College Algebra

Chapter 1: Numbers and Points

1.1 Fields

The word number typically refers to a quantity, which may be definite as in ``the number of people in line'', indefinite as in ``a number of students took the class''; the dictionary definition typically lists over a dozen common meanings. In this course, a variety of types of numbers will be studied, including the numbers in each of the following sets:

1. The positive integers or natural numbers : 1, 2, 3, ....
2. The integers which includes the positive integers, zero, and the negative integers: -1, -2, -3, ....
3. The rational numbers is the set of numbers which can be written as m/n where m and n are integers with n non-zero. Rational numbers can be expressed as finite or repeating decimals.
4. The real numbers include the rational numbers as well as irrational numbers such as and . Every real number has a possibly infinite decimal expansion.
5. The complex numbers which include the real numbers as well as the . Every complex number can be expressed in the form where and are real numbers.

On each of these sets, is defined the binary operations addition and multiplication. For each ordered pair of numbers from one of these sets, there are well defined numbers and called the sum and the product of and in the same set. One expresses this property by saying that the sets are closed under addition and multiplication.

These operations have simple geometric interpretations. For example, if two line segments are of length x and y respectively, then connecting the two together end-to-end yields a line segment of length x + y. Similarly, the area of a rectangle with length x and width y is precisely xy.

By repeatedly combining numbers via addition and multiplication, one can make complicated expressions such as . By substituting various values of x and y into this expression, one gets numerical values for the expression. One finds that regardless of which values of x and y you use, the numerical value is the same as that obtained by the expression . The process of verifying this is one of the skills that you have already mastered in earlier algebra courses. The idea is that you can use a number of properties of numbers to successively simplify the first expression until you get to the second one. Amongst these properties are:

1. Addition and Multiplication are commutative: a + b = b + a and ab = ba.
2. Addition and Multiplication are associative: (a + b) + c = a + (b + c) and (ab)c = a(bc).
3. Multiplication distributes over addition: a(b + c) = ab + ac.

Before going on, we should clear up some ambiguity. We said that was obtained by a succession of additions and multiplications. This is true, but there are many different ways of doing this, e.g. after one gets the value of , , and , one could add the third to the sum of the first two or add the first to the sum of the last two. Of course, the result would be the same, and it is the associative law for addition that guarantees this. For this reason, one typically does not even bother to specify the order by adding in parentheses. Similarly, one didn't put parentheses to indicate the order of evaluation of the product of the three factors in . Another ambiguity occurs in the expression . Does this mean add the product of 4 and x to y or does it mean multiply 4 times the sum of x and y? This is more serious because the two ways of evaluating the expression give different answers. This is resolved by the rules for order of evaluation of expressions. In this case, the rules say that you evaluate multiplications before additions. So, the meaning of the expression is and not . The rules for order of evaluation will be stated in detail at the end of the next section.

Exercise 1.1: Using the geometric interpretation of addition and multiplication given above for positive real numbers, give geometric interpretations of the commutative, associative, and distributive laws. For example, the commutative law for addition might be illustrated by saying that if you join a line segment of length x end-to-end with a line segment of length y, then the length of the result is the same whether you measure it from one end or from the other.

Here are two more important properties of the rational numbers , the real numbers , and the complex numbers :

1. There is a number 0 such that a + 0 = a for all numbers a and there is a number 1 not equal to 0 such that for all numbers a. Any such numbers are called additive or multiplicative identities.
2. For any number , there is a number such that . Such a number is called an additive inverse of . Similarly, if is number other than 0, then there is a number called a multiplicative inverse or reciprocal of such that .

The numbers 0 and 1 are unique:

Proposition 1.1: There is at most one additive identity and at most one multiplicative identity.

Proof: Suppose and are two additive identities. Letting play the role of in the definition of an additive identity , we have . Similarly, one has by letting play the role of in the definition of being an additive identity. Because of the commutative law, one has The same argument shows that multiplicative identities are also unique.

We can define subtraction by and division by if b is non-zero. Alternative ways of denoting division are: .

Exercise 1.2:

1. Which of the five sets of numbers , , , , and are closed under subtraction? Which are closed under division?
2. Are subtraction and/or division commutative and/or associative?
3. Does addition distribute over multiplication? Does division distribute over addition?
4. Do the natural numbers and/or the integers have additive and/or multiplicative identities? What about inverses?

Any set closed under addition and multiplication which satisfies the commutative, associative, and distributive laws as well as has identities and inverses is referred to as a field. Here is the formal definition:

Definition 1.1: A field is a set F together with with two binary operations '+' and '·' called addition and multiplication which satisfy the following conditions:

1. F is closed under addition and multiplication.
2. Addition and multiplication are commutative. This means that and for all a and b in F.
3. Addition and multiplication are associative. This means that and for all a, b, and c in F.
4. Multiplication distributes over addition. This means that for all a, b, and c in F.
5. (Identities) There is an element 0 in F such that a + 0 = a for all . There is an element 1 in F different from 0 such that for all .
6. (Inverses) For every in F, there is an element denoted in F such that . For every in F other than 0, there is an element denoted in F such that .

