# The Real Numbers

### Five Basic Axioms

We already have considered some examples of numerical statements, some true and some false. 7 + 3 = 2+ 8 is a true numerical statement, while 3+6=3∙6 is a false statement. We have also considered open sentences such as 2x + 3 = 5x which are neither true nor false until a replacement is made for the variable.

Now let us consider some examples of **algebraic statements**, some true and some false. The algebraic statement ‘‘For every x {is-in} {2,3,4}, x+1=1+x” is true since each replacement of x by a member of {2, 3, 4} results in a true numerical statement. That is, 2+ 1 = 1+ 2, 3+1=1+3, and 4+1 = 1+ 4 are all true numerical statements. On the other hand, the algebraic statement, ‘‘For every x {is-in} {1,2}, 2x+3=5x ” is false. When x is replaced by 1 a true numerical statement results, but when z is replaced by 2, we obtain the false statement, 2∙2+3=5∙2. Hence it is false that 2x + 3 = 5x for every x {is-in} {1,2} .

Let us next consider the algebraic statement, “For every x {is-in} N, x+1=1+x.” Since the replacement set for x is infinite, we cannot make every possible replacement for x. Our experience may lead us to believe that for every replacement we might make, a true numerical statement results. Therefore we will assume that this is a true algebraic statement. Similarly, we will assume that “For every x {is-in} N, x+2=2+x”, “For every x {is-in} N, x+3=3+x” etc. are all true statements.We shall call this general assumption about such algebraic statements the **commutative axiom of addition** of natural numbers. We shall represent this axiom symbolically by writing, “For every x {is-in} N and y {is-in} N, X+Y=Y+X.” Thus we are assuming that the sum of any two natural numbers does not depend on which number we consider first.

We shall make the similar assumption that the product of any two natural numbers does not depend on which number we write first. We will call this assumption the **commutative axiom of multiplication** of natural numbers and symbolize it by writing, “For every x {is-in} N and y {is-in} N, xy=yx".

Another algebraic statement which we may easily show to be true is, “For every y {is-in} {1,2,3,...,10}, (5+y)+2=5+(y+2).” For example, if y is replaced by 7 we have (5+7)+2=5+(7+2), a true numerical statement. Recall that the symbols of grouping indicate that in 5+(7+2) we first add 7 and 2. Thus 5+(7+2) = 5+9 = 14. And also (5+7)+2 = 12+2 = 14. We cannot verify that “For every y {is-in} N, (5+y)+2=5+(y+2),” but our experience leads us to believe that it is a true statement and we will assume that it is true. More generally we will assume that “For every x {is-in} N, y {is-in} N and z {is-in} N, (x+y)+z=x+(y+z)” is a true statement and call this the **associative axiom of addition** of natural numbers. Similarly, we will assume the truth of “For every x {is-in} N, y {is-in} N and z {is-in} N, (xy)z = x(yz),” and we will call this assumption the **associative axiom of multiplication** of natural numbers.

**Example 1** Decide whether or not the expressions are equivalent over N.

(a) 2+3a and 3a+2 (b) 5(a+2) and (a+2)5 (c) 2(a+3) and 2a+3

**Solution**

(a) Equivalent. In the commutative axiom of addition, x+y =y+x, replace x by 2 and y by 3a.

(b) Equivalent. In the commutative axiom of multiplication, xy=yx, replace x by 5 and y by (a+2).

(c) Not equivalent. Replace a by 1 in the given expressions.

In the last example we see that 2(a+3) is not equivalent to 2a+3, since the respective values of the expressions are not the same for every natural number replacement of a. If a is replaced by 2, for example, 2(a + 3) becomes 2(2+3)=2∙5=10 , while 2a+3 becomes 2∙2+3=4+3=7. Note that when finding the value of 2(a+3), the last operation performed is multiplication, 2∙5, while the last operation performed when finding the value of 2a+3 is addition, 4+3. For this reason we shall call 2(a+3) a **product** and 2a+3 a **sum**. Further, we shall say that 2(a+3) is a product whose **factors** are 2 and (a+3), while 2a+3 is a sum whose **terms** are 2a and 3.

**Example 2** Identify the terms of each sum and the factors of each product.

(a) 4x+4∙2 (b) 4(x+2)

**Solution**

(a) This is a sum whose terms are 4x and 4∙2. Also 4 and x are factors of 4x; 4 and 2 are factors of 4∙2.

(b) This is a product whose factors are 4 and (x+2). Also x and 2 are terms of x + 2.

