Outline Goals and Assumptions Real Numbers Rational and Irrational. L11: Numbers. Alice E. Fischer

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1 L11: Numbers Alice E. Fischer CSCI 1166 Discrete Mathematics for Computing March 5-8, 2018

2 1 Goals and Assumptions 2 Real Numbers 3 Rational and Irrational

3 Assumptions We rely the following assumptions: The student is familiar with the laws of basic Algebra. They are listed in Appendix A of the textbook. numbers, equality is reflexive, symmetric, and transitive. x = x, x = y y = x if x = y and y = z then x = z. numbers, the binary relations <,,, and > are defined. numbers, the binary relations +,, and are defined and operator precedence is defined in the usual way. numbers, division by any number except 0 is defined.

4 Closure Definition: A set of numbers is closed under an operation if the result of the operation is always a member of the original set. Z (the integers) are closed under addition, subtraction, and multiplication. The integers are NOT closed under division. Division by 0 is not defined at all, and often the result of a division is not an integer. R (the reals) is closed under addition, subtraction, and multiplication. R is NOT closed under division because you cannot divide by 0. However, x, y R, y 0 x/y R R is NOT closed under exponentiation because R 1/2 (the square root) is not a real number.

5 Lemmas Definition: A lemma is a small, easily proven proposition that is defined, proven, and used to simplify a longer proof.

6 The Continuum Real numbers correspond to points on an infinitely long number line, where the points corresponding to integers are equally spaced Between every two real numbers, there is another real. Any real number can be described by a possibly infinite decimal representation such as The reals are uncountable, that is, there are infinitely more reals than integers.

7 More Kinds of Numbers The rational numbers (Q) include all the integers and any number that can be expressed as a quotient of two integers. The real numbers (R) include all the rational numbers. Many reals cannot be expressed as fractions and are called irrational numbers. Example: the square root of 2. Some irrationals are algebraic numbers: the roots of some non-zero polynomial function with rational coefficients. The remaining irrational numbers are called transcendental numbers. There are many, but the ones we use are π and e.

8 Floor and Ceiling An integer can be derived from a real in four ways: The floor of a real number r is the unique integer, n, such that n r < n + 1 (closer to ). The ceiling of a real number r is the unique integer, n, such that n 1 < r n ( closer to + )). The result when you truncate a real number, r, is the integer, n, formed by removing the fractional part of r. When you round a real number, r, the result is the nearest integer, n. When rounding a.5 fraction, round toward +.

9 Some examples Floor Ceiling Truncate Round 5.7 = = 6 trunc(5.7) = 5 rnd(5.7) = = = 3 trunc(2.5) = 2 rnd(2.5) = = = 1 trunc(0.3) = 0 rnd(0.3) = = = 0 trunc( 0.3) = 0 rnd( 0.3) = = = 2 trunc( 2.5) = 2 rnd( 2.5) = = = 5 trunc( 5.7) = 5 rnd( 5.7) = 6

10 Theorem 1: the Floor of a Sum x R and m Z, x + m = x + m 1 Define n = x. 2 n Z and n x < n + 1. (Definition of floor.) 3 (n + m) (x + m) < (n + m + 1). (Add m to all parts.) 4 n + m is an integer. (The integers are closed under addition.) 5 Line 3 defines n + m as the x + m. 6 n + m = x + m (Add m to both sides of line 1.). 7 x + m = x + m. (Lines 5, 6, and transitivity of =.)

11 Something that is not true Disprove: x, y R, x + y = x + y Find a counterexample: 1 Let x = 3.6 and let y = 2.5 (Select numbers that will work.) 2 Then x = 3, and y = 2 (Definition.) 3 Then x + y = = 5 (Arithmetic.) 4 Also, x + y = 6.1 (Arithmetic.) 5 x + y = 6 (Definition.) 6 In this example, the sum of the floors of x and y(line3) the floor of their sum (line 5). 7 The statement is not universally true, it is false. 8 Why? the sum of the fractional parts of x and y can be greater than 1, while the fractional part of (x + y) cannot.

12 Getting floored Camp Daniel is a military base with 1,370 soldiers. On July 4, the soldiers all have the opportunity to take a bus to town for the evening. Each bus holds 40 people. If everyone goes, how many busses are needed? 1370/40 = = 35 The base commander announced that he would only send full busses, and the last several people to arrive at the bus stop would have to stay on the base. How many soldiers were grounded? (They were floored!) 1370/40 = = 34 busses went to town * 40 = = 10 soldiers were grounded.

