Construction and Categoricity of the Real Number System Using Decimals
aa r X i v : . [ m a t h . G M ] J un Construction and Categoricity of the Real Number System Using Decimals
Arindama SinghDepartment of MathematicsIndian Institute of Technology MadrasChennai-600036, IndiaEmail: [email protected] 27, 2019
Abstract
In this expository article, the real numbers are defined as infinite decimals. After definingan ordering relation and the arithmetic operations, it is shown that the set of real numbers is acomplete ordered field. It is further shown that any complete ordered field is isomorphic to theconstructed set of real numbers.
Keywords: Real numbers, Decimals, Categoricity
The modern definition of the system of real numbers posits that it is a complete ordered field. Twoquestions arise, whether such a theory is consistent, and whether it is categorical. The demand of con-sistency is met by constructing real numbers from the rational numbers using Dedekind cuts [2] or byequivalence classes of Cauchy sequences of rationals [1]. An alternative algebraic construction usesthe integers leading to Eudoxus reals; see [5]. Besides these popular constructions of real numbers, abrief survey of other constructions may be found in [7]. Once a construction is accepted, the questionof categoricity is addressed by proving that a complete ordered field is isomorphic to the system ofconstructed objects.While working with real numbers, one thinks of them as infinite decimal numbers. Indeed, therehave been various attempts at constructing a model for the theory of real numbers by using infinitedecimals. The infinite decimals come with the inherent difficulty of defining addition and multiplica-tion. For instance, if two infinite decimals are added, then it is not clear what the m th digit of the resultwill be, due to the problem of carry digits. In fact, if two infinite decimals are added (or multiplied)and a digit is eventually changed by a carry digit, then it would never change again by another carrydigit. Though this fact requires a proof, even accepting it does not lead to determining the m th digitof the result.An attempt to define real numbers using decimals was taken by K. Weirstrass but it remainedsomewhat incomplete and unpublished; see [6]. A later published account may be found in [3], whichformalized the notions developed by S. Stevin in [4]. In this approach real numbers have been definedas formal decimals without bothering whether the rational numbers correspond to a restricted versionof these formal decimals; and there was no concern for the modern approach to real numbers viaaxiomatics. Our approach is similar to this, but we use the decimals as actual decimal numbers andthereby connect to the modern axiomatic definition of real numbers. For this purpose, we assume thatwe know the following sets along with the usual arithmetic operations and the order < : N = { , , , . . . } , the set of natural numbers. Z = { . . . , − , − , − , , , , , . . . } , the set of integers. Q = { p / q : p ∈ Z , q ∈ N , p , q have no common factors } , the set of rational numbers.1e know that N ( Z ( Q and each rational number can be represented as a recurring decimalnumber using the digits , , , , , , , , , . (dot)and the minus sign − . In this representation, any m ∈ Z is identified with ± m . · · · , and a decimalnumber with trailing 9s is identified with one having trailing 0 s . that is, ± a . a a · · · a k − ( a k + ) and ± a . a a · · · a k · · · for 0 ≤ a k ≤ , are considered equal. Here, a ∈ Z and 0 ≤ a i ≤ i ≥ , and the sign ± means that either there is no sign or there is a negative sign. The integer − . A caution about notation: a is a finite sequence of digits representing an integer,whereas other a i s are digits.With these identifications, the decimal representation of any rational number is unique. In theprocess, we assume the usual arithmetic of natural numbers, specifically, the algorithms for computingaddition and multiplication of two integers. Further, the notion of less than, < , in N can be specifiedin terms of digits using the basics such as 0 < < · · · < < . Consider two unequal natural numberswritten in decimal notation, say, a = a a · · · a k and b = b b · · · b m . Then a < b iff one of the followingconditions holds:1. k < m . k = m and a < b . k = m , a = b , . . . , a j = b j and a j + < b j + for some j , ≤ j < k . Next, the less than relation is extended from N to Z by requiring the following conditions:4. 0 < a a · · · a k . − a a · · · a k < b b · · · b m . − a a · · · a k < − b b · · · b m iff b b · · · b m < a a · · · a k . We extend the set of rational numbers to the set of real numbers by formally defining a real numberas an infinite decimal, including all terminating, recurring, and non-recurring ones. Our plan is notto extend the rationals to reals but to extend a subset of rational numbers, namely, the terminatingdecimals to reals. For this purpose, we review certain notions involved with terminating decimals.We abbreviate ‘if and only if’ to ‘iff’ as usual. Further, we do not use the pedagogic tool ‘Defini-tion’; the defined terms are kept bold faced.
We assume the usual way of writing integers in decimal notation as explained earlier. A terminatingdecimal is an expression of the form a a · · · a k . b b · · · b m or − a · · · a k . b b · · · b m satisfying thefollowing properties:1. Each a i and each b j is a digit.2. a = k = . For instance, 51 , . , . , . , . , − , − . , − . , . , − . . , . , − . , − . equal iff either they are identical or one is obtainedfrom the other by adjoining a string of 0s at the right end. For two terminating decimals a and b ,
2e write a = b when they are equal. That is, given two terminating decimals, we look at how manydigits are there in each of them after the decimal point. First, we adjoin necessary number of 0s atthe end of the one which has less number of digits after the decimal point so that both of them nowhave the same number of digits after the decimal point. Next, we remove the decimal points, andalso possible occurrences of the digit 0 in the beginning, to obtain two integers. Now, we declarethat the two terminating decimals are equal iff the obtained integers are equal. That is, keeping00 · · · a a · · · a k = a a · · · a k in the background, we have1. a a · · · a k . b b · · · b m = a a · · · a k . b b · · · b m a a · · · a k . b b · · · b m = a a · · · a k . b b · · · b m · · · a a · · · a k . · · · = a a · · · a k . = a a · · · a k − = . − a a · · · a k . b b · · · b m = − c c · · · c k . d d · · · d m iff a a · · · a k . b b · · · b m = c c · · · c k . d d · · · d m . This means that formally, we consider = as an equivalence relation on the set of terminating decimalsand then update the set of terminating decimals to the equivalence class of = . To talk uniformly about terminating decimals, we regard any integer a ∈ Z as a terminating dec-imal a . b b · · · b m with m = . Notice that the integer a can be in one of the forms 0 , a a · · · a k or − a a · · · a k , where a = . This creates no confusion due to the relation of equality as defined above.We write the set of terminating decimals as T . This is how
