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Direct Proof and Counterexample II Lecture 12 Section 3.2 Thu, Feb 9, 2006.

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1 Direct Proof and Counterexample II Lecture 12 Section 3.2 Thu, Feb 9, 2006

2 Rational Numbers A rational number is a number that equals the quotient of two integers. Let Q denote the set of rational numbers. An irrational number is a number that is not rational. We will assume that there exist irrational numbers.

3 Direct Proof Theorem: The sum of two rational numbers is rational. Proof: Let r and s be rational numbers. Let r = a/b and s = c/d, where a, b, c, d are integers, where b, d > 0. Then r + s = (ad + bc)/bd.

4 Direct Proof We know that ad + bc is an integer. We know that bd is an integer. We also know that bd  0. Therefore, r + s is a rational number.

5 Proof by Counterexample Disprove: The sum of two irrationals is irrational. Counterexample:

6 Proof by Counterexample Disprove: The sum of two irrationals is irrational. Counterexample: Let α be irrational. Then -α is irrational. (proof?) α + (-α) = 0, which is rational.

7 Direct Proof Theorem: The sum of two odd integers is an even integer; the product of two odd integers is an odd integer. Proof:

8 Direct Proof Theorem: The sum of two odd integers is an even integer; the product of two odd integers is an odd integer. Proof: Let a and b be odd integers. Then a = 2s + 1 and b = 2t + 1 for some integers s and t.

9 Direct Proof Then a + b = (2s + 1) + (2t + 1) = 2(s + t + 1). Therefore, a + b is an even integer. Finish the proof.

10 Direct Proof Theorem: Between every two distinct rationals, there is a rational. Proof: Let r, s  Q. WOLOG *, WMA † r < s. Let t = (r + s)/2. Then t  Q. (proof?) * WOLOG = Without loss of generality † WMA = We may assume

11 Proof, continued We must show that r < t < s. Since r < s, it follows that 2r < r + s < 2s. Then divide by 2 to get r < (r + s)/2 < s. Therefore, r < t < s.

12 Other Theorems Theorem: Between every two distinct irrationals there is a rational. Proof: Difficult. Theorem: Between every two distinct irrationals there is an irrational. Proof: Difficult.

13 An Interesting Question Why are the last two theorems so hard to prove? Because they involve “negative” hypotheses and “negative” conclusions.

14 Positive and Negative Statements A positive statement asserts the existence of a number. A negative statement asserts the nonexistence of a number. It is much easier to use a positive hypothesis than a negative hypothesis. It is much easier to prove a positive conclusion than a negative conclusion.

15 Positive and Negative Statements “r is rational” is a positive statement. It asserts the existence of integers a and b such that r = a/b. “α is irrational” is a negative statement. It asserts the nonexistence of integers a and b such that α = a/b.

16 Positive and Negative Statements Is there a “positive” characterization of irrational numbers?

17 Irrational Numbers Theorem: Let  be a real number  and define the two sets A = iPart({1, 2, 3, …}  (  + 1)) and B = iPart({1, 2, 3, …}  (  -1 + 1)). Then  is irrational if and only if A  B = N and A  B = . Try it out: Irrational.exeIrrational.exe


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