13 Absolute Value Definition: x R, the absolute value of x, x, is the unique non-negative number y such that y = x if x 0 and y = x if x < 0 f(x)= x -2 2 f(x) = x

14 Lemma 1 and a proof by diagram. Lemma 1. x = x a. Choose an arbitrary non-negative number and call it x. b. Diagram x and x as blue points on the number line. c. The vertical dashed lines at x and x intersect the graph of x at the orange points. d. The orange points are the same distance above the axis. e. The proof proceeds identically if we start with a number < 0. -x x -x x

15 Lemma 2 and a proof by division into cases. Lemma 2. x x x Case 1: x 0 a. x 0 so x = x (By definition) b. So x 0 (Definition of negation) c. And x x (< is transitive.) x x x (Lines c, a.) Case 2: x < 0 a. x > 0 (By definition) b. So x < x (Case 2 and line a) c. x = x (By definition) d. x = ( x) = x (Negate both sides, double negative law) e. x = x x (Lines d, b) x x x

16 Theorem 2: Absolute Value Satisfies the Triangle Inequality Theorem 2: x + y x + y Proof by division into cases: Case 1: x + y 0. 1 x + y = x + y (Definition) 2 x x and y y (Lemma 2) 3 So x + y = x + y x + y (Algebra) Case 2: x + y < 0. 1 x + y = (x + y) (Definition) 2 (x + y) = ( x) + ( y) (Algebra) 3 x x = x and y y = y (Def., Lemma 1) 4 So x + y = ( x) + ( y) x + y (Lines 1 3, Algebra)

17 Heron s algorithm for approximating square root. This algorithm is named after Heron of Alexandria, who first described it 2000 years ago. Each guess will be either too large or too small, but the guesses get closer to the true square root on each iteration. To find the square root of a real number, n, 1 Start with an arbitrary guess, g 0, in the range 1 n. The closer to the square root of n, the fewer the iterations that will be needed to achieve the desired precision. On iteration k, 2 Let g k+1 = (g k + n/g k )/2 3 Repeat from step 2, until the accuracy of g k+1 is adequate.

18 Practice with Heron s algorithm for square root. Calculate the square root of 391 to 4 decimal places of precision. n Guess Quotient Average So the square root of 391 is approximately

19 More Practice with Heron s algorithm for square root. What happens if you don t start with a good guess? It takes more iterations. Calculate the square root of 391 to 4 decimal places of precision. n Guess Quotient Average

20 Rational and Irrational Numbers

21 Rational and Irrational A real number is rational if and only if it can be expressed as a quotient of two integers with a nonzero denominator. r is rational a, b Z, r = a/b and b 0. A real number that is not rational is irrational.

22 Theorem 3. Every Integer is rational. Suppose you are given an arbitrary integer, n. (Premise) 1 is a non-zero integer. (Definitions of 0, 1, and integer.) n/1 is a rational number. (Definition of rational number.)

23 Practice: Rational and Irrational is -10/7 rational? Why or why not? Is zero rational? Is 2/0 rational? is rational? Is π rational? (No, but we can approximate it with a rational number.)

24 Practice: Infinite Repeating decimals Is rational? Yes! Let z = Then 10z = (10z - z) = 9z = ( ) 9z = 8 z = 8/9 This technique can be used (with the appropriate power of 10) to find the rational number that corresponds to any repeating decimal.

25 More Practice: Infinite Repeating decimals Is rational? Yes! Let z = Then 100z = (100z - z) = 99z = ( ) 99z = 12 z = 12/99

26 Theorem 4: The Product of two Rational Numbers Theorem 4: If a and b are rational, then a b is rational. a = n1 d1 So n2 and b = d2, where n1, d1, n2, and d2 are all integers, ab = n1 n2 d1 d2 We know that the result of multiplying two integers is an integer. (The integers are closed under multiplication.) so for some integers n3, d3 ab is rational. ab = n3 d3

27 Theorem 5: The sum of two rational numbers is rational. a = n1 d1 So n2 and b = d2, where n1, d1, n2, and d2 are all integers, a + b = n1 d1 + n2 d2 (n1 d2) + (n2 d1) = d1 d2 We know that the result of adding or multiplying two integers is an integer. (The integers are closed under addition and multiplication) a + b is rational.