Remark: If all the conditions except for the requirement that multiplicative inverse exist are true, then F is called a commutative ring with identity.

Many of the common rules of algebra follow from the fact that one is working in a field. One very important example is:

Proposition 1.2: Let F be a field. If is in F, then . If and in F satisfy , then either or are zero.

Proof: Let be an arbitrary element of 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 1.1: If a is an element of a field F, then -a = (-1)a.

Proof: .

Corollary 1.2: 0 has no multiplicative inverse.

Proof: If this were false, then one would have But, Combining these, we conclude that . But this contradicts the property that 0 and 1 are not equal.

Because the field properties imply most of the rules of algebra, they help us understand why these rules are true. For example, you know that the product of two negative numbers is always positive. But why should this be true? Why not make a new rule, e.g. that the product of two negative numbers should be negative. The answer is that you could do this, but then one of the field properties would no longer be true.

Exercise 1.3: Show that in a field, one has (-a)(-b) = ab.

Hint: Fill in reasons for each of the following steps:

Here are some more common rules of algebra:

Proposition 1.3: Let F be a field containing a, b, c, and d where b and d are non-zero. Then

1. 
2. if and only if
3. 
4. 
5. 
6. If c is also non-zero, then

Proof: (i) By the definition of multiplicative inverse, one must show that Starting from the left hand side, one can simplify the expression as shown here.

Each of the above steps involves one of the fundamental properties of fields; make sure that you can justify each step with the appropriate property.

(ii) This like any "if and only if" statement is really two assertions:

1. If then
2. If then

To prove assertion (a), assume that To see that , multiply both sides of by to get . Now simplify each side:

and

Combining results, one gets (As before, make sure that you can justify each step.)

Now, let's show assertion (b). Assuming that and that both and are non-zero, we know that both and have multiplicative inverses. So, we need only multiply both sides of by and simplify. Here are the details:

and

(Label each line with the property that justifies the step.)

(iii) Simplify starting from the left hand side:

Did you justify each step?

(iv) One has:

(v) This is the usual rule for adding fractions. Notice that property (iv) allows one to convert the fractions so that they have a common denominator . Here are the detailed steps:

(vi) This is the rule for simplifying fractions of fractions. It is the same as

Using property (ii), this is the same as showing

But, we can see this by simplifying each side:

and

This completes the proof of Proposition 1.3.

Exercise 1.3: (i) Redefine addition on the real numbers:

With the new definition of addition and the usual multiplication, do the real numbers still form a field? Which of the field properties are true and which are not.

We defined multiplication by making be the area of a rectangle with sides of length m and n respectively. Suppose that we based things on the area of a triangle with given base and height instead. With the new definition of multiplication (and the usual definition of addition), do the real numbers form a field? What is the multiplicative identity?

1.2 Induction and Order of Operations

1.2.1 Mathematical Induction

The positive integers are the numbers 1, 2, 3, .... Every positive integer n has a successor n' = n + 1. Every positive integer n except 1 is the successor of a unique positive integer, viz. n - 1. The most important property of the set of positive integers is

Principle of Mathematical Induction: If is any property of the positive number such that

1. is true.
2. If is true for a positive integer , then is true for the successor of

This principle can be used for both definitions as well as for theorems. For example, one can use it to define the usual operations on the positive integers using only the successor operation:

Definition 2.1: The sum m + n of two positive integers m and n is defined by:

1. m + 1 is defined to be the successor m' of m.
2. m + n' is defined to be the successor (m + n)' of m + n.

Let be the property of the positive integer n that m + n is defined. Then (i) assures us that is true and (ii) assures us that is true of n' if it is true of n. So, is true for all positive integers n. Similarly, one sees that we can define multiplication by:

Definition 2.2: The product of two positive integers and is defined by:

1. for all positive integers
2. If is defined, then

One can also use the principle of mathematical induction to prove theorems. For example, we can verify that our addition operation is associative:

Proposition 2.1: Addition is associative.

Proof: Let be the property of the positive integer n that (a+ b) + n = a + (b + n) for all positive integers a and b. In the case where n = 1, is true because

(Why is each step true?)

Assuming that is true for n, let's show it for n'. One has

So, must be true for all positive integers

Proposition 2.2: Addition is commutative.

Proof: Let be the property of the positive integer n that a + n = n + a for all positive integers a. We will show that is true by using the principle of mathematical induction. Let be the property of the positive integer a that a + 1 = 1 + a. We know that is true because 1 + 1 = 1 + 1. If is true for some positive integer a, then

and so is true for all positive integers and so is true.

Now suppose that is true for n. Then

and so is true for all positive integers n. (Did you justify each step?)

Proposition 2.3Multiplication distributes over addition.

Proof: Let be the property of the positive integer n that a(b + n) = ab + an for all positive integers a and b. Then is true. because

Suppose that is true for n. Then it is also true for n' because

and so is true for all positive integers n.

