Fundamentals of Mathematics (MATH 1510)

Instructor: Lili Shen Email: [email protected]

Department of Mathematics and Statistics York University

September 11, 2015 About the course

Name: Fundamentals of Mathematics, MATH 1510, Section A. Time and location: Monday 12:30-13:30 CB 121, Wednesday and Friday 12:30-13:30 LSB 106. Please check Moodle regularly for course information, announcements and lecture notes: http://moodle.yorku.ca/moodle/course/view. php?id=55333 Be sure to read Outline of the Course carefully. Textbook

James Stewart, Lothar Redlin, Saleem Watson. Precalculus: Mathematics for Calculus, 7th Edition. Cengage Learning, 2015. Towards real numbers

In mathematics, a set is a collection of distinct objects. We write a ∈ A to denote that a is an element of a set A.

An empty set is a set with no objects, denoted by ∅.

Starting from the empty set ∅, we will construct the set N of natural numbers: 0, 1, 2, 3, 4,... , the set Z of integers: ..., −4, −3, −2, −1, 0, 1, 2, 3, 4,... , the set Q rational numbers: all numbers that can be m expressed as the quotient of integers m, n, where n 6= 0, n and ﬁnally, the set R of real numbers. Natural numbers

Natural numbers are deﬁned recursively as: Deﬁnition (Natural numbers) 0 = ∅, n + 1 = n ∪ {n}. Natural numbers

In details, natural numbers are constructed as follows: 0 = ∅, 1 = 0 ∪ {0} = {0} = {∅}, 2 = 1 ∪ {1} = {0, 1} = {∅, {∅}}, 3 = 2 ∪ {2} = {0, 1, 2} = {∅, {∅}, {∅, {∅}}}, ...... n + 1 = n ∪ {n} = {0, 1, 2,..., n}, ...... Integers and rational numbers

Four basic operations in arithmetic: + − × ÷

Natural numbers

Subtraction −

Integers

Division ÷

Rational numbers Integers and rational numbers

Theorem Integers are closed with respect to addition, subtraction and multiplication. Rational numbers are closed with respect to addition, subtraction, multiplication and division.

In the terminologies of modern algebra, the set Z of integers is a ring, while the set Q of rational numbers is a ﬁeld. Rational numbers as decimals

The decimal representation of a rational number either terminates or becomes periodic after a ﬁnite number of digits: Terminating decimals:

5 = 0.625. 8 Repeating decimals (or recurring decimals):

8 = 1.142857142857 ··· = 1.142857, 7 13 = 0.2363636 ··· = 0.236. 55 Rational numbers as decimals

Conversely, every repeating or terminating decimal represents a rational number.

Interested students may visit here for the methods of converting repeating decimals to fractions (see Section 5): http://en.wikipedia.org/wiki/Repeating_decimal Number line

Draw a horizontal line continuing indeﬁnitely in each direction, ﬁx a point O as the origin and a distance to O as the unit length:

−3 −2 −1 O 1 2 3

This line is known as the number line. Number line and rational numbers

Question Is there a one-to-one correspondence between rational numbers and points of the numbers line? In other words, can every point in the number line be represented by a rational number? Number line and rational numbers

The answer is negative. √ For example, the length of the diagonal of a unit square, 2, is not a rational number.

√ O 1 2 √ 2 is not a rational number

Theorem √ 2 is not a rational number.

Proof. √ m Suppose that 2 = is an irreducible fraction, i.e., m, n are n integers with no common divisors greater than 1. Then m2 = 2n2 and consequently m2 is an even number, thus m is even. Since m, n have no common divisors greater than 1, n must be an odd number. But m being even means that m = 2p for some integer p, which leads to m2 = 4p2 = 2n2, and it follows that n2 = 2p2 is even, thus n is even, a contradiction. From rational numbers to real numbers

That is to say, although rational numbers are dense in the number line, there are numerous gaps between rational numbers.

In order to ﬁll the whole number line, we need to construct and understand real numbers. Cuts of rational numbers

Suppose A and B are non-empty sets of rational numbers. If they satisfy a < b for all a ∈ A and b ∈ B, A ∪ B = Q, We say that A and B form a cut of rational numbers.

