## Chapter 2: The Real Numbers## 2.1 IntroductionAlthough 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: - 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. - 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. - 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.
- 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.
- 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 Given two sets S and T we can define the By a 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 . 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 where t is the mother of s. A
- Let N be the set of positive integers. The successor function 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.
- Let R be the set of real numbers. The set of all pairs where x is in R is a function. We normally write it as .
- Let R be the set of real numbers. The set of all pairs 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.
- Let S be the set of non-negative real numbers. The set of all pairs (a, b) where is not a function. For example, (4, 2) and (4, -2) are two pairs in the set with the same first coordinate.
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 A
## 2.3 FieldsThe 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:
- (Commutativity) For every a and b in S, one has a + b = b + a and ab = ba.
- (Associativity) For every a, b, and c in S, one has (a + b) + c = a + (b + c) and (ab)c = a(bc).
- (Distributive Law) For every a, b, and c in S, one has a(b + c) = ab + ac.
- (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.
- (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 such that .
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:
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.
Suppose ab = 0. If a is zero, there is nothing more to prove. On the other hand, if , then a has a multiplicative inverse c and so .
We can define the binary subtraction operator: a - b = a + (-b) and, for , the binary division operator . The division operator will also be expressed as .
- If c is also non-zero, then
(ii) Using associativity and commutativity, one can show that . By assertion (i), this is the same as . (iii) (iv) One has where some of the steps have been combined. (v) Using property ii, it is easy to see that . 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
- (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.
- (Transitivity) Let a, b, c be arbitrary elements of F. If a < b and b < c, then a < c.
- If a, b, and c in F satisfy a < b, then a + c < b + c.
- If a, b, and c in F satisfy a < b and 0 < c, then ac < bc.
If F is an ordered field, an element a in F is called
- (Trichotomy) For every a in F, exactly one of the following conditions holds: a is in P, a = 0, and -a is in P.
- (Closure) If a and b are in P, then so are a + b and ab.
ii. An element a of an ordered field F is said to be 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 to mean either a < b or a = b and similarly for . iv. The
v. One now has everything you need to deal with inequalities as we did back in section 1.1.4.
- |-a| = |a|
- (i.e. and )
- (Triangle Inequality)
ii. Again, we can treat three cases: If a > 0 or a = 0, then |a| = a and so . 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 , 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 , then |a + b| = a + b. Since and , we can add b to both sides of the first inequality and |a| to both sides of the second one to get and . Using transitivity, we get as desired. Now suppose that . Then adding -a - b to both sides of the inequality gives -a + (-b) < 0. Applying the result of the last paragraph, we get . But a + b < 0 means that |a + b| = -(a + b) and so where we have used assertion i for the last step. This completes the proof.
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 . This still looks complicated until one thinks of grouping the factors differently to get . Subtracting the left side from both sides of the inequality gives . Now, you may recognize the right side as being a perfect square: factoring we get . 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
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 , we get . Using the distributive law to expand this gives . Adding 2(ad)(bc) to both sides of the inequality yields . Using the commmutative and associative law several times allows us to re-arrange this into . 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 - ,
- 1 is in S, and
- If a is in S, then so a + 1.
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
- ,
- 1 is in , and
- If a is in , then so a + 1.
- The only inductive subset of is .
ii. Since 1 is in every set S which is in T, 1 is in their intersection which is, by definition, . iii. Let a be in . 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 . iv. Suppose S is an inductive subset of . Then S is in T. Since is the intersection of all the S in T, it follows that is contained in S. But then we have and , 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 First, let's see how it works for definitions. Let's do an inductive definition of powers. Let a be in F. We define to be a. Now, suppose we have already defined for some natural number k, then define to be the product . Consider the set S of all natural numbers k for which we have defined the . It contains 1 and if k is in S, so is k + 1. By Proposition 6, it follows that 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: - P(1) is true.
- For every natural number k, if P(k) is true, then so is P(k+1).
- For k = 1, we have by the inductive definition of the powers of a. So P(1) is true.
- 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: - inductive definition of powers
- associativity of multiplication
- P(k)
- inductive definition of powers
- associativity of multiplication
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
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.
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 . 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
Let P(k) be the property that 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 . 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 . But then contrary to assumption. ## 2.6 The Fundamental Theorem of ArithmeticThis entire section will deal with natural numbers.
For example, 4 is a divisor of 20 because . The prime numbers are 2, 3, 5, 7, 11, etc. Let us define by induction the
ii. Every natural number can be written as a product of finitely many primes.
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.
Now, suppose that the result is true for some natural number n. Then n = md + r with . If r = d - 1, then n + 1 = (m+1)d + 0. Otherwise, n + 1 = md + (r + 1). To show uniqueness, suppose that with for i = 1 and 2. Taking differences, we see that 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 . Since the left side is a multiple of d, Proposition 8 implies that it must be zero. So . But then, the right side is also zero and so too. If m and n are natural numbers, then a natural number d is called a
Proposition 9 will allow us to develop the 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 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 . On the other hand, starting from the last equation, we see that divides . The second from the last equation then says that it also divides , and so on. We conclude that divides both n and m. We conclude therefore that . Again starting from the second from the last equation, we can solve to get . Using the previous equation we can solve for and substitute the expression into the right hand side of this last equation to express as a linear combination of and . Repeating the process, we eventually get the greatest common denominator written as a linear combination of m and n. Summarizing the results:
We say that two natural numbers m and n are
## 2.7 Real NumbersAll 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 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 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 , then the part is an integer, the digit means , the digit is in the next place and it represents , 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:
Applying Lemma 3 with a = 1/10 gives and the right side simplifies to . 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 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 . 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.
ii. Every such infinite decimal defines a second sequence of finite decimals where . iii. One says that the infinite decimal
ii. An Archimedean ordered field F is called the ## 2.7.2 Decimal ExpansionsGiven 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 . Then if . Choose to be the decimal digit such that and let , so that again . Assuming that we have already defined for some natural number k, the quantities and with , define by induction the digit so that and let , so that . The infinite decimal
was defined so that
with
. So this
infinite decimal has limit a. We say that this is the
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.