Z ( T . Two terminating decimals which are not equal, are called unequal, as usual. A nonzero terminatingdecimal without the minus sign is called a positive terminating decimal , and one with the minus signis called a negative terminating decimal .Next, two unequal terminating decimals are compared a similar way. First, we adjoin necessarynumber of 0s to make the same number of digits after the decimal points in both. Next, we removethe decimal points to obtain two corresponding integers. We then declare that one is less than theother iff the corresponding integers are in the same relation. That is, let a = ± a a · a k . b b · · · b m and c = ± c c · · · c i . d d · · · d j be two unequal terminating decimals. If m < j , adjoin j − m numberof 0s at the end of a ; if m > j , then adjoin m − j number of 0s at the end of c so that each hasequal number of digits after the decimal point. Thus, we take a ′ = ± a a · · · a k . b b · · · b m and c ′ = ± c c · · · c i . d d · · · d m . We say that a < c iff a ′ < c ′ . We read < as ‘less than’, as usual.Notice that < is the lexicographic ordering on T , where each negative terminating decimal is lessthan 0 , and 0 is less than each positive terminating decimal. Moreover, < is an extension of the samefrom Z to T . As in Z , < is a transitive relation on T . That is, if a < b and b < c , then a < c . Further,we write b > a , when a < b . As usual, a ≤ b abbreviates the phrase ‘ a < b or a = b ’, and a ≥ b abbreviates ‘ a > b or a = b ’. We read these symbols using the phrases, less than, greater than, lessthan or equal to, or greater than or equal to, as appropriate.Next, addition and multiplication on T are defined as in usual arithmetic. Due to the equalityrelation defined earlier (and as we adjoin necessary number of 0s at the end of one of the terminat-ing decimals) suppose a = ± a a · · · a k . b b · · · b m and c = ± c c · · · c i . d d · · · d m be two terminatingdecimals. Construct integers a ′ = ± a a · · · a k b b · · · b m and c ′ = ± c c · · · c i d d · · · d m , without thedecimal points. Then use the popular addition and multiplication algorithms to obtain a ′ + b ′ and a ′ × b ′ . Then a + b is obtained by placing the decimal point preceding the m th digit counted from theright of a ′ + b ′ . Similarly, a × b is obtained from a ′ × b ′ by placing the decimal point preceding the3 m ) th digit in a ′ × b ′ ; if there do not exist 2 m digits in a ′ × b ′ , then we prefix to it adequate numberof 0s before placing the decimal point.It is easy to verify the following properties of addition and multiplication of terminating decimals:1. 0 + a = a and 1 × a = a .
2. If a = c and b = d , then a + c = b + d and a × c = b × d . a + c = c + a , a × c = c × a . a + ( b + c ) = ( a + b ) + c , a × ( b × c ) = ( a × b ) × c . a × ( b + c ) = ( a × b ) + ( a × c ) . Let A and B be nonempty subsets of T . We define their sum and product as follows: A + B = { x + y : x ∈ A , y ∈ B } , A × B = { x × y : x ∈ A , y ∈ B } . Since for terminating decimals x , y , z , we have x + y = y + x , x + ( y + z ) = ( x + y ) + z , x × y = y × x and x × ( y × z ) = ( x × y ) × z , we see that for any nonempty subsets A , B , C , of T , A + B = B + A , A + ( B + C ) = ( A + B ) + C , A × B = B × A , A × ( B × C ) = ( A × B ) × C . An infinite decimal is an expression of the form ± a · · · a k . b b · · · b n · · · , where each a i and b j is adigit.A real number is an infinite decimal of the form ± a · · · a k . b b · · · b n · · · , where a = k = , and for no m , all of b m , b m + , b m + , . . . are 9 . Again, the symbol ± says that either the minussign occurs there or it does not.For instance, 21 . · · · , . · · · , . · · · , . · · · , . · · · , − . · · · , − . · · · , − . · · · , − . · · · are real numbers. In the first and second, the digit 0 is repeated; in the third, the string 46 isrepeated; in the fourth one, 0 is repeated, and in the fifth one, nothing is repeated; the others are similarwith a minus sign. The expressions 21 . , . · · · , . · · · , − . , − . · · · , − . · · · are not real numbers.We say that two real numbers are equal iff they are identical. That is, if a = ± a a · · · a k . b b · · · and c = ± c c · · · c i . d d · · · , we say that a = c iff either both start with − or both do not, k = i , and a j = c j for each j ∈ { , . . . , k } , b ℓ = d ℓ for each ℓ ∈ N . Further we identify a real number which ends with a sequence of 0s with the terminating decimalobtained from it by removing the trailing 0s. That is, the real number ± a a · · · a k . b b · · · b m b m + · · · with b j = j > m is identified with the terminating decimal a a · · · a k . b b · · · b m . In this sense ofequality, each terminating decimal is regarded as a real number. Thus the real numbers − . · · · · · · and 0 . · · · · · · are each equal to (identified with) the number 0 . Formally, this identification bringsin an equivalence relation and the ensued equivalence classes are the real numbers.A nonzero real number is called positive if it does not start with the minus sign; and it is called negative if it starts with a minus sign. We consider 0 as the real number which is neither positive nornegative. A real number is called non-negative if it is either 0 or positive.We write the set of all real numbers as R . Notice that
T ( R . < . First, we define < forpositive real numbers. Let a = a a · · · a k . b b · · · and c = c c · · · c i . d d · · · be unequal positive realnumbers. We say that a < c iff one of the following conditions is satisfied:1. a a · · · a k < c c · · · c i as natural numbers.2. a a · · · a k = c c · · · c i and b < d . a a · · · a k = c c · · · c i , b = d , . . . , b j = d j and b j + < d j + . We extend the relation of < to all real numbers in the following manner:1. If a is negative, then a < a is positive, then 0 < a .
2. If a is negative and c is positive, then a < c .
3. If a = − a ′ and c = − c ′ for some positive real numbers a and c , and c ′ < a ′ then a < c . Again, we say that b > a when a < b ; a ≤ b abbreviates the phrase ‘ a < b or a = b ’; and a ≥ b abbreviates the phrase ‘ a > b or a = b ’. We read these symbols in the usual way. It is easy to see thatthe less than relation is an extension of the same from T to R , and the following statements hold forall a , b , c ∈ R :1. a ≤ a .
