Web Homework College Algebra

Chapter 2: The Real Numbers

2.1 Introduction

Although you have doubtless worked quite a bit with the real numbers, this chapter will start at the beginning introducing you to them again from a perhaps higher viewpoint than that you have seen in the past. This will serve to put your knowledge in a proper mathematical framework so that you can better understand some of the topics of the last chapter. Here are some of the topics:

  1. We begin by defining a field; it is basically a set like the rational numbers or the real numbers where one can operate in the usual way with the four basic arithmetic operations.
  2. We then add in the idea of an order. This allows us to work with inequalities in the usual manner. The Real and rational numbers both have order relations.
  3. In order to distinguish between, say, the rationals and the reals, one needs one further property. This amounts to assuming that infinite decimals correspond to numbers. This will complete our definition of the real numbers.
  4. The set of integers is a particularly important set of numbers. In fact, we could spend the entire semester learning about them; for our purposes, we need to know that integers factor uniquely into products of primes. This result is called the Fundamental Theorem of Arithmetic and it is well worth understanding why it is true.
  5. The real numbers are not adequate to guarantee that every polynomial has a root. To get this property, one needs the field of complex numbers. The basics of complex numbers will be introduced.

2.2 Sets and Functions

A set is simply a collection of objects. We could define axioms for set theory, but, instead, we choose to depend on your intuitive understanding of sets. What can you do with a set? Basically, you can check to see if an object is in the set (we then say that the object is an element of the set). You can also define subsets of the set; a subset is simply a collection of objects, every one of which is in the original set. Every set contains the empty subset (the collection with no elements in it), and the original set itself (the collection of all elements of the original set). Furthermore, we assume you are familiar with basic operations like the union and intersection of subsets of a set.

Given two sets S and T we can define the cartesian product images/real1.png of S and T as the set whose elements are all the ordered pairs (s, t) where s is in S and t is in T. For example, in the last chapter, we defined took the cartesian product of the set of real numbers with itself as the set of points in the plane.

By a relation between the sets S and T, we mean simply any subset of the cartesian product images/real2.png .

This may seem like a strange definition. But it is really what we mean by a relation. For example, the property of two real numbers a and b that a be less than b is a relation. It is a relation between the set of real numbers and itself, and it is completely determined by specifying the set of all pairs of numbers (a, b) such that images/real3.png . As another example, we have the relation of b being the mother of b. Let S be the set of all people and T be the set of all women. Then this relation is simply the set of all pairs (s, t) in images/real4.png where t is the mother of s.

A function from the set S to the set T is simply a relation images/real5.png with the property that for every s in S there is exactly one element in R with first coordinate s. The set S is called the domain of the function and set T is called the range space of the function. We often indicate that R is a function with domain S and range space T by using the notation: images/real6.png . The set of all elements t in T for which there is at least one pair in R with second coordinate t is called the range of the function. We write R(a) = b if (a, b) is in R.

Example 1:

  1. Let N be the set of positive integers. The successor function images/real7.png is the function with s(k) = k + 1. The function is the set of all pairs (k, k + 1) where k is a positive integer. The domain of s is N, its range space is N, but it range is the set of integers greater than or equal to 2.
  2. Let R be the set of real numbers. The set of all pairs images/real8.png where x is in R is a function. We normally write it as images/real9.png .
  3. Let R be the set of real numbers. The set of all pairs images/real10.png where x is in R is not a function from R to R. This is because there is no pair with first coordinate a negative number. You could consider it to be a function of from the set of non-negative real numbers to R.
  4. Let S be the set of non-negative real numbers. The set of all pairs (a, b) where images/real11.png is not a function. For example, (4, 2) and (4, -2) are two pairs in the set with the same first coordinate.

Warning: Many textbooks define a function f from S to T as a rule which assigns to each element s in S exactly one element t in T. They then say that the set of all points (s, f(s)) is called the graph of the function. Although this approach corresponds more closely to one's intuitive notion of a function, we have avoided this approach because it is not clear what one means by a rule.

In the case where the domain and range space are both the set R of real numbers, we see that the function is the set of points in what we would have called the graph of the function. The condition that there be exactly one point in the graph with an given first coordinate amounts to saying that every vertical line intersects the graph in exactly one point. For this reason, the condition is often referred to as the vertical line test.

A binary operator on a set S is a function from images/real12.png to S. A unary operator on a set S is a function from S to S.

Example 2: Addition and Multiplication on the real numbers are both binary operators. Negation is a typical unary operator. Binary operators are usually written in infix notation like 5 + 7 rather than in functional notation like +(5,7). Similarly, unary operators are usually written in prefix notation like -3 rather than functional notation like -(3).

2.3 Fields

The collection of elements out of which we will be making algebraic expressions will be referred to as a field. More precisely, a field is set endowed with two binary operators which satisfy some simple algebraic properties:

Definition 1: A field F is a consisting of a set S and two binary operators called addition and multiplication which satisfy the following the properties:

  1. (Commutativity) For every a and b in S, one has a + b = b + a and ab = ba.
  2. (Associativity) For every a, b, and c in S, one has (a + b) + c = a + (b + c) and (ab)c = a(bc).
  3. (Distributive Law) For every a, b, and c in S, one has a(b + c) = ab + ac.
  4. (Identities) There is an element 0 in S such that for all a in S, one has a + 0 = a. There is an element 1 different from 0 in S such that for all a in S, one has acdot 1 = a.
  5. (Inverses) For every a in S, there is an element denoted -a in S such that a + (-a) = 0. For every a in S other than 0, there is an element denoted images/real13.png such that images/real14.png .
We usually will denote the set S with the same symbol F as is used for the field.

Example 3: The set of rational numbers with the usual addition and multiplication operators is a field. The same is true of the set of real numbers with the usual addition and multiplication operators. The set of complex numbers (yet to be defined) will also be a field.

Example 4: Consider the set of expressions of the form p(x)/q(x) where p and q are polynomials with real coefficients and q is non-zero. We say that images/real15.png if images/real16.png (where the multiplication is the usual multiplication of polynomials). Addition and multiplication of these expressions are defined in the usual manner. One can show that these expressions form a field called the field of rational functions.

Warning: We are treating rational functions here as simply expressions, not as functions. In particular, the definition of equality does not correspond to equality of functions.

Example 5: There is basically only one way to make a field with only two elements 0 and 1. See if you can make up the appropriate addition and multiplication table and verify the field properties.

The definition of a field includes only the most basic algebraic properties of addition and multiplication. We will see, however, that all the usual rules for manipulating algebraic expressions are consequences of these basic properties. First, let use begin by noting that the definition of a field only assumes the existence of identities and inverses. In fact, it follows that they are in fact unique:

Proposition 1: If F is a field, then the identity elements 0 and 1 as well as the additive and multiplicative inverses are unique.

Proof: Suppose that 0 and 0' are additive identities. Then, since 0 is an identity, we have 0' + 0 = 0'. Similarly, since 0' is an identity, we have 0 + 0' = 0. Since addition is commutative, we conclude that 0 = 0'.

