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3.1N_Proving_Universal_Statements

3.1N_Proving_Universal_Statements - Math 280 Strategies of...

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Math 280 Strategies of Proof Fall 2007 Class Notes 3.1 Proving Universal Statements The method of proof used to prove a mathematical statement is determined by the statement’s logical structure. We can determine the logical structure by writing the statement in symbolic form, so this is the first step in the proof of any mathematical statement. Almost all of the statements which you’ll be concerned with proving in your mathematics courses will involve one or more quantifiers. So we want to begin our examination of proof techniques by discussing how to prove the general universal statement and how to prove the general existential statement. The overwhelming majority of mathematical theorem statements are universally quantified statements of the general form x p ( x ), where the propositional function p ( x ) may be either simple or compound. We may also have more than one variable involved. The universal statement x p ( x ) is true provided that the propositional function p ( x ) is true for all elements x in the domain D . So to prove the statement x p ( x ) is true, we need to show that, for each element x in the domain D , the propositional function p ( x ) is true. There are two basic approaches to proving a universal statement: proof by exhaustion and proof by arbitrary element. Proof by Exhaustion When the domain D is a finite set, we can prove that x p ( x ) is true by systematically checking that p ( x ) is true for each x D . This method of proof is called proof by exhaustion . Here are two examples of proofs by exhaustion. Example. Prove: For all positive integers n with 1 n 10, n 2 n + 11 is a prime number. Solution. Define the propositional function p ( n ) : n 2 n + 11 is a prime number. The statement has symbolic form n p ( n ) , where the domain for n is the set D = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 } . Proof. For a proof by exhaustion of n p ( n ), we need to show that p ( n ) is true for each value of n in the domain D . Substituting each value of n into the propositional function, we have p (1) : 1 2 1 + 11 = 11 is a prime number. p (2) : 2 2 2 + 11 = 13 is a prime number. p (3) : 3 2 3 + 11 = 17 is a prime number. p (4) : 4 2 4 + 11 = 23 is a prime number. p (5) : 5 2 5 + 11 = 31 is a prime number. p (6) : 6 2 6 + 11 = 41 is a prime number. p (7) : 7 2 7 + 11 = 53 is a prime number. p (8) : 8 2 8 + 11 = 67 is a prime number. p (9) : 9 2 9 + 11 = 83 is a prime number. p (10) : 10 2 10 + 11 = 101 is a prime number. Since each of these statements is true. then n p ( n ) is true, which proves this statement. 55
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56 Chapter 3 Proof Techniques Example. Let α be an irrational number. A rational number a 0 /b 0 is called a good approximation to α provided that for every rational number a/b with 1 b < b 0 , Ø Ø Ø Ø α a 0 b 0 Ø Ø Ø Ø < Ø Ø Ø Ø α a b Ø Ø Ø Ø . This condition means a 0 /b 0 is closer to α than every other rational number with a smaller denom- inator. We assume all rational numbers are in lowest terms.
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