2. If a ≤ b and b ≤ a then a = b .
3. If a ≤ b and b ≤ c , then a ≤ c . a < b or a = b or b < a .
5. If a < b and b < c , then a < c .
6. If a < b , then there exists c ∈ T such that a < c < b . The last statement above requires some explanation. Suppose a < b . First, if a < < b , then take c = . a = b is positive, then in b = b . b b · · · , let m be the least index such that b m = . Construct c = . b b · · · b m − . Then 0 < c < b . Third, if 0 < a and 0 < b , then a = a . a a · · · a m a m + · · · and b = b . a a · · · a m b m + · · · , where a and b are non-negative integers written in decimal notation, and all other a i s are digits. Noticethat no infinite decimal ends with repetition of the digit 9 . Either a < b or there exists m ≥ a i = b i for 0 ≤ i ≤ m and a m + < b m + . If a < b , then let k be the least nonzero index such that a k < . Then c = a . a · · · a k − a = . · · · · · · , and b = . · · · · · · , then c = . a = . · · · · · · and b = . · · · · · · , then c = . . Otherwise, there exists m ≥ a i = b i for 0 ≤ i ≤ m and a m + < b m + . Find out the firstoccurrence of a digit less than 9 after the ( m + ) th digit in a . Change that digit to 9 and chop-off therest to obtain c . For instance, if a = . · · · and b = . · · · , then m = . Wechange this 8 to 9 and chop-off the rest to obtain c = . . Clearly, a < c < b . a and b are negative, then let a ′ and b ′ be the positive real numbers obtained from a and b , respectively, by removing the minus signs. Now, b ′ < a ′ . By the first case, construct c ′ suchthat b ′ < c ′ < a ′ . Then a < − c ′ < b . Let A ⊆ R . We say that A is bounded above iff there exists b ∈ R such that x ≤ b for each x ∈ A . In such a case, b is called an upper bound for A . For instance, the set of all real numbers less than 2is bounded above, for 2 is one of its upper bounds. Suppose a , a , . . . , a k are fixed digits. Then theset of all real numbers of the form ± a a · · · a k . b b · · · is bounded above since a a · · · a k N = { , , , . . . } is not bounded above.Clearly, if b is an upper bound of a set A of real numbers, then any real number c > b is alsoan upper bound of A . However, a real number smaller than b need not be an upper bound of A . Forinstance, The infinite set { . , . , . , . . . } of terminating decimals has an upper bound 1 . Butno real number smaller than 1 is an upper bound of this set. Reason? Any real number less than1 = . · · · is in the form 0 . b b · · · . If it is an upper bound of the set, no b i is less than 9 . Then theinfinite decimal must be 0 . · · · . However, it is not a real number!If a = a a · · · a k . b b · · · is a non-negative real number, write [ a ] = a a · · · a k ; andif a = − a a · · · a k . b b · · · is a negative real number, write [ a ] = − a a · · · a k − . This is how for each real number a , there exists an integer [ a ] such that [ a ] ≤ a < [ a ] + . Here, weuse the known operation of addition of integers by writing an integer as a concatenation of digits withor without the minus sign. Such an integer [ a ] , which is equal to the real number [ a ] . · · · , is calledthe integral part of a . For a nonempty subset A of R which is bounded above, we define its supremum , denoted bysup ( A ) , by considering two cases. Case 1 : Suppose A has at least one non-negative real number. Let A be the set of integral parts of realnumbers in A . Notice that A contains at least one non-negative integer. Since A is bounded above,there exists a maximum of A . Write the maximum as m . Now, m is a non-negative integer. In A , thereare real numbers whose integral part is m . Let A be the set of all real numbers in A whose integralpart is m . Each real number in A is in the form m . b b · · · . Look at the first decimal digit b in allsuch real numbers. There exists one maximum digit, say m . Let A be the set of all real numbers in A which are in the form m . m c c · · · . Then choose the maximum possible digit c among these realnumbers, and call it m . Continuing in this fashion, we obtain a sequence of digits m , m , m , . . . . Then we form the infi-nite decimal m . m m m · · · . This infinite decimal may or may not be a real number. We consider twocases.
Case (1A): Suppose there is no such i such that m j = j ≥ i . That is, the infinite decimal doesnot end with a sequence of 9s. Then, m . m m m · · · is a real number; and we call this real number assup ( A ) . Case (1B): Suppose there exists i such that m j = j ≥ i . That is, the i th digit onwards all digitsare equal to 9 . Then consider the terminating decimal m . m m m · · · m i and add to it 0 . · · · i th digit after the decimal as 1 and all previous digits 0 . The result is a terminating decimal. Thisterminating decimal is a real number, which we call sup ( A ) . Case 2 : Suppose that all elements in A are negative. Let A ′ = { x : − x ∈ A } . That is, we consider realnumbers in A after removing the minus signs. We follow the construction in Case 1, but this time bytaking minimum instead of maximum everywhere, and then put a minus sign in the result. The detailsfollow.Notice that if − u is an an upper bound of A , then u is less than or equal to each element of A ′ . Let A ′ be the set of integral parts of (non-negative) real numbers in A ′ . There exists a smallest non-6egative integer among them, that is, the minimum of A ′ . Write the minimum as n . In A ′ , there arereal numbers whose integral part is n . Let A ′ be the set of all real numbers in A ′ whose integral part is n . Each real number in A ′ is in the form n . d d · · · . Look at the first decimal digit d in all such realnumbers. There exists one minimum digit, say n . Let A ′ be the set of all real numbers in A ′ whichare in the form n . n e e · · · . Then choose the minimum possible digit e among these real numbers,and call it n . Continuing in this fashion, we obtain a sequence of digits n , n , n , . . . . Then we form the infinitedecimal n . n n n · · · . This infinite decimal may or may not be a real number. We consider two cases.
Case (2A): Suppose there is no such i such that n j = j ≥ i . That is, the infinite decimaldoes not end with a sequence of 9s. Then, n . n n n · · · is a real number. We call the real number − n . n n n · · · as sup ( A ) . Case (2B): Suppose there exists i such that n j = j ≥ i . That is, i th digit onwards all digits areequal to 9 . Then consider the terminating decimal n . n n n · · · n i and add to it 0 . · · · i thdigit after the decimal as 1 and all previous digits 0 . The result is a terminating decimal. Prefix thisterminating decimal with a minus sign, and cal, it sup ( A ) . Example 1
Compute the supremum of the following subsets of R :A = { . , . , . · · · · · · , . · · · · · · , . · · · · · · } , B = { . , . , . , . , . , . , · · · , . · · · , · · · } , C = {− . , − . , − . , − . , − . , − . , − . , · · · , − . · · · , · · · } . D = {− . , − . , − . , − . , · · · , − . · · · , · · · } . For A , the set of integral parts is A = { , } . So m = . Next, A = { . , . · · · · · · } . So, m = . A = { . , . · · · · · · } . Then m = . Next, A = { . , . · · · · · · } . We use the equal-ity of real numbers and see that 2 . = . · · · · · · . So, m = . Then A = { . , . · · · · · · } , and m = . Next, A = { . · · · · · · } . So, m = . This point onwards we get each m j to be equalto 1 . Hence sup ( A ) = . · · · · · · . For B , m = , m = , m = , and all succeeding digits are 9 . Then we have the infinite deci-mal 0 . · · · · · · . We see that i = . So, we consider the decimal 0 . . ( B ) = . For C , the set C ′ = { . , . , . , . , . , . , . , · · · , . · · · , · · · } . Then n = , C ′ = { . , . , . , . , . , . , · · · , . · · · , · · · } . So that n = . Now, C ′ = { . } , so that n = , n = , . . . . So, sup ( C ) = − . . For D , the set D ′ = { . , . , . , · · · , . · · · , · · · } . Then n = , n = , . . . . Therefore,sup ( D ) = − . · · · · · · = . ⋄ As we see, each set of real numbers which is bounded above has a unique supremum, which isalso a real number. We give a characterization of the supremum.