Let a be in the field. Suppose that both b and c are additive inverses of a. The a + b = 0 and a + c = 0. We can now calculate b = b + 0 = b + (a + c) = (b + a) + c = (a + b) + c = 0 + c = c + 0 = c. (Please be sure that you understand why each of the steps of this calculation are true.)

To complete the proof, you should make similar arguments for multiplicative identities and inverses.

Proposition 2: Let F be a field. If a is in F, then images/real17.png . If a and b are in F satisfy images/real18.png , then either a or b are zero.

Proof: Let a be in F. Then images/real19.png . Let b be the additive inverse of images/real20.png . Then applying it to the last equation, we get


Suppose ab = 0. If a is zero, there is nothing more to prove. On the other hand, if images/real22.png , then a has a multiplicative inverse c and so images/real23.png .

Corollary 1: If a is an element of a field F, then -a = (-1)a.

Proof: images/real24.png .

We can define the binary subtraction operator: a - b = a + (-b) and, for images/real25.png , the binary division operator images/real26.png . The division operator will also be expressed as images/real27.png .

Proposition 3: Let F be a field containing a, b, c, and d where b and d are non-zero. Then

  1. images/real28.png
  2. images/real29.png
  3. images/real30.png
  4. images/real31.png
  5. If c is also non-zero, then images/real32.png

Proof: (i) Do this as an exercise - it is a matter of simplifying images/real33.png using the commutative and associativity properties of multiplication to see that the product is equal to 1.

(ii) Using associativity and commutativity, one can show that images/real34.png . By assertion (i), this is the same as images/real35.png .

(iii) images/real36.png

(iv) One has


where some of the steps have been combined.

(v) Using property ii, it is easy to see that images/real38.png . But then assertion v follows from assertion ii. This completes the proof of the proposition.

The remaining rules in section 1.1.1 on simplifying expressions are now easy to verify, i.e. they are properties of any field. The material in section 1.1.2 and 1.1.3 on solving linear equations or systems of linear equations are also properties of fields. On the other hand, the material on solving quadratics does not hold for arbitrary fields, both because it uses the order relation of real numbers as well as the existence of square roots.

2.4 Ordered Fields

Definition 2: An ordered field F is a field (i.e. a set with addition and multiplication satisfying the conditions of Definition 1) with a binary relation < which satisfies:

  1. (Trichotomy) For every pair of elements a and b in F, exactly one of the following is true: a < b, a = b, and b < a.
  2. (Transitivity) Let a, b, c be arbitrary elements of F. If a < b and b < c, then a < c.
  3. If a, b, and c in F satisfy a < b, then a + c < b + c.
  4. If a, b, and c in F satisfy a < b and 0 < c, then ac < bc.

Fact: If F is an ordered field, then 0 < 1.

Proof: By Definition 1, images/real39.png and so by trichotomy, if the the fact were wrong, then we would have a field F with 1 < 0. By property iii, we would have 1 + (-1) < 0 + (-1) and so 0 < -1. But then using property iv, we would have images/real40.png . By Proposition 2, the left side is 0 and so images/real41.png . This contradicts trichotomy and so the assertion must be true.

If F is an ordered field, an element a in F is called positive if 0 < a.

Proposition 4: The set P of positive elements in an ordered field F satisfy:

  1. (Trichotomy) For every a in F, exactly one of the following conditions holds: a is in P, a = 0, and -a is in P.
  2. (Closure) If a and b are in P, then so are a + b and ab.

Proof: (i) By property i of the Definition 2, exactly one of a < 0, a = 0, and 0 < a must be true. If a < 0, then by property iii of Definition 2, we have a + (-a) < 0 + (-a) and so 0 < -a. Conversely, if 0 < -a, adding a to both sides gives a < 0. So the three conditions are the same as -a is in P, a = 0, and a is in P.

  • Suppose a and b are in P. Then 0 < a and by property iii of Definition 2, we have 0 + b < a + b and images/real42.png . Since 0 < b and b < a + b, transitivity implies that 0 < a + b. Since images/real43.png by Proposition 2, we have 0 < ab.

    Remarks: i. In one of the exercises, you will show that, if a field has a set P of elements which satisfy the conditions of Proposition 4, then the field is an ordered field assuming that one defines a < b if and only if b - a is in P.

    ii. An element a of an ordered field F is said to be negative if and only if a < 0.

    iii. It is convenient to use the other standard order relations. They can all be defined in terms of <. For example, we define a > b to mean b < a. Also, we define images/real44.png to mean either a < b or a = b and similarly for images/real45.png .

    iv. The absolute value function is defined in the usual way:


    v. One now has everything you need to deal with inequalities as we did back in section 1.1.4.

    Proposition 5: Let a and b be elements of an ordered field F.

    1. |-a| = |a|
    2. images/real47.png (i.e. images/real48.png and images/real49.png )
    3. (Triangle Inequality) images/real50.png

    Proof: i. By Trichotomy, we can treat three cases: a > 0, a = 0, and a < 0. If a > 0, then -a < 0 and so |a| = a and so |-a| = -(-a) = a. If a = 0, then -a = 0 and so |a| = 0 = |-a|. If a < 0, then -a > 0 and so |a| = -a and |-a| = -a. In all three cases, we have |a| = |-a|.

    ii. Again, we can treat three cases: If a > 0 or a = 0, then |a| = a and so images/real51.png . If a < 0, then adding -a to both sides gives 0 < -a and so a < -a by transitivity. In this case we have |a| = -a and so a < |a|.

    We could argue the other inequality the same way, but notice that we could also use our result replacing a with -a. (Since it holds for all a in F, it holds for -a.) The result says images/real52.png , where we have used assertion i. Adding a - |a| to both sides of the inequality gives the desired inequality.

    iii. Once again, do this by considering cases: If images/real53.png , then |a + b| = a + b. Since images/real54.png and images/real55.png , we can add b to both sides of the first inequality and |a| to both sides of the second one to get images/real56.png and images/real57.png . Using transitivity, we get images/real58.png as desired.

    Now suppose that images/real59.png . Then adding -a - b to both sides of the inequality gives -a + (-b) < 0. Applying the result of the last paragraph, we get images/real60.png . But a + b < 0 means that |a + b| = -(a + b) and so images/real61.png where we have used assertion i for the last step. This completes the proof.

    Example 6: Let a, b, c, and d be elements of an ordered field F. Then images/real62.png ( Cauchy -Schwarz Inequality).

    This is an example of how to deal with a more complicated inequality. It is not clear how to begin. In such a case, it is often useful to simply work with the result trying to transform it into something easier to prove. So, suppose the result were true. We could then expand out the expression using the distributive law to get:


    which simplifies to images/real64.png . This still looks complicated until one thinks of grouping the factors differently to get images/real65.png . Subtracting the left side from both sides of the inequality gives images/real66.png . Now, you may recognize the right side as being a perfect square: factoring we get images/real67.png . It is still complicated, but do you expect that it is true? The right hand side is the square of a complicated expression -- and the assertion is that it is non-negative. We have not yet proved this, but it is a common property of real numbers, so you might make the

    Conjecture: If a is any element of an ordered field, then images/real68.png .