If a = r/s is rational with r and s integers, then is a rational number with denominator (a factor of ) s. Furthermore, since , if is rational with denominator s, then so is . By induction, it follows is rational with denominator s for every k. Since lies between 0 and 1 and is rational with denominator s, it follows that there are at most s possible values for . The following principle is called the By the pigeonhole principle, there are subscripts i and j with such that . As indicated at the beginning of the proof, it follows that the sequence of digits starting from must be the same as the sequence of digits starting from and so the decimal repeats over and over again the cycle of values . 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.
Let b be the limit of the infinite decimal, and be the values of the corresponding finite decimals. Then we have and and so it is reasonable to expect that . This is in fact true. Using the identity for geometric series, we see that: . But then the triangle inequality gives where C is a positive constant which does not depend on k. Since this holds for all positive integers k, it follows that . ## 2.7.3 Limits of Sequences A sequence
of real numbers is said to For example, if , is an infinite decimal, and for , then converges to .
- Except for a finite number of terms, the signs 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
- Because there are only finitely many integer values between and B, the values of must all be identical for all for some which we can assume is no smaller than . Let be this common value.
- Assume by induction that we have defined and that for all for some no smaller than . Since the sequence is increasing, the for 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 for for some no smaller than .
Let Then for all because the decimal expansion of all such b_i agrees with that of b up to the decimal digit. So b is the limit of the . This completes the proof.
ii. Let the form a decreasing sequence of positive real numbers which converge to zero. Define for . Then the sequence converges to a real number b. (The value b is said to be the limit of the infinite sum .
ii. This follows from the first assertion using the intervals . ## 2.8 Complex NumbersIf a is a positive element of any ordered field, we know that because the set of positive numbers is closed under multiplication. Since we also have , 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 . We would like to have a field where all polynomial equations
have a root. We will define a field
called the field of
complex numbers which contains the field of rational numbers and which
also has a root, denoted i, of the equation
. In a later
chapter, it will be shown that, in fact,
contains a root
of any polynomial with coefficients in
. This result is
called the ## 2.8.1 The Field of Complex NumbersLet 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: and where we have used the assumption that . 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 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 . 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 . It is now an easy matter to check that this does indeed work as a multiplicative inverse.
We have already seen that the field cannot be ordered. Nevertheless, we can define an absolute value function by .
- |w| = |-w|
- |wz| = |w||z|
- |z| = 0 if and only if z = 0.
- .
- 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 .
## 2.8.2 Geometric Interpretation of Complex OperationsWe 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.
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
separately: in this case, the two sides are coincident.)
This result is the so-called Part of multiplication is easy: In Proposition 15, we have already
seen the property
where the
absolute value
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 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
from the positive x-axis to the segment from O to z. This angle
is called the The figure below shows the relationship between z = a + bi, r = |z|, and . In particular, we see that: , , , and (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 have length and argument for j = 1, 2 (where we are assuming that neither 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:
- There are (at least) n complex numbers which are
-roots of 1. In
fact,
for k = 0, 1, ..., n - 1 all satisfy . - More generally, every non-zero
complex number z has (at least) n distinct
roots. In fact, if
z has length r and argument
, then the following are all
-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
-roots, but it is convenient to defer the proof
of this until a later chapter. The quantities
of the first part of
Corollary 9 are called ## 2.8.3 ApplicationsThe ability to do arithmetic operations on the points in the plane makes a number of topics in geometry much simpler. ## 2.8.3.1 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 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, is a parabola with vertex at (2, 3). One can also do The third type of transformation is a rotation about the origin. We know that we can rotate z = x + iy through an angle by multiplying it be the complex number 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 . So, if the original set is the set of (x, y) where f(x, y) = 0, then the rotated set of (x, y) where . For example, suppose we want to rotate the hyperbola xy = 1 counter-clockwise 45 degrees. Then use and the equation of the rotated figure is or .
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: ## 2.8.3.2 Angles and TrigonometryIn 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
ranges from 0 to
radians,
decreases from 1 to -1. So, for each real value v between -1 and 1,
there is a unique
such that cos(theta) = v. This uniquely
defined
is called the Next we need some definitions for complex numbers. If z = x + iy is
a non-zero complex number, then the For any non-zero complex number z, define arg(z) to be arg(u) where 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 and where we say that two numbers are equal mod if their difference is an integer multiple of . 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 One can now prove many of the results of synthetic geometry. For example,
where the equality is mod . Now, if the sum of the three angles is not equal to , it must be equal to because each of the angles is in the interval . But this could only happen if all three angles were equal to , in which case would not be a triangle. Notice how easy it is to get the addition formulas:
- and
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. All contents © copyright 2001 K. K. Kubota. All rights reserved |