Theorem 1
Let A be a nonempty subset of R which is bounded above and let s ∈ R . Then, s = sup ( A ) iff s is an upper bound of A and one of the following conditions is satisfied:1. If t ∈ R is an upper bound of A , then s ≤ t .
2. If p ∈ R and p < s , then there exists q ∈ A such that p < q ≤ s . Proof.
Suppose S is a finite set of negative integers, say, S = {− n , − n , . . . , − n k } . Let S ′ = { n , n , . . . , n k } . Clearly, − n is the minimum of S iff n is the maximum of S ′ . Moreover, suppose a set X of real num-bers contains at least one non-negative real number. Let Y be the set of all non-negative real numbers7hat belong to X . Then clearly, sup ( X ) = sup ( Y ) . Therefore, all the cases discussed in the constructionof the supremum reduce to the case where a set has only non-negative real numbers. So, without lossof generality, assume that A is a set of non-negative real numbers.From our construction it follows that sup ( A ) is an upper bound of A . (1) Let t be an upper bound of A but t = sup ( A ) . First, the integral part of each element of A is lessthan or equal to that of t . Next, for each x ∈ A , the m th digit of x is less than or equal to the m th digitof t . Since t = sup ( A ) , some m th digit of t is greater than the m th digits of all real numbers in A . However, the m th digit of sup ( A ) is the m th digit of some real number in A . Therefore, the m th digitof sup ( A ) is less than the m th digit of t . So, sup ( A ) < t . Therefore, sup ( A ) ≤ t . Conversely, suppose that s is an upper bound of A and that s is less than or equal to every upperbound of A . As sup ( A ) is an upper bound of A , s ≤ sup ( A ) . However, by what we have proved in theabove paragraph, sup ( A ) ≤ s . So, s = sup ( A ) . (2) Let p be a real number such that p < sup ( A ) . If no real number in A is larger than p , then p is anupper bound of A . By (1), sup ( A ) ≤ p which is a contradiction.Conversely, suppose that s is an upper bound of A and that if p is a real number with p < s , then there exists a real number q ∈ A such that p < q . If s = sup ( A ) , we have sup ( A ) < s . Then with p = sup ( A ) , we have a real number q ∈ A such that sup ( A ) < q . This is a contradiction to the factthat sup ( A ) is an upper bound of A . Once p < q , q ∈ A and s is an upper bound of A , it follows that p < q ≤ s . ✷ The above theorem says that if A is a nonempty set of real numbers bounded above, then sup ( A ) isthe least of all upper bounds of A . Some useful properties of the supremum are proved in the followingtheorem.
Theorem 2
Let A be a nonempty subset of R which is bounded above.1. Let B be a nonempty subset of R that is bounded above. If for each x ∈ A , there exists y ∈ Bsuch that x ≤ y , then sup ( A ) ≤ sup ( B ) . In particular, if A ⊆ B , then sup ( A ) ≤ sup ( B ) .
2. If a , b ∈ R , a < b , then sup { x ∈ R : x < b } = b = sup { x ∈ R : a < x < b } .
3. If c ∈ R , then c = sup { x ∈ T : x < c } . sup { sup { t ∈ T : t < x } : x ∈ A } = sup ( A ) = sup { t ∈ T : t < sup ( A ) } . Proof. (1) Due to our assumption, for each x ∈ A , x ≤ sup ( B ) . Hence sup ( B ) is an upper bound of A . Since sup ( A ) is less than or equal to each upper bound of A , we have sup ( A ) ≤ sup ( B ) . (2) Let B = { x ∈ R : x < b } . Clearly, b is an upper bound of B . If u ∈ R is such that u < b , then thereexists v ∈ T such that u < v < b . Now, v ∈ B . Hence b = sup ( B ) . For the second equality, let a ∈ R and a < b . Write C = { x ∈ R : a < x < b } . Notice that for each x ∈ C , there exists y ∈ B such that x ≤ y ; and for each y ∈ B , there exists x ∈ C such that y ≤ x . Henceby (1), sup ( C ) = sup ( B ) = b . (3) Let c ∈ R , D = { x ∈ R : x < c } and let E = { x ∈ T : x < c } . If x ∈ D , then there exists y ∈ T such that x < y < c . That is, x < y for some y ∈ E . Moreover, E ⊆ D . Hence by (1)-(2), sup ( E ) = sup ( D ) = c . (4) By (3) and (2), sup { sup { t ∈ T : t < x } : x ∈ A } = sup { x : x ∈ A } = sup ( A ) = sup { t ∈ T : t < sup ( A ) } . ✷
8e define addition and multiplication of two real numbers a and b as follows.1. If a = , then a + b = b and a × b = .
2. If b = , then a + b = a and a × b = .
3. If a = b = , then a + b = sup ( A + B ) , where A = { x ∈ T : x < a } and B = { x ∈ T : x < b } .
4. If 0 < a and 0 < b , then a × b = sup ( A × B ) , where A = { x ∈ T : 0 < x < a } and B = { x ∈ T :0 < x < b } .
5. If 0 < a and b < , then b = − y , < y so that a × b = − ( a × y ) .
6. If a < < b , then a = − x , < x so that a × b = − ( x × b ) .
7. If a < b < , then a = − x , b = − y , < x , < y so that a × b = x × y . Notice that in the cases (3)-(4), the sets { x + y : x ∈ A , y ∈ B } and { x × y : x ∈ A , y ∈ B } are boundedabove. For instance, if 0 < a = a a · · · a k . c c · · · and 0 < b = b b · · · b i . d d · · · , then upper boundsof A + B and A × B can be given as a a · · · a k + b b · · · b i + ( a a · · · a k + ) × ( b b · · · b i + ) , respectively. Similar constructions work for other possibilities of a and b . Hence addition andmultiplication of real numbers are well defined.
Theorem 3
Let A be a nonempty subset of R that is bounded above and let b ∈ R . Then sup { b + x : x ∈ A } = b + sup ( A ) and sup { b × x : x ∈ A } = b × sup ( A ) . Proof.