As in Example 2 the expression 3x+6 is a sum of two terms. The first term has factors 3 and x, while the second term can be thought of as having factors 2 and 3. Also the expression 3(x+2) is a product of two factors. The second factor has x and 2 as terms. The words “term” and “factor” will be important in our study of algebra, and we must be careful to use them correctly. Notice that they are words which describe one expression as it relates to another expression. Thus, it makes no sense to say that 2x is either a term or a factor, but it does make sense to say that 2x is a term of the expression 2x + 3 or that 2x is a factor of the expression 2x(x+3).

**Example 3** Identify all of the terms and factors in the expression 3x+4(5x+1).

**Solution** 3x and 4(5x+1)are terms of 3x+4(5x+1).

3 and x are factors of 3x.

4 and (5x+1) are factors of 4(5x+1).

5x and 1 are terms of 5x+1.

5 and x are factors of 5x.

It would be time consuming, but possible, to show that 4(x+2) and 4x+4·2 are equivalent over {1, 2, 3, ..., 100}. For example, if x is replaced by 9, then 4(x+2)=4(9+2)=4∙11=44. Also 4x+4∙2=4∙9+4∙2=36+8=44. Let us assume that for every x {is-in} N, 4(x+2) and 4x+4·2 are equivalent expressions. Similarly, we shall assume that the statement, “For every x {is-in} N, y {is-in} N and z {is-in} N, x(y+z) = xy+xz,” and call this assumption the **distributive axiom** of natural numbers. Notice that x(y+z) is a *product* whose factors are x and (y+z), while xy+xz is a *sum* each of whose terms has the factor x. Thus the distributive axiom assumes that certain products are equivalent to certain sums. The sum, xy+xz, must always have a factor repeated in each term and that same factor, x, must also be one of the factors of the product, x(y+z).

**Example 4** Use the axiom named to find an equivalent expression over N.

(a) 3(x + 5), commutative axiom of addition

(b) 3(x + 5), commutative axiom of multiplication

(c) 3(x + 5), distributive axiom

**Solution**

(a) 3(x + 5) = 3(5 + x) (b) 3( x+ 5) = (x + 5)3 (c) 3( x + 5) = 3x + 15

**Example 5** Which axioms can be used to show that for every x {is-in} N, 4x+3x=7x?

**Solution**

4x+3x=x∙4+x∙3 [Commutative axiom of multiplication]

= x(4 + 3) [Distributive axiom]

= x∙7 [Simplifying]

= 7x [Commutative axiom of multiplication]

Example 5 suggests that the form of the distributive axiom can be modified in several ways. We shall not try to carefully establish the fact that the following true statement is a consequence of the distributive and commutative axioms. For every x {is-in} N,y {is-in} N,z {is-in} N,(x+y)z=xz+yz. In fact, we shall often refer to this statement as ‘‘the distributive axiom” even though it is merely a similar statement which follows from other assumptions.

While they have caused us no trouble, expressions such as 3 + 2 + 7 and 2∙3∙5 are, in a way, meaningless. This is because both addition and multiplication are **binary operations**—they connect a *pair* of numbers. Thus, the sum of the two numbers 3 and 2 is a third number, 5. But 3 + 2 + 7 indicates a sum of three numbers! Since (3 + 2) + 7 and 3 + (2 + 7) have the same value, let us agree that 3 + 2 + 7 will mean either (3 + 2) + 7 or 3 + (2+ 7). And we will agree that a product of three numbers such as 2∙3∙5 means either (2∙3)5 or 2(3∙5), since they have the same value. In general let us agree that x + y + z means either (x+y) +z or x+ (y+2), since the associative axiom of addition assumes that these latter two expressions have the same value for every natural number replacement of the variables. Similarly, we will agree that xyz means either (xy)z or x(yz) since the associative axiom of multiplication assumes that the latter two expressions have the same value for every natural number replacement of the variables.

Let us now summarize the five basic axioms we have assumed. Each of these is true for every natural number replacement of the variables.

### Other Basic Axioms

Let us consider a special set, {0, 1}. This set has only two members, the numbers 0 and 1. We can find all possible products using these two numbers, 0∙0=0, 0∙1=0, 1∙0=0 and 1∙1=1. Note that in every case the product is again a member of the original set {0, 1}. Because this is so, we shall say that the set {0, 1} is **closed** with respect to multiplication.

Consider again {0, 1}, but this time let us find all possible sums using the members of the set. 0 + 0 = 0, 0+1=1, 1+0=1 and 1 +1=2. Here we see that the sum of two members of the set {0, 1} is not always another *member* of that set. For this reason we shall say that {0, 1} is **not closed** with respect to addition.