    The conjecture looks like it should be simple enough to prove. In fact, you should go ahead and try an prove by considering cases as we have done before. Once it is proved, is the Cauchy-Schwartz Inequality proved? No. We had assumed that it was true and shown that it a result which would be true provided the conjecture were true. That is no proof of the inequality. But, all is not lost, the idea is that we might be able to trace back through our steps in reverse order and reach the desired inequality.

    Assuming that the conjecture is true, let's see that the Cauchy-Schwartz inequality must also be true. We know that the square of any field element is non-negative. So, applying this to the element images/real69.png , we get images/real70.png . Using the distributive law to expand this gives images/real71.png . Adding 2(ad)(bc) to both sides of the inequality yields images/real72.png . Using the commmutative and associative law several times allows us to re-arrange this into images/real73.png . Adding appropriate terms to both sides and again using associativity and commutativity takes us back to the step:


    Finally, we can factor the sides to get the Cauchy-Schwartz Inequality.

    Only later, will you see that the Cauchy-Schwartz inequality is useful. But already there is a lesson here. If we had just presented the last paragraph as a proof, you would have no idea of why one was doing each of the steps -- you would know that the inequality was true, but have no intuitive grasp of why or how one ever came up with the proof. In this case, the idea is that the proof is obtained by non-deductive means -- we simply worked backward from the result we wanted to prove until we got to an assertion we could prove. We proved the assertion and then used it to work forward to the desired result. When reading any proof you should always be asking yourself whether or not the proof is something you could have come up with yourself. If not, then you need to work more with the material until you hopefully will understand enough to be able to do it yourself.

    2.5 The Natural Numbers and Induction

    The so-called natural numbers (Are the others un-natural?) are the numbers 1, 2, 3, etc. But expressing this is a bit complicated. Assume for the whole section that we are operating within a particular ordered field F. The set S we want to describe satisfies the conditions:

    1. images/real75.png ,
    2. 1 is in S, and
    3. If a is in S, then so a + 1.
    Any such set S will be called inductive.

    Now consider the set T of all inductive sets S. For example, the set F is in T as well as the set of all positive elements of F. The set of natural numbers would appear to be contained in any of the sets in T. So, one way to define the set we want is

    Definition 3: The set images/real76.png of natural numbers is the intersection of all inductive sets, i.e. a is a natural number provided that it is an element of every set S in T.

    Proposition 6: (Induction) The set images/real77.png of natural numbers satisfy the conditions

    1. images/real78.png ,
    2. 1 is in images/real79.png , and
    3. If a is in images/real80.png , then so a + 1.
    4. The only inductive subset of images/real81.png is images/real82.png .

    Proof: i. Let a be in images/real83.png . Then a is an element of every set S in T. If S is in T, then images/real84.png an so a is an element of F.

    ii. Since 1 is in every set S which is in T, 1 is in their intersection which is, by definition, images/real85.png .

    iii. Let a be in images/real86.png . If S is in T, then a must also be in S. But then a + 1 is also in S. Since this is true for every S in T, a + 1 is in the intersection of all the S in T, i.e. a + 1 is in images/real87.png .

    iv. Suppose S is an inductive subset of images/real88.png . Then S is in T. Since images/real89.png is the intersection of all the S in T, it follows that images/real90.png is contained in S. But then we have images/real91.png and images/real92.png , which means that the two sets must be equal. This completes the proof.

    The importance of Proposition 6 is that it is the basis of a method of definition and of proof called mathematical induction which we will normally refer to simply as induction.

    First, let's see how it works for definitions. Let's do an inductive definition of powers. Let a be in F. We define images/real93.png to be a. Now, suppose we have already defined images/real94.png for some natural number k, then define images/real95.png to be the product images/real96.png . Consider the set S of all natural numbers k for which we have defined the images/real97.png . It contains 1 and if k is in S, so is k + 1. By Proposition 6, it follows that images/real98.png is defined for all natural numbers k.

    We can also use induction in proofs. Here is the general scheme: Suppose that for every natural number k, we have a property P(k). Assume furthermore that:

    1. P(1) is true.
    2. For every natural number k, if P(k) is true, then so is P(k+1).
    Then the set of natural numbers for which P(k) is true must be the set of all natural numbers since it is inductive. So, P(k) is true for every natural number k.

    Example 7: Let's prove that images/real99.png whenever m and n are natural numbers. We use induction with the property P(k) being the condition on k that for all natural numbers m, we have images/real100.png .

    1. For k = 1, we have images/real101.png by the inductive definition of the powers of a. So P(1) is true.
    2. Suppose that we know that P(k) is true for some particular k. Let m be a natural number. We have


      For the record, the justification for each of the steps in the above series of equalities is:

      1. inductive definition of powers
      2. associativity of multiplication
      3. P(k)
      4. inductive definition of powers
      5. associativity of multiplication
      So, P(k + 1) is true. By Proposition 6, it follows that the P(k) is true for all natural numbers k.

    Suppose we wanted to show the same result for all integers n which greater or equal to zero. There are two possibilities: either show the special case of n = 0 separately. Or, you could define P(k) to be the property which we were calling P(k - 1); either approach is equally valid.

    The set of integers is defined to be the set of all elements in F which are either natural numbers, 0, or whose negative is a natural number. If P(k) is defined for all integers k, then you can sometimes prove that P(k) is true for all integers k by using two induction proofs, the first showing that it is true for all non-negative integers and the second showing that it is true for all negative integers.

    Warning: From the exercises, you will see that proof by induction is an extremely powerful tool. If one is dealing with inductively defined quantities like positive integer powers of a number, then induction is both natural and leads to a good understanding. On the other hand, it is often the case that even though you can prove things by induction, you are left with the feeling that you still do not have any intuitive understanding of why the result should be true. So, whereas induction may lead to a quick and easy proof, the result can be less than fully satisfying.

    Lemma 1: images/real103.png for every natural number b.

    Proof: Let P(k) be the property that images/real104.png . Clearly P(1) is true. If for some natural number k, we have P(k) true, then images/real105.png . Since we have already seen that 1 is positive, we have 0 < 1. Adding k to each side, we get k = k + 0 < k + 1. By transitivity, it follows that images/real106.png , and so P(k+1) is true. Therefore, P(k) is true for all natural numbers k.

    Lemma 2: If k is a natural number there is no natural number m with k < m < k + 1.