When b = , the result is obvious. For 0 < b , using the definition of + , × and Theorem 2(4),we obtainsup { b + x : x ∈ A } = sup { sup { y + t : y ∈ T , t ∈ T , y < b , t < x } : x ∈ A } = sup { y + t : y ∈ T , t ∈ T , y < b , t < sup ( A ) } = b + sup ( A ) . sup { b × x : x ∈ A } = sup { sup { y × t : y ∈ T , t ∈ T , y < b , t < x } : x ∈ A } = sup { y × t : y ∈ T , t ∈ T , y < b , t < sup ( A ) } = b × sup ( A ) . Similarly, the case b < ✷ In particular, if A ⊆ T that is bounded above, then the conclusion of the above theorem also holds.We now proceed to prove the required properties of real numbers. Theorem 4
Let x , y , z ∈ R and let S be a nonempty subset of R . Then the following are true:1. x + y = y + x . ( x + y ) + z = x + ( y + z ) . + x = x .
4. Corresponding to x there exists u ∈ R such that x + u = .
5. x × y = y × x . ( x × y ) × z = x × ( y × z ) . . × x = x .
8. Corresponding to x = , there exists w ∈ R such that x × w = .
9. x × ( y + z ) = ( x × y ) + ( x × z ) .
10. Exactly one of the conditions x < y or x = y or y < x is true.11. If x < y and y < z , then x < z .
12. If x < y , then x + z < y + z .
13. If x < y and < z , then z × x < z × y .
14. If there exists b ∈ S with x ≤ b for each x ∈ S, then there exists s ∈ R such that for eachx ∈ S , x ≤ s; and for each u ∈ R , if x ≤ u , then s ≤ u . Proof.
Let A , B , C be the corresponding sets of terminating decimals less than x , y , z , respectively. Bydefinition, x + y = sup ( A + B ) and etc.(1) Since A + B = B + A , we have x + y = sup ( A + B ) = sup ( B + A ) = y + x . (2) On the contrary, suppose x + ( y + z ) < ( x + y ) + z . Let d ∈ T be such that x + ( y + z ) < d < ( x + y ) + z . Now, d < ( x + y ) + z implies that there exist p , q ∈ T such that d < p + q , p < x + y , q < z . Again, p < x + y implies that there exist r , s ∈ T such that p < r + s , r < x , s < y . Then d < ( r + s ) + q = r + ( s + q ) . But s ∈ T , s < y , and q ∈ T with q < z . So, s + q < y + z . Since r ∈ T and r < x , we see that r + ( s + q ) < x + ( y + z ) . Hence d < x + ( y + z ) . This is a contradictionsince x + ( y + z ) < d . Similarly, ( x + y ) + z < x + ( y + z ) leads to a contradiction. Therefore, x + ( y + z ) = ( x + y ) + z . (3) It follows by definition.(4) We show that x + ( − x ) = . Let A = { s ∈ T : s < x } and let B = { t ∈ T : t < − x } . Clearly, A , B and A + B are nonempty, and A , B are bounded above. Now, if s ∈ A and t ∈ B , then s < x < − t sothat s + t < t + ( − t ) = . That is, A + B is bounded above by 0 . Also, if r ∈ T and r < , then withany s ∈ A we have r + ( − s ) < + ( − s ) = − s < − x . That is, with t = r + ( − s ) , we see that s + t = r . Itimplies that each r ∈ T with r < A + B . Hence, A + B = { y ∈ T : y < } . By Theorem 2(4), x + ( − x ) = sup ( A + B ) = . (5)-(6) Proofs of these are similar to (1)-(2).(7) By Theorem 3, 1 × x = sup { × t : t ∈ T , t < x } = sup { t : t ∈ T , t < x } = x . (8) Let 0 < x . Write A = { t ∈ T : x × t < } . If x = a a · · · a k . b b · · · , then take u = . · · · , with k + x × u < . So, A = ∅ . Further, if x ≥ , then 1 is an upper boundof A . If 0 < x < , then suppose there are m number of 0s in x that immediately follow the decimalpoint. Then 10 m + , that is, the natural number 1 followed by m + A . Hence A is bounded above. Thus, it has a supremum. Let w = sup ( A ) . Notice that 0 < w . ByTheorems 3 and 2 (2), x × w = x × sup ( A ) = sup { x × t : t ∈ A } = sup { x × t : x × t < } = . Next, if x < , let x = − y , where 0 < y . By what we have just proved, there exists z ∈ R , < z such that y × z = . Then x × ( − z ) = ( − y ) × ( − z ) = y × z = . (9) This follows from a similar equality for terminating decimals by taking supremum.(10)-(11) These properties follow directly from the definition of < . (12) Let x < y . Write A = { r ∈ T : r < x } , B = { s ∈ T : s < y } and C = { t ∈ T : t < z } . Now A ⊆ B
10o that A + C ⊆ B + C . By Theorem 2 (1), x + z = sup ( A + C ) ≤ sup ( B + C ) = y + z . If x + z = y + z , then by (1)-(4), we get x = y , which is not possible. Therefore, x + z < y + z . (13) Let x < y and 0 < z . Write A = { r ∈ T : r < x } , B = { s ∈ T : s < y } and C = { t ∈ T : 0 < t < z } . Let u ∈ A × C . Then u = r × t , where r , t ∈ T , r < x < y and 0 < t < z . Thus u ∈ B × C . Hence x × z = sup ( A × C ) ≤ sup ( B × C ) = y × z . If x × z = y × z , then using (6)-(8) we would get x = y , which is a contradiction. Therefore, x × z = y × z . (14) Due to Theorem 1, this statement may be rephrased as: if a nonempty subset S of R is boundedabove, then s = sup ( S ) ∈ R . This is obvious from the construction of supremum. ✷ The modern definition of R uses the properties of R as axioms. In such a treatment, we define R as aset containing at least two distinct symbols 0 and 1 , where two binary operations, denoted as + and × , and a binary relation called less than, denoted by <, are assumed given; and + , × , < are such thatthe fourteen conditions specified in Theorem 4 with R replaced by R are satisfied. These conditionsare called axioms of R . It can be shown that an element u ∈ R corresponding to any x ∈ R in Axiom (4) is unique, andwe write it as − x . Similarly, it can be shown that an element w ∈ R corresponding to a nonzero x ∈ R in Axiom (8) is unique, and we write it as 1 / x . In fact, any set with at least two elements, called 0 and 1 , satisfying the first nine axioms is calleda field . A field where a binary relation, denoted as < is defined, satisfying the axioms (10)-(13) iscalled an ordered field . In any ordered field, an element x is called positive if 0 < x ; and x is called negative if x < . The element 0 is neither positive nor negative. Further, it can be shown that 0 < . Notice that both R and Q are ordered fields. To distinguish R from Q the fourteenth condition isassumed. The fourteenth axiom of R concerns not every element of R but every subset of R . Thiscondition is worth reformulating introducing new simpler phrases. Suppose S is a nonempty subsetof R . A number b ∈ R is called an upper bound of S iff every element of S is less than or equal to b . A number s ∈ R is called a least upper bound of S in R iff s is an upper bound of S and no upperbound of S is less than s . We abbreviate the phrase ‘least upper bound’ to lub . Axiom (14) assertsthat every subset of R having an upper bound has an lub in R . It is called the completeness axiom.The following characterization of the lub holds in R . Theorem 5
Let S be a nonempty subset of R which is bounded above and let s ∈ R . Then, s = lub ( S ) iff s is an upper bound of S, and if p ∈ R and p < s , then there exists q ∈ S such that p < q ≤ s . Proof.