Exercise 2.1:Use induction to prove the following results:

1. Multiplication is associative.
2. (a + b)c = ac + bc for all positive integers a, b, and c.
3. Multiplication is commutative.

Exercise 2.2: (Laws of Exponents)

1. Use induction to define where a is number (real or complex) and n is a positive integer.
2. Use induction to prove that where a is a number and m and n are positive integers.
3. Use induction to prove that where a is a number and m and n are positive integers.
4. Assuming that you want these properties to hold for all integers m and n, what are the only possible definitions for ? What about where n is a negative integer?

Exercise 2.3: (Sums of Powers)

1. Show by induction that the sum of the first n positive integers is precisely n(n+1)/2.
2. Show by induction that the sum of the squares of the first n positive integers is precisely n(n + 1)(2n + 1)/6.

1.2.2 Binomial Theorem

It will often be useful to expand powers of a binomial a + b. For small positive integer value of n, you can do this using the distributive law. For example,

Exercise 2.4: Show that

It should be clear that for any given positive integer n, we can obtain a formula for ; but also that it is becoming more and more tedious as n increases. The goal is to find a general formula. Here is a first approximation:

Proposition 2.4: (Binomial Theorem) For each positive integer n, one can write

where the coefficients are all positive integers. In fact, one has for k = 1, ..., n-1 and

Proof: Define by induction using the formulas in the statement of the proposition. Let be the property of the positive integer n which is true provided that the can be expressed as in the statement of the proposition. Then is clearly true because

Assume that is true for n so that

So is true for all positive integers n.

The equations defining the coefficients appear quite complicated, however, if we write them down in the form of a triangle with the row containing for k = 0, 1, ..., n from left to right:

                    1
                  1   1
                1   2   1
              1   3   3   1
            1   4   6   4   1
          1   5  10   10  5   1
       1    6  15  20   15  6   1

Exercise 2.5: Define n! (pronounced n factorial) by 0! = 1! = 1 and (n+1)! = (n+1)n!. So, n! is the product of all positive integers less than or equal to n.

Define for k = 0, 1, ..., n. Show by induction that for every positive integer n and k = 0, 1, ..., n.

Definition 2.3: The numbers are called binomial coefficients.

With this notation, the binomial theorem can be stated as:

Another interpretation of the number is the number of combinations of objects taken at a time which is a formal way of saying the number of subsets of elements which can be formed from a given set of elements.

Exercise 2.6: In Pascal's triangle, note that each row reads the same from left to right as from right to left, i.e. is palindromic. Express this relation as an equation involving the . Explain why the relation is true.

1.2.3 Order of Operation

The rules for order of operations are used to determine the precise order in which operations are to be carried out when evaluating an expression. Using them allows one to write expressions with far fewer parentheses making them both more readable and less error prone. On the other hand, the rather large number of operators makes for a rather complicated set of rules. The goal is to make your expressions readable and correct; sometimes it is better and clearer to add a set of parentheses even though the rules indicate that they are not really necessary.

Here are the rules. You should read through the rules now and start using them. Because they are complicated, you will need to refer back to them a few times before they become clear. I am purposely not specifying exactly which numbers we are talking about because we want the material to apply to any of several types of algebraic quantities. Variables like x or y are letters that stand for numbers. By parentheses we mean any of the usual types of parentheses seen in algebraic expressions; they include rounded parentheses (), square brackets [], and curly brackets {}; in the rules below, we will use the rounded parentheses, but any other kind can be used as well.

Definition 2.4: An expression is a particular kind of string of numbers, variables, operators, and parentheses. The following rules can be used to determine if such a string is an expression:

1. Every number and every variable is an expression.
2. If E is an expression, then so is ( E ).
3. If E is an expression, then so is -E.
4. If E₁ and E₂ are expressions and op is any of the binary relations '+', '-', '·', '÷', and '^'then E₁ op E₂ is also an expression.
5. The only strings which are expressions are those which can be shown to be expressions by applying the above rules a certain number of times.

Some fine points:

1. Sometimes we will omit the multiplication operator ·. You can handle these expressions either by adding in the operator or by defining a new operator which is simply a space character.
2. One can add the alternative slash '/' character as an alternate for the division symbol '÷'.
3. Division is also often indicated with a horizontal bar. To handle this, simply replace the expression with the numerator between parentheses followed by ÷ and the denominator between parentheses.
4. Sometimes, one does not assume that the exponentation operator is evaluated from left to right. In such cases, you need to be sure to put in additional parentheses so that the order of evaluation is explicit.

Any expression has a well defined value. To guarantee this property, one needs to agree on the order in which an expression is evaluated. For example, if one evaluates the addition in 4 + 5 · 3 before the multiplication, the value is 27; but evaluating the multiplication first gives a value of 19. In order to avoid this kind of ambiguity, expressions are always evaluated using the so-called order of operations. The rule is:

1. Evaluate parenthesized subexpressions starting from the first innermost parenthesized expression. If E has value v, then the value of (E) is also v. Use the remaining rules in order to evaluate a subexpression with no parentheses.
2. The value of a number is the number itself; the value of a variable is the number which it represents.
3. Powers should be evaluated first, from left to right.
4. Unary minus operations should be evaluated next starting from the innermost.
5. Multiplications and divisions should be done next; do this left to right.
6. Additions and subtractions should be done next; again do this left to right.