    Proof: Suppose that there is a natural numbers n and m with n < m < n + 1. Let S be the set of all natural numbers except m. Then S is a subset of the set of natural numbers and 1 is in S. (If 1 were not in S, then we would have to have 1 = m because m is the only natural number not in S; but then n < m = 1 contrary to Lemma 1.) Further, if k is any natural number in S such that k + 1 is not in S, then k + 1 = m since m is the only natural number not in S. But then n < m = k + 1 < n + 1 implies n - 1 < k < n. So there is a natural number lying strictly between n - 1 and n.

    Now let P(k) be the property that there be no natural number m lying strictly between k and k + 1. Proceeding by induction, let us note that P(1) is true. If not, then letting n = 1 in the last paragraph, we see that 0 = n - 1 < m < n = 1. This contradicts Lemma 1 since m is a natural number.

    Now suppose that P(k) is true but P(k + 1) is false. So there is a natural number m with k + 1 < m < k + 2. But the result of the first paragraph of the proof with n = k + 1 then shows that there is a natural number lying strictly between k and k + 1. But this contradicts the assumption that P(k) is true. We conclude therefore that if P(k) is true, then so is P(k + 1). By induction, it follows that P(k) is true for all natural numbers k, and so Lemma 2 is proved.

    Proposition 7: (Descent) Every non-empty set S of natural numbers contains a smallest element, i.e. there is an a in S such that images/real107.png for all b in S.

    Proof: Suppose S is a non-empty set of natural numbers that does not have a smallest element. Let S' be the set of all natural numbers smaller than all elements of S. The element 1 must be in S' because otherwise 1 would be the smallest element of S by Lemma 1. Let k be any natural number in S' such that k + 1 is not in S'. Since k + 1 is not in S', there must be a natural number n in S for which images/real108.png . Now n must be greater than k since k is in S'. We cannot have k < n < k + 1 by Lemma 2 and so images/real109.png , which means that n = k + 1.

    We will show that n is the smallest element in S. For suppose m is any element of S. We have k < m because k is in S'. We cannot have k < m < k + 1 = n by Lemma 2. So we have images/real110.png . So n is the smallest element of S. Since we assumed that S had no smallest element, we have a contradiction. This proves Proposition 7.

    The value of Proposition 7 is that it is the basis for another proof technique called infinite descent which is another variant on induction. Here is an example.

    Example 8: Prove that for every a in F and for every natural number n and m, we have: images/real111.png .

    Let P(k) be the property that images/real112.png for all a and m. If P(k) were not true for all natural numbers k, then the set S of natural numbers for which it were false would be non-empty. By Proposition 7, there is a smallest natural number k for which P(k) is false. Now, k cannot be 1, since images/real113.png . So k - 1 must also be a natural number and P(k) must be true (otherwise k would not be the smallest element of S). So images/real114.png . But then images/real115.png contrary to assumption.

    2.6 The Fundamental Theorem of Arithmetic

    This entire section will deal with natural numbers.

    Definition 4: A natural number d is a divisor of the natural n if and only if there is a natural number m such that n = dm. A divisor d of n is said to be proper if it is other than 1 and n itself. A natural number p is said to be prime if it is not equal to 1 and it has no proper divisors.

    For example, 4 is a divisor of 20 because images/real116.png . The prime numbers are 2, 3, 5, 7, 11, etc.

    Let us define by induction the product of n numbers. A product of 1 number is defined to be itself. Assuming that we have defined the product of k numbers. A product of k + 1 numbers is the product of the product of the first k of the numbers and the last number. For convenience, we also stipulate that the product of zero numbers is 1. If images/real117.png for images/real118.png are n numbers, then the product of these n numbers is denoted images/real119.png . Of course, if all the numbers are equal to a, we can still denote the product of a with itself n times as images/real120.png . (One defines the sum of n numbers images/real121.png in an analogous manner; the notation for the sum of the n numbers images/real122.png for images/real123.png is images/real124.png .)

    Proposition 8: i. Every divisor of a natural number n satisfies images/real125.png . Every proper divisor satisfies 1 < d < n.

    ii. Every natural number can be written as a product of finitely many primes.

    Proof: i. Suppose dm = n. Then images/real126.png and so images/real127.png . We also have 1 leq d since d is a natural number. So, images/real128.png . In particular, if d is a proper divisor of n, then 1 < d < n.

    ii. We will prove the second assertion by infinite descent. If the assertion is false, then there is a smallest natural number n which is not expressible a product of primes. Then n is not a prime or else it would be the product of a single prime. Since n is not a prime, we can write n = dm where d is a proper factor of n. But then m is also a proper factor of n. (Why?) In particular, both d and m are natural numbers smaller than n. Since n was the smallest number not expressible as a product of primes, both d and m can be expressed as a product of primes. But then by taking the product of all the factors in the expressions of both d and m, we see that n is a product of primes also. This contradiction proves the result.

    Proposition 9: (Division Theorem) If n and d are natural numbers, there there are unique non-negative integers m and r such that n = md + r and images/real129.png .

    Proof: Let us prove the result by induction on n. If n is 1 and d is also 1, then m = 1 and r = 0 is the unique solution. On the other hand, if n = 1 and d > 1, then m = 0 and r = 1 is the unique solution.

    Now, suppose that the result is true for some natural number n. Then n = md + r with images/real130.png . If r = d - 1, then n + 1 = (m+1)d + 0. Otherwise, n + 1 = md + (r + 1). To show uniqueness, suppose that images/real131.png with images/real132.png for i = 1 and 2. Taking differences, we see that images/real133.png where we have ordered the terms so that the right hand side is non-negative (If it were not, just swap the subscripts). But then the left side is also non-negative. So, we must have images/real134.png . Since the left side is a multiple of d, Proposition 8 implies that it must be zero. So images/real135.png . But then, the right side is also zero and so images/real136.png too.

    If m and n are natural numbers, then a natural number d is called a common divisor of m and n if it is both a divisor of m and also a divisor of n. The largest common divisor of m and n is called the greatest common divisor of m and n and is denoted gcd(m, n).

    Proposition 9 will allow us to develop the Euclidean Algorithm for calculating gcd(m, n). Repeatedly apply Proposition 9 to obtain


    where one stops with as soon as one obtains the first zero remainder.

    If d is a common divisor of m and n, then d also divides images/real138.png by the first equation. But then, the second equation shows that d divides r_1, and so on. We finally determine that d divides all the remainders. In particular, it divides the last remainder images/real139.png . On the other hand, starting from the last equation, we see that images/real140.png divides images/real141.png . The second from the last equation then says that it also divides images/real142.png , and so on. We conclude that images/real143.png divides both n and m. We conclude therefore that images/real144.png .

    Again starting from the second from the last equation, we can solve to get images/real145.png . Using the previous equation we can solve for images/real146.png and substitute the expression into the right hand side of this last equation to express images/real147.png as a linear combination of images/real148.png and images/real149.png . Repeating the process, we eventually get the greatest common denominator images/real150.png written as a linear combination of m and n. Summarizing the results:

    Proposition 10: (Euclidean Algorithm) If m and n are natural numbers, the Euclidean algorithm described above calculates the greatest common divisor of m and n. Furthermore, it allows one to find integers a and b such that gcd(m, n) = am + bn.

