positive integer (a general element of the domain). For a direct proof, assume that 1 ≤ n ≤ 4. Goal : WTS that n2 + 1 ≥ 2n
Since we assume that n is an integer between 1 and 4, there are only four possible cases. We check that the conclusion is true in each one.
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Universal conditionals
Claim: is true. Proof: WTS Let n be a positive integer (a general element of the domain). For a direct proof, assume that 1 <= n <= 4. Goal : WTS that n2 + 1 >= 2n
Case 1: Assume n=1. Then LHS is 12+1 = 2; RHS is 21=2 so LHS >=RHS. J Case 2: Assume n=2. Then LHS is 22+1 = 5; RHS is 22=4 so LHS >=RHS. J Case 3: Assume n=3. Then LHS is 32+1 = 10; RHS is 23=8 so LHS >=RHS. J Case 4: Assume n=4. Then LHS is 42+1 = 17; RHS is 24=16 so LHS >=RHS. J
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Universal conditionals
Claim: is true. Proof: WTS Let n be a positive integer (a general element of the domain). For a direct proof, assume that 1 <= n <= 4. Goal : WTS that n2 + 1 >= 2n
Case 1: Assume n=1. Then LHS is 12+1 = 2; RHS is 21=2 so LHS >=RHS. J Case 2: Assume n=2. Then LHS is 22+1 = 5; RHS is 22=4 so LHS >=RHS. J Case 3: Assume n=3. Then LHS is 32+1 = 10; RHS is 23=8 so LHS >=RHS. J Case 4: Assume n=4. Then LHS is 42+1 = 17; RHS is 24=16 so LHS >=RHS. J
Exhaustive proof (cf. page 93)
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Reminder: perfect squares Rosen p. 83 An integer a is a perfect square iff there is some integer b such that
a = b2
Which of the following is not a perfect square? A. 0 B. 1 C. 2 D. 4 E. More than one of the above
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Reminder: perfect squares Rosen p. 83 An integer a is a perfect square iff there is some integer b such that
a = b2
Which of the following is not true? A. If n is a perfect square, so is n2. B. If n2 is a perfect square, so is n. C. If n2 is not a perfect square, neither is n. D. For each n, n2 is a perfect square. E. More than one of the above
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Reminder: perfect squares Rosen p. 83 Theorem: For integer k>1, then 2k-1 is not a perfect square. Proof: Let k be an integer.
What would we assume in a direct proof? A. k > 1 B. k <= 1 C. 2k-1 is not a perfect square D. 2k-1 is a perfect square E. More than one of the above
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Reminder: perfect squares Rosen p. 83 Theorem: For integer k>1, then 2k-1 is not a perfect square. Proof: Let k be an integer.
What would we assume in a proof by contraposition? A. k > 1 B. k <= 1 C. 2k-1 is not a perfect square D. 2k-1 is a perfect square E. More than one of the above
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Reminder: perfect squares Rosen p. 83 Theorem: For integer k>1, then 2k-1 is not a perfect square. Proof: ???
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Tangent, for a moment
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Reminder: evens and odds An integer a is even iff there is some integer b such that
a = 2b
Which of the following is equivalent to definition of a being even? A. a/2 B. a div 2 is an integer. C. a mod 2 is zero. D. 2a is an integer. E. More than one of the above.
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Reminder: divisibility Rosen p. 238-239 For integers a, b with a nonzero means
, a.k.a. "a divides b" "b is an integer multiple of a"
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Reminder: evens and odds An integer a is even if there is some integer b such that
a = 2b
Which of the following numbers is not even? A. 0, even though 0 = 2(0) B. 0.5, even though 0.5 = 2 (0.25) C. -2, even though -2 = 2 (-1) D. I don't know.
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Reminder: evens and odds An integer a is even if there is some integer b such that
a = 2b. An integer a is odd iff • it's not even or equivalently • there is some integer b such that
a = 2b + 1.
Why equivalent?
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Flexing proof muscles Theorem: If n is even, then so is n2. Proof:
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Flexing proof muscles Theorem: If n is odd, then so is n2. Proof:
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Flexing proof muscles Theorem: If n2 is even, then so is n. Proof:
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Flexing proof muscles Theorem: If n2 is odd, then so is n. Proof:
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Back to: perfect squares Rosen p. 83 Theorem: For integers k>1, then 2k-1 is not a perfect square. Proof: ???
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Back to: perfect squares Rosen p. 83 Theorem: For integers k>1, then 2k-1 is not a perfect square. Proof: Try proof by contraposition again… Let k be an integer. Assume: 2k-1 is a perfect square
Note: k must be nonnegative because perfect squares are integers. WTS: k <=1.
Note: using above, what we want to show is that k=0 or 1.
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Back to: perfect squares Rosen p. 83 Theorem: For integers k>1, then 2k-1 is not a perfect square. Proof: Try proof by contraposition again… Let k be an integer. Assume: 2k-1 is a perfect square
Note: k must be nonnegative because perfect squares are integers. By definition of perfect square, there is c such that 2k-1 = c2.
Either k=0 (i.e. c=0) or 2k-1 is a positive integer. Is this integer even or odd?
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Overall strategy
• Do you believe the statement? • Try some small examples.
• Determine logical structure + main connective. • Determine relevant definitions. • Map out possible proof strategies.
• For each strategy: what can we assume, what is the goal? • Start with simplest, move to more complicated if/when get stuck.
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Proof by contradiction Rosen p. 86
Idea: To prove P, instead, we prove that the conditional is true. But, the only way for a conditional to be true if its conclusion is false is for its hypothesis to be false too. Conclude: is false, i.e. P is true!
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Next up • Using proofs in the context of sets