28 Theorem 5, proved more formally. This is a more formal proof of the same theorem as the prior slide. x, y Q, x + y Q a, b Z b 0 x = a/b (Definition of rational number.) c, d Z d 0 y = c/d (Definition of rational number.) x + y = a/b + c/d (Substitute = s.) x + y = (ad + cb)/bd (Arithmetic.) bd is an integer (Definition of integer multiplication.) bd 0 (Zero product property.) x + y is rational. (Definition of rational)

29 Theorem 6: 2 is Irrational The number, 2, is irrational. The proof of this theorem depends on Theorem 7 from Lecture 12. That theorem could be proved first, since it does not depend on this one. However, the theorems are presented in this order so that facts about irrational numbers and facts about integers are not mixed together. Proof by contradiction. 1 Suppose that 2 is rational. 2 Then, by definition, 2 = a/b, for some integers a and b 0, 3 If let a and b have any common factors, reduce the fraction until they don t.

30 Proof of Theorem 6: 2 is Irrational 4 Now 2 = 2 2 = (a/b) (a/b) = a 2 /b 2. 5 So 2 = a 2 /b 2. (Line 4 and Transitivity of =) 6 So a 2 = 2b 2, which means that a 2 is even since b 2 is an integer. (Closure. of *) 7 From Lecture 12, theorem 7, a 2 is even a is even. So a = 2k, for some integer, k. 8 Now a 2 = (2k) 2 = 4k 2 and a 2 = 2b 2. (Algebra and line 6) 9 So 2b 2 = 4k 2. (Line 8 and Transitivity of =) 10 And b 2 = 2k 2 and b 2 is even. (Algebra, definition of even) 11 So b is even. (Lecture 12, theorem 7) 12 Finally, both a and b are even, which means they have a common factor of But Line 3 says that a and b have no common factor. 14 This contradiction means the original theorem is true.

31 Rational/Irrational sums The sum of a rational number and an irrational number is irrational. Suppose that there is a rational number, r, and an irrational number, s, such that the sum of the two is rational. By definition, r = a/b, for integers a and b, b 0. By definition, r + s = c/d, for integers c and d, d 0. Now, r + s = a/b + s = c/d. So s = c/d a/b = (bc ad)/bd. Since bc ad and bd are integers via closure, and bd 0 by the zero product property, s is by definition rational. But we stated that s was irrational. This contradiction means the original theorem is true.

32 Irrational Sums If a and b are both irrational and a b, is a + b irrational? No, the sum can be rational or irrational. Examples: Irrational: 2 is irrational = 2 2, which is irrational. Rational: 2 and 1 2 are irrational but their sum is 1, a rational number.

33 Rational/Irrational products The product of any non-zero rational number and an irrational number is irrational. Try proving this yourself.

34 Rational/Irrational products The product of any non-zero rational number and an irrational number is irrational. Try proving this yourself. Suppose that there is a rational number, r, and an irrational number, s, such that the product of the two is rational. By definition, r = a/b, for integers a and b, b 0 and a 0 (non-zero). By definition, r s = c/d, for integers c and d, d 0. Now, r s = a/b s = c/d. So s = c/d b/a = bc/ad. Since bc ad and ad are integers via closure, and ad 0 by the zero product property, s is by definition rational. But we assumed that s was irrational. This contradiction means the original theorem is true.

35 Irrational products Is the product of any two irrational numbers going to be irrational? Try to find a counterexample yourself.

36 Irrational products Is the product of any two irrational numbers going to be irrational? Try to find a counterexample yourself. Give a counterexample to disprove the theorem. Let a = 2 be an irrational number. Let b = a. b is irrational, since b = 1 a, the product of a rational number and an irrational number. Now a b = a ( a) = 2 ( 2) = 2. But -2 is a rational number, therefore it is not true that the product of any two irrational numbers is irrational.

MA 180 Lecture. Chapter 0. College Algebra and Calculus by Larson/Hodgkins. Fundamental Concepts of Algebra

MA 180 Lecture. Chapter 0. College Algebra and Calculus by Larson/Hodgkins. Fundamental Concepts of Algebra 0.) Real Numbers: Order and Absolute Value Definitions: Set: is a collection of objections in mathematics Real Numbers: set of numbers used in arithmetic MA 80 Lecture Chapter 0 College Algebra and Calculus