Let s = lub ( S ) . Then s is an upper bound of S . Let p ∈ R and p < s . Then p is not an upperbound of S . So, there exists q ∈ S such that p < q . As s is an upper bound of S , we also have q ≤ s . That is, such a q ∈ S satisfies p < q ≤ s . Conversely, suppose that s is an upper bound of S and that if p ∈ R satisfies p < s , then thereexists q ∈ S such that p < q ≤ s . Now, if r ∈ R satisfies r < s , then r is not an upper bound of S . Hence s = lub ( S ) . ✷ It follows that lub of a set, which is bounded above, is unique. The axioms assert that R is acomplete ordered field. We give an application of the completeness axiom of R , which will be usefullater. In what follows, we will use the commonly accepted abbreviations such as writing x × y as xy , x × x as x , x + ( − y ) as x − y , x × ( / y ) as x / y , and the precedence rules, where ( x × y ) + ( z × w ) isabbreviated to xy + zw , and etc. 11 heorem 6 Let r ∈ R be positive. Then there exists a unique positive s ∈ R such that s = r . Proof. If r = , then take s = s = . If r > , then 1 / r < . If there exists t > t = / r , then s = / t satisfies s = r . So, without loss of generality, suppose 0 < r < . Write A = { x ∈ R : 0 < x ≤ r } . Now, r ∈ A and A is bounded above by 1 . So, let s = lub ( A ) . Notice that s ≤ r < . If r < s , then take ε = s / . Now, ε > , and r = s − ε < s − s ε < ( s − ε ) . That is, s − ε isan upper bound of A . This contradicts the fact that s = lub ( A ) . If s < r , then take ε = min n r − s s + , o . Now, s + ε > s and ( s + ε ) = s + s ε + ε ≤ s + s ε + ε = s + ε ( s + ) ≤ s + r − s s + ( s + ) = r . That is, s + ε ∈ A . It contradicts the fact that s = lub ( A ) . Therefore, s = r . To show uniqueness of such a positive s , suppose t ∈ R , < t and t = r . If s < t , then 0 < s < st < t contradicts the assumption that s = r = t . Similarly, t < s is not possible. That is, s = t . ✷ We write the unique positive s ∈ R that satisfies s = r as √ r , and call it the positive square root of the given positive element r ∈ R . We have not assumed that our known sets of numbers such as N , Z , or Q are contained in R . Reason? They are already inside R in some sense. How?Let R be a complete ordered field. Let us look at N first. The number 1 of N is in R . Of course,the 1 inside R is just a symbol, like 1 inside N . Well, we make a correspondence of 1 inside N to 1inside R . Now, 1 + N corresponds to 1 + R . Given a sum of n number of 1s inside N now corresponds to that inside R . In this sense, N is inside R , that is, a copy of N is inside R . But isthe induction principle of N available to this copy of N inside R ? The answer is affirmative, and weshow it as follows.Let S ⊆ R . We call S an inductive subset of R iff 1 ∈ S and for each x ∈ S , x + ∈ S . For example, R is an inductive subset of itself. Now, let N be the intersection of all inductive subsets of R . Itfollows that if S is any inductive subset of R , then N is a subset of S . Further, let A ⊆ N . If A is aninductive set in R , then using what we have just proved, we obtain A = N . Thus, we have proved thefollowing:Let S be a subset of N satisfying (i) 1 ∈ S ; and (ii) if x ∈ S , then x + ∈ S . Then S = N . This is exactly the induction principle in N . Therefore, a proper copy of N inside R is this set N , theintersection of all inductive subsets of R . Recall that the induction principle implies the well ordering principle, which states thatEach nonempty subset of N has a least element.Then N is extended to Z by taking Z = N ∪ { } ∪ {− n : n ∈ N } with the relation of < ,the operations + and × as in R . Next, Q is obtained from Z by taking Q = { p / q : p ∈ Z , q ∈ N , and p , q have no common factors } . Now, the sets N , Z , Q are copies of N , Z , Q , inside R ,respectively. We show that Q 6 = R by using the axioms of R and the construction of Q inside R . Theorem 7 Q is an ordered field, but it is not complete. roof. It is obvious that Q is an ordered field. On the contrary, suppose that Q is complete. Noticethat Theorem 6 holds in every complete ordered field, and in particular, for Q . Now that 2 ∈ Q , we seethat there exists s ∈ Q such that s = . Then s = pq , where p ∈ Z and q ∈ N do not have any commonfactor. Then s = p = q . Since p is an even number and a square, it follows that p is a multiple of 4 (by induction). In that case, p = m for some m ∈ Z . That is, 4 m = q . It gives2 m = q . With a similar argument, we see that q = n for some n ∈ N . This is a contradiction since p and q do not have any common factor. ✷ The above shows, essentially, that √
6∈ Q . Notice that N and its copy N inside R are isomorphic, in the sense that if x , y ∈ N are writtenrespectively as x ′ , y ′ ∈ N , then x + y and x × y are renamed as x ′ + y ′ and x ′ × y ′ , respectively. Here,the + and × are the same operations as they are given in R . Same way, Z and its copy Z inside R are isomorphic. It can be shown that Q is isomorphic to Q inside R . Further, the isomorphism of Q with Q is an extension of the said isomorphisms of N with N and of Z with Z . In addition, thisisomorphism of Q to Q also preserves the relation of < . Reason is, the same conditions (10)-(13)hold in Q as well as in Q . R Notice that R satisfies all the axioms of R . Hence R is a model of the theory R . Our construction of R assumes the constructions of N and T . Again, T is constructed from N . Hence it shows that if N is a consistent theory, then so is R . We will show that R is a categorical theory, in the sense that theaxioms (1)-(14) define the object R uniquely up to an isomorphism, by proving that R is isomorphicto R . In fact, the decimal representation of any element of R provides such an isomorphism. For acomplete treatment, we give the details in the following. Theorem 8 (Archimedean Principle): Let x , y ∈ R with x > . Then there exists n ∈ N such thatnx > y . Proof . Suppose nx ≤ y for all n ∈ N . Then y is an upper bound of the nonempty set A = { nx : n ∈ N } . Let s = lub ( A ) . Now, there exists z ∈ A such that s − x < z . That is, there exists m ∈ N such that s − x < mx . So that s < ( m + ) x . This contradicts the fact that s is an upper bound of A . ✷ The following corollary to the Archimedean principle helps in defining the integral value functionin R . Theorem 9
Corresponding to each real number x there exists a unique integer n such that n ≤ x < n + . Proof.