Note that the above conditions guarantee that every rational number is either in A or B, but cannot be in both A and B. Cuts of rational numbers

Logically, there are four cases for cuts of rational numbers: (1) A has a greatest element, and B has no least element. (2) A has no greatest element, and B has a least element. (3) A has no greatest element, and B has no least element. (4) A has a greatest element, and B has a least element. Cuts of rational numbers

In fact, the fourth case cannot happen. If A has a greatest element a0 and B has a least element b0, the deﬁnition of cuts shows a0 < b0, therefore a + b a < 0 0 < b . 0 2 0 a + b That is, the rational number 0 0 is neither in A nor in B, 2 contradicting to the hypothesis A ∪ B = Q.

Furthermore, we will combine the ﬁrst two cases as they make no difference in the construction of real numbers. Dedekind cuts

Deﬁnition A Dedekind cut, written as A/B, is a partition of the rational numbers into two non-empty sets A and B, such that a < b for all a ∈ A and b ∈ B, A ∪ B = Q, A contains no greatest element. Dedekind cuts

Now we have two cases of Dedekind cuts:

(1) A has no greatest element, and B has a least element b0, e.g.

A = {a ∈ Q | a < 0}, B = {b ∈ Q | b ≥ 0}.

(2) A has no greatest element, and B has no least element, e.g.

2 A = {a ∈ Q | a < 2 or a < 0}, 2 B = {b ∈ Q | b > 2 and b > 0}. Irrational numbers

For the ﬁrst case, we say that the Dedekind cut A/B determines a rational number b0.

For the second case, the Dedekind cut A/B does not determine any rational number; that is, there is a gap between A and B. Therefore, we must introduce a new number, i.e., an irrational number, to ﬁll this gap. In√ the above example, the irrational number ﬁlling the gap is 2. Irrational numbers

Deﬁnition Let A/B be a Dedekind cut of rational numbers. If A has no greatest element and B has no least element, we say that A/B determines an irrational number c. c is greater than any rational number in A and less than any rational number in B. Irrational numbers

√ For example, the irrational number 2 determined by the Dedekind cut

2 A = {a ∈ Q | a < 2 or a < 0}, 2 B = {b ∈ Q | b > 2 and b > 0}

is greater than any rational number in A and less than any rational number in B.

The decimal representation of an irrational number neither terminates nor inﬁnitely repeats but extends forever without regular repetition. For example, √ 2 = 1.41421356 ..., π = 3.14159265 .... Real numbers

Deﬁnition The set R of real numbers consists of all the rational numbers and all the irrational numbers determined by Dedekind cuts of rational numbers. Cuts of real numbers

We may deﬁne cuts of real numbers in a similar way to Dedekind cuts of rational numbers: Deﬁnition A cut A/B of real numbers consists of two non-empty sets A and B, such that a < b for all a ∈ A and b ∈ B, A ∪ B = R, A contains no greatest element. Cuts of real numbers

Theorem Let A/B be a cut of real numbers. Then B must contain a least element.

This theorem states that there is no gap in the set of real numbers. Equivalently speaking, every point in the number line can be represented by a real number. Therefore, the number line is usually called the real line. 0.9 = 1?

Question Is the repeating decimal 0.9 = 0.999... equal to 1? 0.9 = 1?

An elementary solution. Let a = 0.999..., then 10a = 9.999.... Thus 9a = 10a − a = 9.999... − 0.999... = 9, and therefore a = 1. 0.9 = 1?

A rigorous proof. Let x = 0.9, form two Dedekind cuts of rational numbers:

A1 = {a ∈ Q | a < x}, B1 = {b ∈ Q | b ≥ x}; A2 = {a ∈ Q | a < 1}, B2 = {b ∈ Q | b ≥ 1}. In order to prove x = 1, it sufﬁces to show that A1/B1 and A2/B2 are the same Dedekind cuts; or equivalently, A1 = A2.

First, if a rational number a ∈ A1, then a < x, and it is clear that a < 1, thus a ∈ A2. This means A1 ⊆ A2. 0.9 = 1?

Second, if a rational number a ∈ A2, then a < 1. Suppose m a > 0 and a = , where m, n are positive integers. Then n m < n, and it follows that m 1 1 − ≥ > 0 n n 1 1 =⇒ There exists a positive integer k such that ≥ > 0 n 10k m 1 1 =⇒ a = ≤ 1 − ≤ 1 − = 0. 99 ... 9 < x. n n 10k | {z } k

That is, a < x, i.e., a ∈ A1, and consequently A2 ⊆ A1.

Therefore A1 = A2, completing the proof.