Chapter 2: The Real Numbers
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:
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 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 .
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 function from the set S to the set T is simply a relation 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: . 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.
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 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).
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:
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 if (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 . If a and b are in F satisfy , then either a or b are zero.
Proof: Let a be in F. Then . Let b be the additive inverse of . 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 , then a has a multiplicative inverse c and so .
Corollary 1: If a is an element of a field F, then -a = (-1)a.
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 .
Proposition 3: Let F be a field containing a, b, c, and d where b and d are non-zero. Then
Proof: (i) Do this as an exercise - it is a matter of simplifying 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 . By assertion (i), this is the same as .
(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.
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:
Fact: If F is an ordered field, then 0 < 1.
Proof: By Definition 1, 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 . By Proposition 2, the left side is 0 and so . 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:
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.
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 to mean either a < b or a = b and similarly for .
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.
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 . 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
Conjecture: If a is any element of an ordered field, then .
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.
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:
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 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 of natural numbers satisfy the conditions
Proof: i. Let a be in . Then a is an element of every set S in T. If S is in T, then 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, .
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 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 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:
Example 7: Let's prove that 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 .
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: for every natural number b.
Proof: Let P(k) be the property that . Clearly P(1) is true. If for some natural number k, we have P(k) true, then . 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 , 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 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 . Now n must be greater than k since k is in S'. We cannot have k < n < k + 1 by Lemma 2 and so , 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 . 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: .
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.
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 . 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 for are n numbers, then the product of these n numbers is denoted . Of course, if all the numbers are equal to a, we can still denote the product of a with itself n times as . (One defines the sum of n numbers in an analogous manner; the notation for the sum of the n numbers for is .)
Proposition 8: i. Every divisor of a natural number n satisfies . 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 and so . We also have 1 leq d since d is a natural number. So, . 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 .
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 . 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 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 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:
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 , , and . We conclude that gcd(310, 464) = 2. Furthermore, we have and . Substituting gives .
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 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 is a factor of n. Corollary 4 implies that must divide some . Since is a prime, it follows that 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 . Let n = . Clearly n is a natural number not divisible by contrary to the Fundamental Theorem.
Corollary 7: is irrational.
Proof: Suppose that 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 . Since 2 divides the left side, 2 divides and, since 2 is a prime, we have 2 divides m. Dividing our equation by 2, we get . So 2 divides the right side and 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 is rational must have been false. This completes the proof.
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 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
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 . For example, we have .
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:
Lemma 3: (Geometric Series) If , then
Proof: This is actually a familiar factorization of which the first few cases are: and . 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 less the product of a and the original expression, i.e. . Notice that almost all the terms are repeated in the second expression. The one left out is 1 and we have one additional one . So the difference is just , which shows the result.
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.
Definition 5: i. An infinite decimal is an expression of the type , where is an integer, and 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 where .
iii. One says that the infinite decimal represents the number r (or has limit r) if 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.
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 . 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 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 is completely determined by the value of .
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 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 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.
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 where at each step one adds another zero.
Proposition 12: Every a > 0 in the field of real numbers has a positive -root for every natural number n, i.e. there is a real number b with .
Proof:It is easy to show by induction that, if , then for every natural number n. So the function 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 . By descent, it follows that there is a smallest natural number m such that . Let so that the -root of a must lie between and . Next evaluate 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 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 such that the -power of the finite decimal differs from a by no more than .
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 .
A sequence of real numbers is said to converge to a real number b (or to have limit b) if all the are as close to b as desired as long as k is sufficiently large. More formally, this means that for every (regardless of how small), there is a (possibly quite large) N > 0 such that for all k larger than N. The sequence is said to be bounded above by a real number B if for all .
For example, if , is an infinite decimal, and for , then converges to .
Proposition 13: Let be a sequence of real numbers bounded above by a real number B. If the sequence is increasing, i.e. , then the sequence converges to some real number b.
Proof: Since each is a real number, it has an decimal expansion where is a sign, either plus or minus. Because the sequence is increasing, one has:
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.
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 of real numbers which is bounded below converges to a real number. (To see this, simply apply the Proposition to the sequence .)
Corollary 8: i. Let for be closed intervals with . If the lengths converge to 0, then there is precisely one real number c contained in all the intervals such that the sequence of the as well as the sequence of the converge to c.
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 .
Proof: i. The sequence of the is increasing and bounded above by every . So, the sequence converges to a real number a which is no larger than any of the . Clearly, for all . Similarly, the sequence of the is decreasing and bounded below by every . So, the sequence converges to a real number b no smaller than any of the , and for all . We cannot have or else the distance between them would be positive; but this cannot be true because both lie in for all j and the sequence of the converges to zero.
ii. This follows from the first assertion using the intervals .
If 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 Fundamental Theorem of Algebra.
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 .
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.
Proposition 14: The set of all expressions a + bi, where a and b are real and i behaves like , is a field if we define operations as shown above.
We have already seen that the field cannot be ordered. Nevertheless, we can define an absolute value function by .
Proposition 15: Let w and z be complex numbers. Then
Proof: These are all left as exercises.
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 for j = 1, 2, then . In our 1-1 correspondence we have the four numbers corresponding to the four points , , , . 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 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 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 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 from the positive x-axis to the segment from O to z. This angle is called the argument of z and is denoted . Note that arg(z) is determined only up to a multiple of 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 .
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:
Proposition 16:(De Moivre) If a non-zero complex number z = a + bi has length r = |z| and argument and if n is any natural number, then: .
Corollary 9: Let n be a natural number.
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 -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 -roots of unity.
The ability to do arithmetic operations on the points in the plane makes a number of topics in geometry much simpler.
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 . Then adding (2, 3) to each of these points gives the set of points (x + 2, y + 3) where . 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 .
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 reflections across the y-axis by replacing x with -x. For example, the set of (x, y) where is the upper half of a parabola having the x-axis as its axis. Its reflection across the y-axis has equation . 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 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 .
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 .
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 and are transformed by a finite number or translations, rotations, and reflections, the distance between the two points does not change.
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 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 arccosine of v and is denoted either or .
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 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 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 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 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 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 in the interval such that . Note that the angles are directed in the sense that . 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 for undirected angles.
One can now prove many of the results of synthetic geometry. For example,
Proposition 18: If 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 . 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:
Proposition 19: (Addition Formulas) If A and B are two angles, then
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 ) and -A = arg(1/u) = arg( ). One has and . 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 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.
All contents © copyright 2001 K. K. Kubota. All rights reserved