1.3 Unique Factorization

1.3.1 Order

For the positive integers , the order relation is defined by

Definition 3.1: If and are positive integers, then is smaller than (or is larger than ) if there is a positive integer d such that . One writes this as or .

Here are the main properties of the order relation:

Proposition 3.1: Let and be any postive integers.

1. (Trichotomy) Exactly one of these is true: or
2. (Transitivity) If and , then
3. If then
4. If then

Proof: The hard part is to prove trichotomy. So, let's assume that we have proven it and use it to prove the other assertions.

(ii) If and , then there arepositive integers and with and . But then

and so

(iii) If then there is a positive integer with But then using the associative and commutative laws, one has

and so

(iv) If then there is a positive integer with But then the commutative and distributive laws shwo that

and so

(i) Now, let's tackle the problem of showing that trichotomy is true. Let's start with:

Lemma 3.1: Let r be a positive integer. There is no positive integer m such that

Proof: We prove this by induction using the property which is true of m provided that We know that is true because 1 is not the successor of any positive integer. Now suppose that is true for If were not true for its successor then But then

and so because each positive integer is the successor of at most one positive integer. This contradiction shows that must be true for and so the lemma is true.

We can use the lemma to show that no two of the conditions , , and can be simultaneously true. There are three cases:

1. If and , then
2. If and , then
3. If and , then
   

It remains to show that at least one of the three conditions must always be true. Let m be a fixed positive integer and be the property of a positive integer n that is true provided that at least one of and is true. is true because either or else for some positive integer p; in the second case, and so

If is true for the positive integer n, then there are three cases:

1. If then there is an r with But then
   

   and so

2. If then and so .
3. If there is a positive integer with If then On the other hand, if is not 1, then it is the successor of some positive integer, say and so One has
   

   and so

Exercise 3.1: Prove that if then

Definition 3.2: If then the difference is defined to be

1.3.2 Descent

Descent is a variant on the principle of mathematical induction:

Descent Principle: Let be a property which may or may not be true of the positive integer n. If is false for at least one positive integer then there is a smallest integer for which is false, i.e. is false for and true for all with .

Proof: Suppose the descent principle is false for some property . Let be the property of the positive integer which is true provided that is true for all positive integers less than or equal to must be true; otherwise, would be false and clearly n = 1 would be the smallest number for which would be false because 1 is smaller than every positive integer except 1.

Now suppose that is true for If were not true, then would be false but would be true for all numbers k less than . But then would be the smallest positive integer for which was false which contradicts the assumption that the descent principle is false for .

1.3.3 Unique Factorization

Definition 3.2:Let and be positive integers.

1. is a factor of (or is a multiple of ) if there is a positive integer such that A factor of is a proper factor if
2. is prime if it is not equal to 1 and has no proper factors.

Examples of primes are 2, 3, 5, 7, 11,....

Exercise 3.2: Replace the order relation in the statement of Proposition 3.1 with the property that be a factor of Which of the conclusions of the Proposition are still true? Which are false?

Exercise 3.3: Show that if is a factor of then

Proposition 3.2 Every positive integer can be factored into a product of primes. In other words, if is a positive integer, then there are prime positive integers ..., (where is a non-negative integer) such that

Proof: This is a proof by descent. If the Proposition is false, then there is a smallest positive integer for which the assertion is false. cannot be equal to 1 or prime and so it can be written as product of proper factors: Since each of the factors is smaller than , one has and where the and are primes. But then

Proposition 3.3 (Euclid) There are infinitely many primes.

Proof: If not, then there are only finitely many. Let ..., be the complete list of them. Consider the positive integer Let be any prime factor of we know that for some = 1, 2, ..., Then But then which means that is a prime factor of 1. By Exercise 3.3, it follows that is less than or equal to 1 and so must be equal to 1 contrary to the assumption that is prime.

Clearly, positive integers might be factored in more than one way as product of primes, e.g.

As it turns out, the factorization is unique except for the order of the factors (and the way it is parenthesized):

Proposition 3.4 (Division Theorem) If and are positive integers, then there are non-negative integers and such that and Furthermore, the numbers and are unique.

Proof: (i) Existence: Let and be positive integers for which there are no such numbers and For the fixed value choose to be the smallest such that and do not exist. We cannot have because that would allow for Also, one cannot have as this would allow for By trichotomy, it follows that and so for some positive integer But then and so there must be non-negative numbers and such that and and But then

contrary to the assumption that no appropriate and exist.