Let x ∈ R . If x ∈ Z , then we take n = x . If x > x
6∈ Z , the Archimedean principle implies that there exists an m ∈ N such that m > x . Let k be the least natural number such that k > x ; so that k − < x < k . If x < x
6∈ Z , then − x > − x
6∈ Z . By what we have just shown, there exists a naturalnumber j such that j − < − x < j . Then − j < x < − ( j − ) = − j + . For uniqueness suppose that n − ≤ x < n and m − ≤ x < m for n , m ∈ Z . Assume that n < m . Then n − < m − n − < m − ≤ x < n < m . It implies that m − n − n , and m − m < n leads to a contradiction. Therefore, n = m . ✷ x n , where n ≤ x < n + , defines a function from R to Z . We call itthe integral value function , and write it as [ x ] . That is, [ x ] = the largest integer less than or equal to x for x ∈ R . Once integral value function is defined, we can have a decimal representation of elements of R . Let x ∈ R . If x = , then its decimal representation is itself.Suppose x > . Take x = [ x ] . Then 0 ≤ y = x − x < . Now, 0 ≤ y ≤ . Take x = [ y ] . With y = y − x , we have 0 ≤ y ≤ . Take x = [ y ] . Continuing this process n times for any n ∈ N gives rise to an n -places decimal approximation of x such as x + x + x + · · · + n x n = x . x x · · · x n . This process leads us to think of elements of R being placed on a straight line. We think of a positiveelement x of R as a point on this line to the right of the point marked 0 . To approximate it, we takethe largest integer less than or equal to x ; this is x . Next, we divide the line segment x to x + x lies in the i th interval, then x is this i . This process continues to obtain the n th digit x n for each n . Notice that at any stage n , | x − x . x x · · · x n | ≤ − n . For each n , x is represented by x . x x · · · x n up to n decimal places.In case the digit 9 repeats after a certain stage, we need to update the decimal representation. Thedecimal representation so obtained is in the form a . a a · · · , where a is a non-negative integer andother a i s are digits. If a i = i ≥ , then we replace the infinite decimal with the natural number a + . If a i = i > m ≥ , then we replace the infinite decimal with the terminating decimal a . a · · · a m − b , where b = a m + . Leaving these cases, all other infinite decimals are kept as they are.The updated decimal, whether terminating or infinite, is referred to as the decimal representation of x . Next, suppose x < . Then x = − y for some y > . If y . y y · · · is the decimal representation of y , then the decimal representation of x is − y . y y · · · . As we know, if x ∈ Q , then such a decimal representation recurs (possibly with all x i = x
6∈ Q , then such a decimal representation is not recurring.Thus the decimal representation defines a function from R to R . We define the function φ : R → R by φ ( x ) = x . x x · · · ∈ R for x ∈ R . We will see that φ is an isomorphism. Towards this, we start with the following result. Theorem 10
Let x , y ∈ R . Then, x < y iff φ ( x ) < φ ( y ) . Proof.
First, we look at the case where 0 < x < y . We use the obvious facts such as φ ( n ) = n for every n ∈ N , and 0 < < < · · · < , which are used in defining < on R . Then φ ( x ) < φ ( y ) follows fromthe construction of φ ( z ) for any z ∈ R with 0 < z . Next, suppose x < < y . Then φ ( x ) starts with a minus sign whereas y does not. So, φ ( x ) < φ ( y ) . Finally, if x < y < , then 0 < − y < − x . In this case, φ ( − y ) < φ ( − x ) . As it is defined in R , − φ ( − x ) < − φ ( − y ) . However, − φ ( − x ) = φ ( x ) and − φ ( − y ) = φ ( y ) . Therefore, φ ( x ) < φ ( y ) . Thisproves that if x < y , then φ ( x ) < φ ( y ) . For the converse, notice that if x = y , then φ ( x ) = φ ( y ) . From what we have just proved, It followsthat if x > y , then φ ( x ) > φ ( y ) . This proves that if x < y , then φ ( x ) < φ ( y ) . ✷ heorem 11 Let A be a nonempty subset of R bounded above. Then { φ ( x ) : x ∈ A } is a nonemptysubset of R bounded above, and φ (cid:0) lub ( A ) (cid:1) = sup { φ ( x ) : x ∈ A } . Proof.
Clearly, { φ ( x ) : x ∈ A } 6 = ∅ . Write α = lub ( A ) . For each x ∈ A , x ≤ α , and if β ∈ R issuch that for each x ∈ A , x ≤ β , then α ≤ β . By the previous theorem it follows that for each x ∈ A , φ ( x ) ≤ φ ( α ) , and if β ∈ R is such that for each x ∈ A , φ ( x ) ≤ φ ( β ) , then φ ( α ) ≤ φ ( β ) . Therefore, { φ ( x ) : x ∈ A } is bounded above in R and φ ( α ) = sup { φ ( x ) : x ∈ A } . ✷ Theorem 12
Let x ∈ R . Then x = lub { t ∈ R : t < x } = lub { t ∈ R : t < x , φ ( t ) ∈ T } . Proof.