    We say that two natural numbers m and n are relatively prime if gcd(m,n) = 1.

    Corollary 2: If two natural numbers m and n are relatively prime, then there are integers a and b with 1 = am + bn. In particular, if p is a prime and n is not a multiple of p, then p and n are relatively prime and so there are integers a and b with 1 = ap + bn.

    Corollary 3: If a prime p divides a product mn of natural numbers, then either p divides m or p divides n.

    Proof: If it divides neither, then there are integers a, b, c, and d such that 1 = ap + bm and 1 = cp + dn. Taking products, we get 1 = 1cdot 1 = (ap + bm)(cp + dn) = (acp + adn +bmc)p + bd(mn). Since p divides mn, we have mn = ep for some e. Substituting, we see that p divides the right side and so p must divide the left side. But the left side is 1, which is a contradiction. So, it must be that p divides m or p divides n.

    Corollary 4: If a prime p divides a product of any finite number of natural numbers, then p divides at least one of the numbers.

    Proof: This follows by induction from Corollary 3.

    Corollary 5: (Linear Diophantine Equations) The equation ax + by = c where a, b, and c are constants has integer solutions (x, y) if and only if gcd(a,b) divides c.

    Proof: If a or b is zero, then the result is obvious. If not, then we can assume that a and b are natural numbers (by replacing one or both of x and y with their negatives). If gcd(a,b) divides c, then the Euclidean Algorithm gives integers u and v with au + bv = gcd(a,b). Multiplying by d = c/gcd(a,b) shows that x = ud and y = vd are solutions of the equation. On the other hand, if there is a solution, then clearly gcd(a,b) divides ax + by = c.

    Example 9: Let's calculate the gcd(310, 464). One has images/real151.png , images/real152.png , and images/real153.png . We conclude that gcd(310, 464) = 2. Furthermore, we have images/real154.png and images/real155.png . Substituting gives images/real156.png .

    Theorem 1:(Fundamental Theorem of Arithmetic) Every natural number can be represented as a product of zero or more primes. Furthermore, these primes are uniquely determined (including the number of times each prime is repeated) up to order of the factors.

    Proof: We already know that at least one representation exists. If the result is false, then let n be the smallest natural number for which uniqueness is false. Suppose one can write two distinct representations images/real157.png where the factors are all prime. There cannot be a prime p which appears in both factorizations; if there were, then n/p would be a smaller natural number with two distinct representations as products of primes. But, clearly images/real158.png is a factor of n. Corollary 4 implies that images/real159.png must divide some images/real160.png . Since images/real161.png is a prime, it follows that images/real162.png as primes have no proper divisors. This contradiction shows that there is no natural number n for which the factorization is non-unique. This completes the proof.

    Corollary 6: (Euclid) There are infinitely many prime numbers.

    Proof: Suppose that the only prime numbers were images/real163.png . Let n = images/real164.png . Clearly n is a natural number not divisible by images/real165.png contrary to the Fundamental Theorem.

    Corollary 7: images/real166.png is irrational.

    Proof: Suppose that images/real167.png where m and n are positive integers. By dividing m and n by their greatest common divisor, we may assume that they are relatively prime. Squaring both sides of the equation and multiplying through by the denominator, we get images/real168.png . Since 2 divides the left side, 2 divides images/real169.png and, since 2 is a prime, we have 2 divides m. Dividing our equation by 2, we get images/real170.png . So 2 divides the right side and images/real171.png is divisible by 2. Again, this means that 2 is a divisor of n. But this contradicts the assumption that m and n are relatively prime. So, our original assumption that images/real172.png is rational must have been false. This completes the proof.

    2.7 Real Numbers

    All of the material in the previous sections of this chapter applies to any ordered field. In particular, it applies to both the fields of rational numbers as well to the fields of real numbers. The problem with working in the field of rational numbers is that it is relatively sparse; so, when you go to solve equations of degree greater than one, we often find that what would have been a solution is not rational. We have already seen that images/real173.png is not a rational number. It will be convenient to have an algebraic domain in which every polynomial equation has a solution. We will find that the complex numbers fill this role; the real number field will be both useful to construct the complex numbers as well as being important in and of itself.

    Exactly what makes the real numbers special is a rather subtle matter. This section will start that explanation, and the version given here will suffice until we can revisit the question in a later chapter.

    In section 1.2.2, we said that real numbers could be represented as infinite decimals. This is the aspect of real numbers that we will discuss in this

    2.7.1 Infinite Decimals

    First let us start with a a finite decimal. This is a numeral of the form 3.14159. The form of a finite decimal is an optional sign (either plus or minus) followed by a string of decimal digits, a period, and another string of decimal digits. Each of these represents a rational number: If there are n digits to the right of the period (called the decimal point), then the rational number is the quotient of the decimal (with the period removed) divided by images/real174.png . For example, we have images/real175.png .

    Because the denominator is always a power of 10, many rational numbers cannot be represented as a finite decimal. For example, 1/3 cannot be written as a finite decimal. On the other hand, we can write arbitrarily good approximations: 0.3, 0.33, 0.333, 0.3333, etc. of 1/3. So, it is reasonable to say that if we just allowed ourselves to keep writing digits, we would get 1/3. You might write this as 0.333333.... where the ellipsis means to keep repeating the pattern. Another example would be 1/7 = 0.142857142857142857.... These are examples of infinite decimals.

    But in what sense do these represent the rational number? To get a better idea, let's look again at how we convert a finite decimal to a fraction. If the decimal is images/real176.png , then the part images/real177.png is an integer, the digit images/real178.png means images/real179.png , the digit images/real180.png is in the next place and it represents images/real181.png , and so on. So our whole decimal becomes


    Let's go back to our specific examples, we have a succession of improving estimates:


    where there are n digits 3 in the last approximation. Our last one is then


    It is still hard to tell what value we are approaching as n gets larger and larger. Here is the secret to calculating the sum:

    Lemma 3: (Geometric Series) If images/real185.png , then


    Proof: This is actually a familiar factorization of which the first few cases are: images/real187.png and images/real188.png . To understand how this works, just multiply the left side of our general expression by (1 - a). By the distributive law, this is the same as the original expression images/real189.png less the product of a and the original expression, i.e. images/real190.png . Notice that almost all the terms are repeated in the second expression. The one left out is 1 and we have one additional one images/real191.png . So the difference is just images/real192.png , which shows the result.

    Applying Lemma 3 with a = 1/10 gives


    and the right side simplifies to images/real194.png . This is the exact value when there are precisely n digits to the right of the decimal point. Now, what happens when we take more and more digits? The result is always a little less than 1/3, but the error images/real195.png shrinks to zero as n gets arbitrarily large. This is the sense in which we can say that the infinite decimal represents 1/3.