(ii) Uniqueness: Suppose where and are non-negative integers with and If then and so Otherwise, assume that (swapping the roles of q, r and u, v if necessary. Re-arranging, we get which is a contradiction because the right hand side is smaller than n and the left hand side is greater than equal to n.

Theorem 3.1 (Fundamental Theorem of Arithmetic) Every positive integer can be factored as a product of primes and this factorization is unique up to order of the factors.

Proof: By Proposition 3.2, we need only show the uniqueness assertion. Let us begin with

Lemma 3.2 If is a prime factor of the product then is a factor of either or (or both).

Proof: If this is not true in general, then let be the smallest prime for which there is a counter-example. Of all such counter-examples choose one with the smallest possible value for , and of all these choose one with the smallest possible value for n. Using the division theorem, we know that one can write and where and are smaller than Since , we see that is a factor of and cannot be a factor of either or of without being a factor of or respectively. So, we can assume by the choice of and that they are both smaller than Since is a factor of there is a with . Further, and so But then every prime factor of must be smaller than . We have a factor of and so by the minimality of it must be that is either a factor of or of In either case, we could divide both sides of by and get a contradiction with the minimality of the choice of either or So, it must be that has no prime factors and so But then contradicting the fact that is a prime. This proves the lemma.

Corollary 3.1: If is a prime factor of then is a prime factor of at least one of the factors ...,

Proof: Suppose not. Choose a counter-example with the smallest possible value of . Then divides and so Lemma 3.2 tells us that is either a factor of or of . But, it can't be a factor of the second quantity because this would give us a counter-example with a smaller . So, must be a factor of contrary to assumption.

If factorization can be non-unique, then let be the smallest positive integer which has at least two factorizations into products of primes differing other than in the order of the factors. No prime can occur in both factorizations; otherwise, would be a smaller positive integer with at least two factorizations. If is any prime factor of the first factorization, then is a factor of some prime of the second factorization by Corollary 3.1. So for some positive integer Since is a prime, it must be that and so contrary to assumption.

1.4 Numerals and Geometric Series

A numeral is a symbol used to represent a number. The standard numeration system is the Arabic numeral system. A base 10 Arabic numeral is a sequence of digits 0, 1, 2, 3, ..., 9. The numeral where each is a digit represents the number

For example, 3147 means . Decimal numerals are similar, e.g. .

Sometimes it is more convenient to use bases other than 10. A base b Arabic numeral (where b is a positive integer bigger than 1) uses digits 0, 1, ..., b-1 and the numeral where each is a digit represents the number

The commonly used bases other than 10 are 2 and 16. Base 2 numerals are called binary numbers and base 16 numerals are called hexadecimal numbers. Hexadecimal numbers use the letters A through F (or a through f) to denote the digits 10 through 15 respectively. For example, the decimal number 3147 is equal to the Number C4B in hexadecimal because 11 + 16(4) + 16²(12) = 3147. You can check that in binary, the same number is 110001001011. One can also work with decimals in other bases as well; for lack of a better term, base 2 decimals will be referred to as binary decimals.

The usual algorithms for doing operations with base 10 numbers work fine in other bases. However, most people find it easier to convert the numbers to base 10, do the computation in base 10, and convert the answer back to the desired base. Doing base conversions is easy. We have already seen how to convert to base 10. If you wanted to convert to say base 16, then take the number and divide by 16. The remainder is the lowest digit. Dividing the quotient by 16 gives a remainder which is the next digit, etc.

Example 4.1: To convert the base 10 number 3145 to hexadecimal, divide 3145 by 16 to get So the lowest digit is eleven which is denoted B. Next divide 196 by 16 to get and so the next digit is 4. Finally, and so the high order digit is twelve or C. The number 3147 in base 10 is equal to the number C4B in base 16.

Example 4.2:Calculate

1. First, let's calculate it using the same algorithm as one uses in school. Here is the calculation:

             C4B
             C4B
             ---
            8739
           312C
          9384
          ------
          971DF9
   

   To understand this, recall that and so by the distributive law:

   

   The three terms on the right are written above in the three lines between the horizontal bars. Instead of writing multiplying the lines by 10 (hex) and (hex), the numbers are simply written one and two columns to the left of the where they would normally be written.

   To see how one calculates , write this as

   

   Verify that this is precisely what you do when you do multiplication in base 10; then check the other two subproducts and the final addition.

   Now check your work: We know that C4B is 3145 in base 10. So its square in base 10 is in base 10 which converted to base 16 is 971DF9. Check the computation! Of course, if the two answers do not match, there must be an error in the computation.

   Repeating decimals are possibly infinite decimals whose digits eventually start to repeat themselves, numbers like 3.14141414... Such numbers are actually rational numbers. To see this, we will need:

   Proposition 4.1: (Geometric Series) One has

   

   In particular, if x is a number less than 1 in absolute value, then

   

   Proof: Use the distributive law to expand out the left side of the first equation. All but two of the terms subtract out leaving the two terms on the right hand side. To see the second assertion, divide both sides of the first equation by . As grows larger and larger, the term approaches 0 because .