Write A = { t ∈ R : t < x } . Now, x is an upper bound of A , so lub ( A ) ≤ x . If x = lub ( A ) , thenlub ( A ) < x . So, there exists α ∈ R such that lub ( A ) < α < x . Then α ∈ A , and as an upper bound of A , lub ( A ) < α . This is a contradiction. Hence x = lub { t ∈ R : t < x } . For the second equality, let B = { t ∈ R : t < x , φ ( t ) ∈ T } . Clearly, x is an upper bound of B . If x = lub ( B ) , then there exists z ∈ B such that lub ( B ) ≤ z < x . Then φ ( z ) ∈ T and φ ( z ) < φ ( x ) . Then φ ( x ) = sup { w ∈ T : w < φ ( x ) } . This contradicts Theorem 2(4). ✷ Theorem 13 φ : R → R is a bijection.Proof. Since x < y implies that φ ( x ) < φ ( y ) it follows that φ is one-one. To show that φ is an ontomap, we first consider all terminating decimals, and then lift the result to real numbers.Let t = ± a · · · a k . b b · · · b m ∈ T . Write n = ± a · · · a k b b · · · b m and x = − m × n . Consequently, n ∈ Z and x ∈ R . We see that φ ( x ) = t . Next, let t ∈ R \ T . Then t = sup { s ∈ T : s < t } . By what we have just proved, each s ∈ T is φ ( u ) for some u ∈ R . Thus t = sup { φ ( u ) : u ∈ R , φ ( u ) ∈ T , φ ( u ) < t } . By Theorem 12, t = φ (cid:0) lub { u ∈R : φ ( u ) ∈ T , φ ( u ) < t } (cid:1) . Then, with x = lub { u ∈ R : φ ( u ) ∈ T , φ ( u ) < t } , we see that t = φ ( x ) . Therefore, φ is an onto map. ✷ Theorem 14
Let x , y ∈ R . Then φ ( x + y ) = φ ( x ) + φ ( y ) . Proof.
We first show the result for those elements in R whose decimal representations are terminatingdecimals. For any integer n in R , φ ( n ) = n . Thus φ ( x + y ) = φ ( x ) + φ ( y ) holds for integers x and y . Then by multiplying a suitable negative power of 10 , the same equality is proved for terminatingdecimals. The details follow.Let s , t ∈ R be such that φ ( s ) , φ ( t ) ∈ T . If the number of digits after the decimal points in φ ( s ) and φ ( t ) are unequal, then adjoin the necessary number of zeros to one of them so that the numberof digits after the decimal points in both are same. So, let φ ( s ) = ± s · · · s j . a a · · · a m and φ ( t ) = ± t · · · t k . b b · · · b m . Then s = − m × ( ± s · · · s j a a · · · a m ) , t = − m × ( ± t · · · t k b b · · · b m ) and s + t = − m × ( ± c · · · c i d d · · · d m ) for suitable digits c s and d s. It follows that φ ( s + t ) = ± c · · · c i . d d · · · d m = φ ( s ) + φ ( t ) . Here, the signs ± are taken suitably.Next, we show the result for those elements of R whose decimal representations are non-terminatingdecimals. Towards this, let r , s , t , x , y ∈ R . We observe that if r < x and s < y , then r + s < x + y . Fur-ther, if t < x + y , then write x + y − t = ε with some ε > . Take r = x − ε / s = y − ε / . Then r + s = x + y − ε = t . That is, if t < x + y , then there exist r , s such that r < x , s < y and t = r + s . Weuse these observations in the following calculation.15et x , y ∈ R be such that φ ( x ) , φ ( y ) ∈ R \ T . Then φ ( x + y ) = φ (cid:0) lub { t ∈ R : t < x + y , φ ( t ) ∈ T } (cid:1) = sup { φ ( t ) : t ∈ R , t < x + y , φ ( t ) ∈ T } = sup { φ ( r + s ) : r , s ∈ R , r + s < x + y , φ ( r + s ) ∈ T } = sup { φ ( r ) + φ ( s ) : r , s ∈ R , r + s < x + y , φ ( r + s ) ∈ T } = sup { φ ( r ) + φ ( s ) : r , s ∈ R , r < x , s < y , φ ( r ) , φ ( s ) ∈ T } ( It follows from the above observations and Theorem 2 . )= sup { φ ( r ) + φ ( s ) : r , s ∈ R , φ ( r ) < φ ( x ) , φ ( s ) < φ ( y ) , φ ( r ) , φ ( s ) ∈ T } = sup { u + v : u < φ ( x ) , v < φ ( y ) , u , v ∈ T } = φ ( x ) + φ ( y ) . ✷ Theorem 15
Let x , y ∈ R . Then φ ( x × y ) = φ ( x ) × φ ( y ) . Proof.
Clearly, if one of x or y is equal to 0 , the equality holds. For x , y ∈ N , φ ( x ) = x , φ ( y ) = y and φ ( x + y ) = x + y ; thus the equality holds. Next, suppose x , y ∈ R are such that φ ( x ) , φ ( y ) are positiveterminating decimals. Then x > y > . As in the case of addition, we see that x = − m k and y = − n k for n , n ∈ N . Then x × y = − m − n ( k × k ) so that the equality holds.Next, suppose that x , y ∈ R are such that φ ( x ) and φ ( y ) are positive non-terminating decimals.Before proceeding towards the equality, we observe the following:Let r , s , t , x , y ∈ R , where all of these are greater than 0 . If r < x and s < y , then r × s < x × y . Further,if t < x × y , write t / ( x × y ) = ε . Then ε < . Take r = x × √ ε and s = y × √ ε . Then r × s = x × y × ε = t . That is, if t < x × y , then there exist r , s such that r < x , s < y and t = r × s . Using these observations a calculation similar to that in the case of addition can now be given,where we replace all occurrences of + with × to obtain the required equality.When one or both of x , y is (are) less than 0 , we use ( − x ) × y = − ( x × y ) , etc. for proving theequality. ✷ To summarize, we have proved the following for any x , y ∈ R and any nonempty subset A of R :1. If x < y , then φ ( x ) < φ ( y ) . φ ( x + y ) = φ ( x ) + φ ( y ) . φ ( x × y ) = φ ( x ) × φ ( y ) .
4. If A is bounded above, then { φ ( x ) : x ∈ A } is a nonempty subset of R bounded above; and φ ( lub ( A )) = sup { φ ( x ) : x ∈ A } . That is, the following result has been proved.
Theorem 16
The function φ : R → R given by φ ( x ) = the decimal representation of x , is an isomor-phism. Now onwards, we do not distinguish between N and N , between Z and Z , between Q and Q , andbetween R and R . We use the latter symbols instead of the former. That is, we have a unique completeordered field, namely, R and it contains N , Z and Q . Also, each number in R , called a real number,16s an infinite decimal. The rational numbers are those which can be written in the form p / q for p ∈ Z and q ∈ N , where p , q have no common factors. Again, a rational number is an infinite decimalwhere after some finite number of digits, some finite sequence of digits is repeated infinitely often.As usual, a terminating decimal is identified with an infinite decimal with trailing 0s and no infinitedecimal ends with a sequence of the digit 9 . We have seen that the supremum or least upper boundof a nonempty set of real numbers that is bounded above, is unique. In accordance with this, we donot distinguish between sup and lub of such a set. The real numbers which are not rational numbers,called the irrational numbers, are non-recurring infinite decimals. For instance, √ R . References [1] Cantor G. Grundlagen einer allgemeinen mannigfaltigkeitslehre. 1883. In:
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