    Let's repeat the same computation with our second example. In this case, it is inconvenient to use powers of 10 because the pattern repeats itself every 6 digits. But things are easy if we simply use powers of images/real196.png .


    where again we have repeated the pattern exactly n times. Lemma 3 says that this is equal to


    Clearly, as n gets arbitrarily large the right factor approaches 1. Furthermore, if you reduce the fraction, you will see that 142857/999999 = 1/7. Again, we see that the infinite decimal represents 1/7 in the sense that if we take the sequence of numbers we get by taking more and more digits, the limiting value of the elements of the sequence is 1/7.

    Now, let's formalize our discussion.

    Definition 5: i. An infinite decimal is an expression of the type images/real199.png , where images/real200.png is an integer, and images/real201.png is an infinite sequence of decimal digits (i.e. integers between 0 and 9).

    ii. Every such infinite decimal defines a second sequence of finite decimals images/real202.png where images/real203.png .

    iii. One says that the infinite decimal represents the number r (or has limit r) if images/real204.png can be made arbitrarily close to zero simply by taking k sufficiently large.

    Definition 6: i. An ordered field F is said to be Archimedean if, for every positive a in F, there is a natural number N with a < N.

    ii. An Archimedean ordered field F is called the field of real numbers if every infinite decimal has a limit in F.

    2.7.2 Decimal Expansions

    Given any element a in F, we can form an infinite decimal for a. First, we can assume that a is positive, since the case where a = 0 is trivial, and if a < 0, then we can replace a with -a. Next, we see why we needed to add the Archimedean property to the above definition. Without it, we would not know how to get the integer part of a: Since F is Archimedean, the set of natural numbers N with a < N is non-empty and so there it has a smallest element b. Let images/real205.png . Then images/real206.png if images/real207.png . Choose images/real208.png to be the decimal digit such that images/real209.png and let images/real210.png , so that again images/real211.png . Assuming that we have already defined for some natural number k, the quantities images/real212.png and images/real213.png with images/real214.png , define by induction the digit images/real215.png so that images/real216.png and let images/real217.png , so that images/real218.png .

    The infinite decimal images/real219.png was defined so that images/real220.png with images/real221.png . So this infinite decimal has limit a. We say that this is the infinite decimal expansion of the element a in F.

    Proposition 11:i. Every element a in F is the limit of the infinite decimal expansion of a.

    ii. The decimal expansion of every rational number is a repeating decimal, i.e. except for an initial segment of the decimal, the decimal consists of repetitions of a single string of digits.

    iii. Every repeating decimal has limit a rational number.

    Proof: The first assertion has already been proved. For the second assertion, note that the definition of the sequence of digits images/real222.png is completely determined by the value of images/real223.png .

    If a = r/s is rational with r and s integers, then images/real224.png is a rational number with denominator (a factor of ) s. Furthermore, since images/real225.png , if images/real226.png is rational with denominator s, then so is images/real227.png . By induction, it follows images/real228.png is rational with denominator s for every k. Since images/real229.png lies between 0 and 1 and is rational with denominator s, it follows that there are at most s possible values for images/real230.png .

    The following principle is called the pigeonhole principle: If s + 1 objects are assigned values from a set of at most s possible values, then at least two of the objects must be assigned the same value.

    By the pigeonhole principle, there are subscripts i and j with images/real231.png such that images/real232.png . As indicated at the beginning of the proof, it follows that the sequence of digits starting from images/real233.png must be the same as the sequence of digits starting from images/real234.png and so the decimal repeats over and over again the cycle of values images/real235.png .

    The third assertion is easy to prove -- it is essentially the same as our calculation of the limit of the infinite decimal expansions of 1/3 and 1/7. The formalities are left as an exercise.

    Example 10: The field of real numbers contains many numbers which are not rational. All we need to do is choose a non-repeating decimal and it will have as its limit an irrational number. For example, you might take images/real236.png where at each step one adds another zero.

    Proposition 12: Every a > 0 in the field of real numbers has a positive images/real237.png -root for every natural number n, i.e. there is a real number b with images/real238.png .

    Proof:It is easy to show by induction that, if images/real239.png , then images/real240.png for every natural number n. So the function images/real241.png is an increasing function. By the Archimedean property, we know that there is a natural number M > a. Again by induction, it is easy to see that images/real242.png . By descent, it follows that there is a smallest natural number m such that images/real243.png . Let images/real244.png so that the images/real245.png -root of a must lie between images/real246.png and images/real247.png . Next evaluate images/real248.png for integers j from 0 to 10. The values start from a number no smaller than a and increase to a number larger than a. Let images/real249.png be the largest value of j for which the quantity is at most a. Repeating the process, one can define by induction an infinite decimal images/real250.png such that the images/real251.png -power of the finite decimal images/real252.png differs from a by no more than images/real253.png .

    Let b be the limit of the infinite decimal, and images/real254.png be the values of the corresponding finite decimals. Then we have images/real255.png and images/real256.png and so it is reasonable to expect that images/real257.png . This is in fact true. Using the identity for geometric series, we see that: images/real258.png . But then the triangle inequality gives images/real259.png where C is a positive constant which does not depend on k. Since this holds for all positive integers k, it follows that images/real260.png .

    2.7.3 Limits of Sequences

    A sequence images/real261.png of real numbers is said to converge to a real number b (or to have limit b) if all the images/real262.png are as close to b as desired as long as k is sufficiently large. More formally, this means that for every images/real263.png (regardless of how small), there is a (possibly quite large) N > 0 such that images/real264.png for all k larger than N. The sequence is said to be bounded above by a real number B if images/real265.png for all images/real266.png .

    For example, if images/real267.png , is an infinite decimal, and images/real268.png for images/real269.png , then images/real270.png converges to images/real271.png .

    Proposition 13: Let images/real272.png be a sequence of real numbers bounded above by a real number B. If the sequence is increasing, i.e. images/real273.png , then the sequence converges to some real number b.

    Proof: Since each images/real274.png is a real number, it has an decimal expansion images/real275.png where images/real276.png is a sign, either plus or minus. Because the sequence is increasing, one has:

    1. Except for a finite number of terms, the signs images/real277.png must all be identical. (Either they are all +, all -, or change once from - to plus.) Assume that the signs are all identical, say with value s, for images/real278.png
    2. Because there are only finitely many integer values between images/real279.png and B, the values of images/real280.png must all be identical for all images/real281.png for some images/real282.png which we can assume is no smaller than images/real283.png . Let images/real284.png be this common value.
    3. Assume by induction that we have defined images/real285.png and that images/real286.png for all images/real287.png for some images/real288.png no smaller than images/real289.png . Since the sequence is increasing, the images/real290.png for images/real291.png must be increasing (if s is +) or decreasing (if s is -). So, except for a finite number of terms, the numbers must be constant, say equal to images/real292.png for images/real293.png for some images/real294.png no smaller than images/real295.png .

    Let images/real296.png Then images/real297.png for all images/real298.png because the decimal expansion of all such b_i agrees with that of b up to the images/real299.png decimal digit. So b is the limit of the images/real300.png . This completes the proof.