   We can use the Proposition to evaluate our infinite decimal:

   

   A similar argument shows that any repeating decimal represents a rational number. Conversely, if you have a rational number p/q, then you can expand it out into a decimal by using the standard long division algorithm. At each stage in the computation, the remainder is a number between 0 and q - 1. As soon as the remainder repeats, the sequence of digits also repeat; so, the decimal expansion of a rational number is repeating. The formal result is:

   Proposition 4.2: Every repeating decimal represents a rational number and every rational number has a repeating decimal expansion.

   Although we argued in base 10 numerals, everything works anlaogously regardless of which base we work in.

   Corollary 4.1: There exist real numbers which are not rational.

   Proof: Any decimal expansion which is not repeating represents a real number which is not rational.

   Remark 4.1: The square root of 2 is irrational. For otherwise, we could write where and are integers with non-zero. Squaring both sides and multiplying through by , one gets . Now express and as products of primes and substitute these into this last equation. Because integers factor uniquely, we have a contradiction, because the left side of the equation would have an odd number of factors of 2 and the right side would have an even number of factors of 2.

   1.5 Ordered Fields

   Definition 5.1: An ordered field F is a field (i.e. a set with addition and multiplication satisfying the conditions of Definition 2) 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 2, 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 5.1: 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 3, exactly one of a < 0, a = 0, and 0 < a must be true. If a < 0, then by property iii of Definition 3, 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.

   (ii) Suppose a and b are in P. Then 0 < a and by property iii of Definition 3, we have 0 + b < a + b and . Since 0 < b and b < a + b, transitivity implies that 0 < a + b. Since 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 8, 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:

   

   Proposition 5.2: Let a and b be elements of an ordered field F.

   1. |-a| = |a|
   2. (i.e. and )
   3. (Triangle Inequality)

   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.

   1.6 Decimal Expansions

   As noted earlier, one can write numerals in any integer base b > 1. The choice b = 10 has the advantage of being most familiar; but choosing b = 2 often makes the proofs a bit simpler - basically, each additional digit cuts an interval in two equal pieces which is easier to handle than 10 equal pieces. For this section, we will use base 10; but after reading it, you should go back and verify that everything works regardless of the choice of the base. In some later sections, we will use base 2 in order to have simpler proofs. We start by formalizing our notion of infinite decimal.

   Definition 6.1: 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 6.2: 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 6.1: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 2: 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 6.2: 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 . Now consider the set S of all numbers m with . Clearly 0 is in this set. If every successor of an element of S lies in S, then the principle of mathematical induction would imply that S would be the set of all non-negative numbers contrary to the fact that we have already identified a number M not in S. So, let be an element of S such that is not in S. We know 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 .

   1.7 Complex Numbers

   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:

   

   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.

   Proposition 7.1: 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 . The conjugate of a complex number is a + bi is defined to be a - bi and is denoted .

   Proposition 7.2: 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. (Triangle Inequality) .
   5. .
   6. 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 .

   Proof: These are all left as exercises except for the triangle inequality. To prove the triangle inequality, let and where and are real numbers. To show the triangle inequality, it is enough to show that Substituting in the values for and one sees that this will hold provided that

   

   This simplifies to the equivalent inequality

   

   This in turn would be true if the square of the left side were less than or equal to the square of the right side, i.e.

   

   which is equivalent to But subtracting from both sides and factoring gives

   

   which is obviously true. So, the triangle inequality is also true.

   Exercise 7.1 (i) Prove the rest of Proposition 7.2

   Prove that in the triangle inequality, one has equality if and only if one of and is a non-negative real multiple of the other.

   1.8 Solving Equations

   Solving equations of one variable uses two basic principles:

   1. If a = b, then a + c = b + c and a · c = b · c.
   2. If ab = 0, then a = 0 or b = 0.

   Example 8.1: To solve the general linear equation ax + b = c for x where a, b, and c are constants with a non-zero, one can assume that x is a solution so that ax + b = c. Add -b to both sides of the equation to get (ax + b) + (-b) = c + (-b). This simplifies to ax = c - b. Multiplying both sides by a^-1 and simplifying gives x = a^-1(c - b). We have shown that this is the only possible solution. Substituting it into the original equation and simplifying verifies that this value is indeed a solution. So, the equation has the unique solution x = a^-1(c - b).

   Example 8.2: To solve the general quadratic equation ax² + bx + c = 0, we can restrict ourselves to the case where a is non-zero. (Otherwise, the equation is linear and Example 3 applies.) If x is a solution of the equation, then one can divide both sides by a and simplify to get x² + (b/a)x + c/a = 0.

   If 2 is not zero, then one can complete the square to get (x + b/(2a))² = (b/(2a))² - c/a. If the right side was square, then we could solve for x to get

   

   Furthermore, one can check that this actually is a solution of the original equation. This last equation is known as the quadratic formula.

   In particular, if we are working in the field of real numbers, then we have completely solved the quadratic; there are two, one, or no solutions when b² - 4ac is positive, zero, or negative respectively. In the case of the field of complex numbers, things are even simpler as we will show that all complex numbers have square roots.