Let us next return to the set, N, of natural numbers and consider whether N is closed with respect to addition or with respect to multiplication. Since N is an infinite set, we cannot possibly test every sum or product to determine whether or not each is a member of N. But neither can we exhibit a pair of natural numbers whose sum or whose product is not a natural number. This suggests that we make two additional assumptions concerning the set N. We shall assume that the sum of any two natural numbers is, again, a natural number, that the set N is closed with respect to addition. We shall call this assumption the **axiom of closure under addition** of natural numbers and express it symbolically by, “For every x {is-in} N and y {is-in} N, (x+y) {is-in} N.” We shall also assume that the product of any two natural numbers is another natural number, that the set N is closed with respect to multiplication. We shall call this assumption the **axiom of closure under multiplication** of natural numbers and express it symbolically by, ‘‘x {is-in} N and y {is-in} N, (xy) {is-in} N .”

It is also interesting to see whether Q, the set of quotient numbers, is closed with respect to addition and multiplication.

**Example 1** Find the simplest forms of these sums and products of quotient numbers.

(a) 2/3+4/5 (b) (2/3)(4/5)

**Solution**

(a) 2/3+4/5=(2∙5+3∙4)/(3∙5)=22/15 (b) 2/3∙4/5=(2∙4)/(3∙5)=8/15

The example should help us recall the general definitions for the sum and product of any two quotient numbers.

a/b+c/d=(ad+bc)/(bd) a/b∙c/d=ac/bd

As the example suggests, it is possible to use these definitions and the assumptions we have made about closure for natural numbers to *prove* that the sum and product of any two quotient numbers is again a quotient number. That is, we can prove it is true that “For every x {is-in} Q and y {is-in} Q, (x+y) {is-in} Q and (xy) {is-in} Q.” Thus the quotient numbers are closed with respect to both addition and multiplication.

**Example 2 ** Find the simplest form of each sum or product of quotient numbers.

(a) 1/2+2/3 (b) 2/3+1/2 (c) (1/4)(3/2) (d) (3/2)(1/4)

**Solution** (a) 7/6 (b) 7/6 (c) 3/8 (d) 3/8

From the example we see that 1/2+2/3=2/3+1/2 and (1/4)(3/2)=(3/2)(1/4). This suggests that these operations with quotient numbers, as well as with natural numbers, are commutative. In fact it is possible to use the general definitions for addition and multiplication of quotient numbers and the assumptions we have made about natural numbers to *prove* that “For every x {is-in} Q and y {is-in} Q, x+y=y+x and xy=yx.” There are commutative properties of both addition and multiplication for the quotient numbers.

**Example 3 ** Use the order indicated by the symbols of grouping and find the simplest form.

(a) (1/2+2/5)+1/3 (b) 1/2+(2/5+1/3)

**Solution**

(a) (1/2+2/5)+1/3=9/10+1/3=37/30

(b) 1/2+(2/5+1/3)=1/2+11/15=37/30

The example shows that (1/2+2/5)+1/3=1/2+(2/5+1/3) and suggests that addition of quotient numbers may be associative. Actually it is possible, by using the general definitions and the axioms for natural numbers, to prove that there are associative properties of both addition and multiplication with quotient numbers. ‘‘For every x {is-in} Q, y {is-in} Q and z {is-in} Q, (x+y) +z=x+ (y+z) and (xy)z = x(yz).”

**Example 4 ** Use 1/2 and 2/3 as terms and 3/4 as a factor to illustrate the distributive property of quotient numbers.

**Solution **(3/4)(1/2+2/3)=(3/4)(7/6)=7/8 and (3/4)(1/2)+(3/4)(2/3)=3/8+1/2=7/8

Again it is possible to prove that “For every x {is-in} Q, y {is-in} Q and z {is-in} Q, x(y + z) = xy + xz.” There is a general distributive property for quotient numbers.

The number 1 plays a unique role in multiplication. Let us assume that the product of any natural number and 1 is that same natural number again: ‘‘For every x {is-in} N, x∙1=x.’’ We shall describe this special multiplicative property which only the number 1 has by calling 1 the **multiplicative identity**. And we shall call our assumption the **multiplicative identity axiom**. We see at once that it is also true that “For every x {is-in} Q, x∙1=x .” For example, if x = 2/3, we have 2/3∙1=2/3∙1/1=(2∙1)/(3∙1)=2/3

Hence 1 is also the multiplicative identity for the quotient numbers.