    Remark: A sequence is said to be bounded below by a real number B if all the terms of the sequence are no smaller than B. A decreasing sequeence images/real301.png of real numbers which is bounded below converges to a real number. (To see this, simply apply the Proposition to the sequence images/real302.png .)

    Corollary 8: i. Let images/real303.png for images/real304.png be closed intervals with images/real305.png . If the lengths images/real306.png converge to 0, then there is precisely one real number c contained in all the intervals images/real307.png such that the sequence of the images/real308.png as well as the sequence of the images/real309.png converge to c.

    ii. Let the images/real310.png form a decreasing sequence of positive real numbers which converge to zero. Define images/real311.png for images/real312.png . Then the sequence images/real313.png converges to a real number b. (The value b is said to be the limit of the infinite sum images/real314.png .

    Proof: i. The sequence of the images/real315.png is increasing and bounded above by every images/real316.png . So, the sequence converges to a real number a which is no larger than any of the images/real317.png . Clearly, images/real318.png for all images/real319.png . Similarly, the sequence of the images/real320.png is decreasing and bounded below by every images/real321.png . So, the sequence converges to a real number b no smaller than any of the images/real322.png , and images/real323.png for all images/real324.png . We cannot have images/real325.png or else the distance between them would be positive; but this cannot be true because both lie in images/real326.png for all j and the sequence of the images/real327.png converges to zero.

    ii. This follows from the first assertion using the intervals images/real328.png .

    2.8 Complex Numbers

    If a is a positive element of any ordered field, we know that images/real329.png because the set of positive numbers is closed under multiplication. Since we also have images/real330.png , it follows by trichotomy that the square of any element in an ordered field is always non-negative. In particular, such a field cannot contain a solution of images/real331.png .

    We would like to have a field where all polynomial equations have a root. We will define a field images/real332.png called the field of complex numbers which contains the field of rational numbers and which also has a root, denoted i, of the equation images/real333.png . In a later chapter, it will be shown that, in fact, images/real334.png contains a root of any polynomial with coefficients in images/real335.png . This result is called the Fundamental Theorem of Algebra.

    2.8.1 The Field of Complex Numbers

    Let us first define the field of complex numbers. Since it is a field which contains both the field of real numbers and the element i, it must also contain expressions of the form z = a + bi where a and b are real numbers. Furthermore, there is no choice about how we would add and multiply such quantities if we wanted the field axioms to be satisfied. The operations can only be:




    where we have used the assumption that images/real338.png .

    It is straightforward, but a bit tedious to show that these operations satisfy all the field axioms. Most of the verification is left to the exercises. But let us at least indicate how we would show that there are multiplicative inverses. Let us proceed heuristically -- we would expect the inverse of a + bi to be expressed as images/real339.png but this does not appear to be of the desired form because there is an i in the denominator. But our formula from geometric series shows how to rewrite it: We have images/real340.png . This is just what we need:


    Of course, we have proven nothing. But we now have a good guess that the multiplicative inverse might be images/real342.png . It is now an easy matter to check that this does indeed work as a multiplicative inverse.

    Proposition 14: The set of all expressions a + bi, where a and b are real and i behaves like images/real343.png , is a field if we define operations as shown above.

    We have already seen that the field images/real344.png cannot be ordered. Nevertheless, we can define an absolute value function by images/real345.png .

    Proposition 15: Let w and z be complex numbers. Then

    1. |w| = |-w|
    2. |wz| = |w||z|
    3. |z| = 0 if and only if z = 0.
    4. images/real346.png .
    5. If r is a real number, its absolute value is the same as a real number as it is if it is considered to be the complex number images/real347.png .

    Proof: These are all left as exercises.

    2.8.2 Geometric Interpretation of Complex Operations

    We have defined a 1-1 correspondence between the set of complex numbers and the Euclidean plane of pairs of real numbers. The complex numbers are not just a set; they also have addition and multiplication operators. Our next job is to see how these arithmetic operators correspond to geometric operations in the plane.

    The Parallelogram Law Let's start with addition. If images/real348.png for j = 1, 2, then images/real349.png . In our 1-1 correspondence we have the four numbers images/real350.png corresponding to the four points images/real351.png , images/real352.png , images/real353.png , images/real354.png . Here is a picture of the situation.

    From the picture, it certainly looks like the four points are vertices of a parallelogram. Showing that it is true is a simple matter of calculating the slopes of the four sides. For example, the slope of the line containing z_1 and z_1 + z_2 is


    which is the slope of the containing O and z_2. (You need to treat the case where images/real356.png separately: in this case, the two sides are coincident.) This result is the so-called parallelogram rule, which may be familiar to you as addition of forces in physics.

    Part of multiplication is easy: In Proposition 15, we have already seen the property images/real357.png where the absolute value images/real358.png is clearly just the length of the line segment from O to z (by the distance formula, a.k.a. Pythagorean Theorem). This tells us that the length (another word for absolute value) of the product of two complex numbers is the product of the lengths of the factors.

    Besides length, what else is needed to determine the line segment from O to the complex number z? One way to determine it is to use the angle images/real359.png from the positive x-axis to the segment from O to z. This angle is called the argument of z and is denoted images/real360.png . Note that arg(z) is determined only up to a multiple of images/real361.png radians and that it is not defined at all in case z = 0.

    The figure below shows the relationship between z = a + bi, r = |z|, and images/real362.png .

    In particular, we see that: images/real363.png , images/real364.png , images/real365.png , and images/real366.png (unless a = 0). Notice that our definition of the trigonometric functions using the unit circle automatically guarantees that all the signs in these formulas are correct regardless of quadrant in which the point z lies. In particular, we have


    which tells us that multiplying a complex number by a positive real number does not change the argument, but just expands the length by that factor. For example, doubling a complex number makes it twice as long but it points in the same direction.

    Let's calculate products using the argument function. Let images/real368.png have length images/real369.png and argument images/real370.png for j = 1, 2 (where we are assuming that neither images/real371.png is zero, since that case is trivial). The product is:


    where we the final simplification used the addition formulas for both the sine and the cosine function. So, the full story on multiplication is that you multiply the lengths of the factors and add the arguments:


    In the very special case of natural number powers this says:

    Proposition 16:(De Moivre) If a non-zero complex number z = a + bi has length r = |z| and argument images/real374.png and if n is any natural number, then: images/real375.png .

    Corollary 9: Let n be a natural number.

    1. There are (at least) n complex numbers which are images/real376.png -roots of 1. In fact,


      for k = 0, 1, ..., n - 1 all satisfy images/real378.png .

    2. More generally, every non-zero complex number z has (at least) n distinct images/real379.png roots. In fact, if z has length r and argument images/real380.png , then the following are all images/real381.png -roots of z:


      for all k = 0, 1, ..., n - 1.