   Another example is the general system of 2 linear equations in two unknowns x and y: ax + by = e, cx + dy = f. What makes the general system of 2 linear equations appear difficult is that both equations involve both variables. There are two approaches:

   1. One could solve the first equation for one of the variables in terms of the other. Then substitute this into the second equation giving an equation in only the second variable. Then proceed as in the easy case.
   2. One could subtract an appropriately chosen multiple of one equation from the other in order to obtain an equation involving only one variable.
   As before, one needs to check that the possible solutions one obtains do indeed satisfy the original equations.

   Example 8.3: Consider the system: 2x + 3y = 5, 4x - 7y = -3. Assume that x and y are satisfy both equations. One can proceed using either method:

   1. Solving for x using the first equation, gives x = (5/2) - (3/2)y. Substituting this into the second equation gives 4((5/2) - (3/2)y) - 7y = -3, which can be used to find y = 1. Substituting this back into our expression for x yields x = 1. One then checks that the values x = 1, y = 1 do indeed satisfy the original equations.
   2. If one subtracts twice the first equation from the second equation, one gets (4x - 7y) - 2(2x + 3y) = -3 - 2(5) or -13y = -13. This gives y = 1 and substituting this value back into the first equation gives 2x + 3 = 5 or x = 1. As before, we need to check that x = 1, y = 1 does indeed satisfy the original two equations.

   Example 8.4: Find all the solutions of the system of equations: , . Assume that one has a solution x, y. Solving the second equation for y, one gets a value y = 1 -x which when substituted into the firs equation gives: or . Collecting terms, we get a quadratic . Factoring and using the second part of Proposition 1, gives x = 0 or x = 1. Substituting these values into our expression for y, gives two possible solutions (x,y) = (0, 1) and (x,y) = (1, 0). Substituting each of these into the original equations, verifies that both of these pairs are solutions of the original system of equations.

   If we have more than two variables and more equations, we can apply the same basic strategies. For example, if you have three linear equations in three unknowns, you can use one of them to solve for one variable in terms of the other two. Substituting this expression into the two remaining equations gives two equations in two unknowns. This system can be solved by the method we just described. Then the solutions can be substituted back into the expression for the first variable to find all possible solutions. When you have these, substitute each triple of numbers into the original equations to see which of the possibilities are really solutions. One can also use the second approach as is illustrated by the next example.

   Example 8.5: Solve the system of equations: x + y + z = 0, x + 2y + 2z = 2, x - 2y + 2z = 4. Assume that (x, y, z) is a solution. Subtracting the first equation from each of the other two equations gives y + z = 2, -3y + z = 4. Now subtracting the first of these from the second gives -4y = 2 or y = -1/2. Substituting this into the y + z = 2 gives z = 5/2. Finally, substituting these into the first of the original equations gives x = -2. So the only possible solution is (x, y, z) = (-2, -1/2, 5/2). Substituting these values into the original equations shows that this possible solution is, in fact, a solution of the original system.

   1.9 Number Line and Plane

   This section is an informal review of some elementary notions of analytic geometry. Let's start with the number line. Choose a line and mark off two points 0 and 1 on the line. By marking off segments of length equal to that between 0 and 1, one can define points 2, 3, etc. Moving in the other direction, one gets -1, -2, etc. To each rational number, one can associate a point on the line; e.g. 1/2 is the point midway between 0 and 1; 7/4 is the point a quarter of the way between 0 and 7, etc. Real numbers can be represented as infinite decimals; we can associate them with the points obtained as limits of the finite decimals obtained by throwing away the tail end of the decimal. For example, 1.2121212... is the point which is the limit of the points 1, 1.2, 1.21, 1.212, etc. At least intuitively, it appears that we have defined a one-to-one correspondence between the real numbers and the set of points on the line.

   To name the points in the plane, simply use the cartesian product of the real numbers with itself. Geometrically, this corresponds to taking two perpendicular number lines intersecting at 0. These are called the x-axis and y-axis respectively. Each point on the plane can be projected perpendicularly onto each of the two axes. The name the point is the ordered pair (x, y) where x is the number associated with the projection of the point on the x-axis and y is the number associated with the projection of the point on the y-axis. The diagram below shows the point (3, 2):

   

   Recall that the complex numbers can be written in the form a + b i where and a and b are real numbers. So, a complex number is essentially the same thing as an ordered pair (a, b) of real numbers. This allows one to think of the plane as being the set of real numbers. The number a is called the real part and the number b is called the imaginary part of the complex number a + b i. In the above diagram, the point (3, 2) corresponds to the complex number 3 + 2i.

   Proposition 9.1: (Distance Formula) If and are two points in the plane, then the distance between them is .

   

   Proof: Let . Then is a right triangle with right angle at . The legs have length and respectively. The distance formula now follows by the Pythagorean Theorem. (This theorem will be proved in a next chapter.)