So far we have made a total of eight general assumptions about the natural numbers—two closure axioms, two commutative axioms, two associative axioms, a distributive axiom, and a multiplicative identity axiom. And we have shown, by example at least, that these axioms lead to eight analogous properties of addition and multiplication with quotient numbers. We are now ready to make an assumption about the quotient numbers which will not be true for the natural numbers.

It often happens that the product of two quotient numbers is 1, the multiplicative identity. For example, (2/3)(3/2) = 1, (6)(1/6) = 1, etc. When the product of two numbers is 1, we will call each factor the **multiplicative inverse** of the other. Then since (6)(1/6) = 1, we call 1 the multiplicative inverse of 6, and we call 6 the multiplicative inverse of 1/6.

**Example 5** Find the multiplicative inverse.

(a) 4/9 (b) 2(3)/4 (c) 1.25 (d) 1

**Solution**

(a) 9/4 (b) 4/11 (since 2(3)/4=11/4)

(c) 4/5 (since 1.25 = 5/4) (d) 1

Let us assume that every quotient number has a multiplicative inverse which is also a quotient number. We shall call this assumption the **multiplicative inverse axiom** of quotient numbers and represent it symbolically by, “For every x {is-in} Q, there is some 1/x {is-in} Q, such that x∙(1/x)=1.” Note that the natural number 3 has a multiplicative inverse, 1/3, but this multiplicative inverse is not a natural number. It is *false* that ‘For every x {is-in} N , there is some 1/x {is-in} N, such that x∙(1/x)=1.”

Let us now consider the set * I*={..., -2, -1, 0, 1, 2,... }. In the next section we will make general definitions for the sum and product of any two integers, and we will deliberately choose those definitions in such a way that the eight general axioms we have assumed for the natural numbers will lead to eight analogous properties for the integers! Before doing so, however, we need to observe that

*contains a special number, zero, which plays a unique role in adding. Let us assume that “x + 0 =x for every x {is-in} N, x {is-in} Q, or x ∈*

**I****.” That is, we are assuming that the sum of any number (natural number, quotient number, or integer) with zero is that same number again. We shall call zero the**

*I***additive identity**and call our assumption the

**additive identity axiom**.

Another assumption about the integers has already been made when we first discussed the integers in previous tutorial. We called -1 the additive inverse of 1, since 1+-1=0. Whenever the sum of two integers is zero, the additive identity, let us call each one the

**additive inverse**of the other. We shall assume that the additive inverse of each integer is again an integer. “For every x∈

*, there is some -x∈*

**I***, such that x+-x=0.” We shall call this assumption the*

**I****additive inverse axiom**of integers.

**Example 6** For what replacement of a will the following be a true statement? “For every x∈ **I***, *x=x+a.”

**Solution** Since x = x+0 for every -x∈ * I*, we see that the only replacement for a is 0.

The example shows that whenever the sum of an integer and some other number is that same integer again, then the other number must be zero, the additive identity.

Let us assume that the integers have the same basic properties we have been discussing for the natural numbers and show how those properties can be used to prove a theorem about the integers.

**Theorem 1**

For every x∈ ,x∙0=0I |

**Proof** Each of the following is true for every x∈ * I*.

x=x∙1 [Multiplicative identity]

= x(1+ 0) [ 1 =1+0 ]

= x∙1+x∙0 [Distributive property]

= x+x∙0 [Multiplicative identity]

Then x=x+x∙0. Hence we see that the sum of x and some other number, x∙0, is x again. Example 6 shows that this other number, x∙0, must be zero. Therefore, x∙0=0.

Paired with every positive integer there is a negative integer which is its additive inverse. While 1 and -1, for example, are two distinct integers, we may emphasize the fact that they form a pair by saying that they both have the same **absolute value**, namely 1. Similarly, both 2 and -2 can be said to have absolute value 2, and the absolute value of both 3 and -3 is 3. In general we shall say that the absolute value of any positive integer is that integer, while the absolute value of a negative integer is the additive inverse of that integer. We shall indicate the absolute value of -4 in symbols by writing |-4|, and we see that |-4| = 4. Similarly, we have |-5| = 5 and |6| = 6. For the integer zero we shall agree that |0|= 0. As the next example shows, the symbols for absolute value are also considered symbols of grouping and indicate “first.”

**Example 7** Find the simplest form.

(a) |4| + |-3| (b) |4-3|

**Solution** (a) |4| + |-3|=4+3=7 (b) |4-3|=|1|=1

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