    The corollary follows immediately from De Moivre's formulas. In fact, these are the only images/real383.png -roots, but it is convenient to defer the proof of this until a later chapter. The quantities images/real384.png of the first part of Corollary 9 are called images/real385.png -roots of unity. Geometrically, they all lie on the unit circle and are evenly spaced around the around the circle. The second part of the Corollary says that we can obtain the various roots of any number by simply multiplying any one of them by the various images/real386.png -roots of unity.

    2.8.3 Applications

    The ability to do arithmetic operations on the points in the plane makes a number of topics in geometry much simpler. Rotations and Translations

    A geometric figure is a collection of points. We can transform this set by applying operations to each of the points. For example, if you add a complex number to each point, it translates or shifts the figure. For example, if the figure is the unit circle, consisting of all the points (x,y) where images/real387.png . Then adding (2, 3) to each of these points gives the set of points (x + 2, y + 3) where images/real388.png . Letting x' = x + 2 and y' = y + 3, we see that x = x' - 2 and y = y' - 3. So, the shifted circle is the set of all (x', y') where images/real389.png .

    In general, let S be a set of points (x, y) where f(x, y) = 0. If we want to shift this a units to right and b units upward, then the new set of points is the set of (x, y) where f(x - a, y - b) = 0.

    For example, images/real390.png is a parabola with vertex at (2, 3).

    One can also do reflections across the y-axis by replacing x with -x. For example, the set of (x, y) where images/real391.png is the upper half of a parabola having the x-axis as its axis. Its reflection across the y-axis has equation images/real392.png . Similarly, one can reflect across the x-axis by replacing y with -y in the equation of the set. Note that this corresponds to mapping z = x + iy to its complex conjugate.

    The third type of transformation is a rotation about the origin. We know that we can rotate z = x + iy through an angle images/real393.png by multiplying it be the complex number images/real394.png to give the number zu = (xcos(theta) -ysin(theta)) + i(xsin(theta) + ycos(theta)). So, our new point is (x', y') where


    Alternatively, we can get x and y from x' and y' by rotating (x', y') through an angle images/real396.png . So, if the original set is the set of (x, y) where f(x, y) = 0, then the rotated set of (x, y) where images/real397.png .

    For example, suppose we want to rotate the hyperbola xy = 1 counter-clockwise 45 degrees. Then use images/real398.png and the equation of the rotated figure is images/real399.png or images/real400.png .

    Warning! It is notoriously easy to make a mistake in rotating in the wrong direction. You should always check a point afterwards to make sure you have rotated in the direction intended. The formula for doing the rotation is also hard to remember correctly; it is usually best to just remember that you rotate by multiplying by images/real401.png .

    When we set up the plane, we defined it to be the set of pairs (x, y) of real numbers. One then defined the distance between two points as by a formula involving the x and y coordinates of the two points. This means that our notion of distance appears to depend on the choice of coordinate system. In fact, it is independent of the choice of coordinate system as is straightforward to verify:

    Proposition 17: If two points images/real402.png and images/real403.png are transformed by a finite number or translations, rotations, and reflections, the distance between the two points does not change. Angles and Trigonometry

    In Chapter 1, we assumed an intuitive notion of what one meant by an angle -- it was measured as the length of the arc of the unit circle swept out as you traversed the angle. Although intuitive, it is difficult to define exactly what one means by the length of the arc of the circle swept out as you traverse the angle. In this section, we will see how to make this precise assuming that one has the basic properties of the sine and cosine function. In Chapter 5, we will complete the job by rigourously defining the trigonometric functions.

    First you need to remember that as images/real404.png ranges from 0 to images/real405.png radians, images/real406.png decreases from 1 to -1. So, for each real value v between -1 and 1, there is a unique images/real407.png such that cos(theta) = v. This uniquely defined images/real408.png is called the arccosine of v and is denoted either images/real409.png or images/real410.png .

    Next we need some definitions for complex numbers. If z = x + iy is a non-zero complex number, then the direction of z is defined to be u = z/|z|. (Geometrically, it is a complex number of length one pointed in the same direction as z.) For any complex number u = a + bi of length one, define


    For any non-zero complex number z, define arg(z) to be arg(u) where images/real412.png is the direction of z. The principal argument was defined so that


    In fact, it is the unique real number which satisfies this condition. It follows that if w and z are two non-zero complex numbers, then




    where we say that two numbers are equal mod images/real416.png if their difference is an integer multiple of images/real417.png .

    Finally, in this section, we will refer to points (x, y) in the plane as if they are the corresponding complex number z = x + iy. A directed line segment AB is determined by its two endpoints A and B; so a directed line segment is really an ordered pair of points, which is the same thing as an ordered pair of complex numbers. Assuming that r and s are the complex numbers associated with A and B respectively, the complex number images/real418.png is called the direction of the directed line segment AB. Clearly the direction of AB is a complex number of length one, and the direction of BA is -u if u is the direction of AB. (Geometrically, u is a complex number of length one which points in the same direction as AB.) A directed angle images/real419.png is an ordered triple of points A, O, and B where both A and B are distinct from O. The measure of the directed angle is arg(u/v) where u is the direction of OA and v is the direction of OB. So, the measure of an angle is a real number images/real420.png in the interval images/real421.png such that images/real422.png . Note that the angles are directed in the sense that images/real423.png . One needs to be careful about this because many of the theorems in synthetic geometry refer to undirected angles, i.e. they use the absolute value of the measure of the angle so that images/real424.png for undirected angles.

    One can now prove many of the results of synthetic geometry. For example,

    Proposition 18: If images/real425.png is a triangle, then (as directed angles) one has


    Proof: Let the directions of BA, BC, and CA be u, v, and w respectively, Then the sum is the direction


    where the equality is mod images/real428.png . Now, if the sum of the three angles is not equal to images/real429.png , it must be equal to images/real430.png because each of the angles is in the interval images/real431.png . But this could only happen if all three angles were equal to images/real432.png , in which case images/real433.png would not be a triangle.

    Notice how easy it is to get the addition formulas:

    Proposition 19: (Addition Formulas) If A and B are two angles, then

    1. images/real434.png
    2. images/real435.png
    3. images/real436.png and images/real437.png

    Proof: Let u and v be the complex numbers of length 1 such that A = arg(u) and B = arg(v). Then A + B = arg(uv) (mod images/real438.png ) and -A = arg(1/u) = arg( images/real439.png ). One has images/real440.png and images/real441.png . Multiplying these expressions for u and v together givees the first two assertions. The last assertion follows from the definition of the complex conjugate.

    Proposition 20: (Law of Cosines) Let a, b, and c be the lengths of the sides of the triangle images/real442.png opposite angles A, B, and C respectively. Then


    Proof: Let u and v be the directions of CA and CB respectively. Then identifying the points with the corresponding complex numbers, we have B = C + av and A = C + bu. So, one has


    where one has used part iii of Proposition 19.

    Directions can be used to verify the usual properties of similar triangles. The exercises will give further details.

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    Revised: July 7, 2001
    All contents © copyright 2001 K. K. Kubota. All rights reserved