   If (a, b) and (c, d) are two points in the plane, then one can define their sum to be (a + c, b + d). Note that this is precisely the point corresponding to the sum z + w of the complex numbers z = a + bi and w = c + di corresponding to the two points.

   The addition is the so-called parallelogram law Let L be the line segment from the origin (0, 0) to (a, b) and M be the line segment from the origin to (c, d). If we move M parallel to itself so that it starts at (a, b), then its other end-point will be at (a + c, b + d). So the line segments R from (a, b) to (a + c, b + d) and S from (c, d) to (a + c, b + d) combine with L and M to make a parallelogram whose diagonal starts from the origin and ends at the sum of z = (a, b) and w = (c, d). You can also think of this as a triangle rule: To add z and w, start with a line segment from the origin to (a, b), then move the line segment from the origin to (c, d) parallel to itself until it is starting from (a, b); the final end-point is the sum z + w.

   Let and be any two complex numbers with We can think of as defining a line segment M from the origin to If is a positive real number, then corresponds to a line segment obtained by stretching M by a factor of t. If is a negative real number, then the segment is streched by the a factor of but goes in the oppositive direction as does M. When you add to it, the complex numbers in the set

   

   make up a line through and parallel to M.

   Definition 9.1 A line is any set of points of the form

   

   where and are complex numbers with The corresponding set of points in the plane are also referred to as a line.

   Proposition 9.2: (Midpoint Formula) If and are two points in the plane, then the point . is on a line through and and the distance between and either end-point is equal to half the distance between the end-points.

   Proof: The line is the one defined by and Letting t = 1/2 gives the point M. The assertions about the distance are easy to verify using the distance formula.

   1.10 Functions

   Intuitively speaking, a function is a rule which associates with each element of a set an element of a set . The set is called the domain of the function and the set is called the codomain of the function. A function with domain and codomain is often denoted and the number in associated by the rule to the element in is denoted The range of is the set of all elements of that are associated to at least one element of the domain

   For example, function which squares each real number is a function with domain and codomain both equal to the field of real numbers. For each real number one has and the range of the function is the set of non-negative real numbers. On the other hand, the square root function has domain the non-negative real numbers and The codomain might be the set of non-negative real numbers or any set containing this set.

   For any function one can form the set of all ordered pairs where is in the domain of This set

   

   is called the graph of the function. In the special case where the function has domain and codomain contained in the set of real numbers, the graph of f can be thought of as a subset of the number plane.

   Remark 1.10.1 One should think of the graph of a function as a visual representation of the function. In the special case in which the domain and codomain are subsets of the real numbers, one has:

   1. The domain is the set of x-coordinates of points in the graph.
   2. The range is the set of y-coordinates of points in the graph.
   3. Every vertical line intersects the graph in at most one point.
   The last property is often referred to as the vertical line test

   Example 1.10.1 Not every subset of the number plane is the graph of a function. For example, the circle with center at the origin and radius 1 is not the graph of a function. This is because there are vertical lines which intersect the graph in more than one point. For example, the y-axis intersects the circle at which means that there cannot be a rule which associates to 0 a single number and still have both of these points in the graph. On the other hand, the part of the circle which has y-coordinate non-negative is the graph of the function which assigns to every in the closed interval [-1, 1] the value

   Many times the function will be ambiguously specified by simply giving the rule for associating elements of the domain with elements of the codomain, without specifying a domain and codomain. In this case, one normally assumes one is dealing with the largest domain for which the rule makes sense, and this domain is referred to as the natural domain.

   For example, the natural domain of the squaring function is the set of all real numbers, even though there are other functions with domains any specific subset of the real numbers. In some cases, one needs to intuit the meaning of what one means by the largest domain for which the rule makes sense; for example, the squaring function is also defined on the field of complex numbers.

   Remark 1.10.2 Because it is difficult to formalize what one means by a rule, a formal definition of function usually is a definition of the graph of the function. For example, one could say that a function with domain and codomain is any subset of the set of all ordered pairs where and such that for every there is exactly one ordered pair in first coordinate equal to One then writes to indicate that

   1.10.1 Operations on Functions

   Let and be functions with domains some set codomains some field Then one can combine and to form new functions:

   1. The sum of and is defined by
      

   2. The difference of and is defined by
      

   3. The product of and is defined by
      

   4. The quotient of and is defined by
      

   Note that the domain of the first three functions is but the domain of the fourth function is the set of for which

   Finally, if and are two functions, then the composition of the two functions is defined by

   

   Example 1.10.2 If is the squaring function and is the cubing function, then is the function Also, is the function with The function is the reciprocal function, which is only defined for non-zero real numbers. Finally, the composition is the function which raises numbers to their sixth power.

   Exercise 1.10.1

   1. Show that the operations sum and product are commutative but that the other three operations are not.
   2. Show that the operations sum, product, and composition are associative, but the other two are not.

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   Revised: September 7, 2003
   All contents © copyright 2002, 2003 K. K. Kubota. All rights reserved