You don't have to be a mathematician to have a feel for numbers. John Forbes Nash, Jr.

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1 Course Title: Real Analsis Course Code: MTH3 Course instructor: Dr. Atiq ur Rehman Class: MSc-II Course URL: You don't have to be a mathematician to have a feel for numbers. John Forbes Nash, Jr. To understand this chapter, one must deepl know about the different tpe of numbers sstems; especiall Rational numbers Irrational numbers Also one must know about rigorousl ( ). In mathematics, a real number is a value that represents a quantit along a continuous line. The real numbers include all the rational numbers, such as the integer 5 and the fraction 4/3, and all the irrational numbers such as (.44356, the square root of two, an irrational algebraic number) and π ( , a transcendental number). Real numbers can be thought of as points on an infinitel long line called the number line or real line, where the points corresponding to integers are equall spaced. An real number can be determined b a possibl infinite decimal representation such as that of 8.63, where each consecutive digit is measured in units one tenth the size of the previous one. The real number sstem can be describe as a complete ordered field. Therefore let s discusses and understand these notions first. Order Let S be a non-empt set. An order on a set S is a relation denoted b with the following two properties (i) If, S, then one and onl one of the statement,, is true. Available online at

2 Chap - (ii) If,, z S and if, z then z. Ordered Set A set is said to be ordered set if an order is defined on S. Eamples The set,4,6,7,8,9, and are eamples of ordered set with standard order relation. The set a, b, c, d and,,, are eamples of set with no order. Also set of comple numbers have no order. Bound Upper Bound Let S be an ordered set and E S. If there eists a S such that E, then we sa that E is bounded above. And is known as upper bound of E. Lower Bound Let S be an ordered set and E S. If there eists a S such that E, then we sa that E is bounded below. And is known as lower bound of E. Eample Consider S,,3,...,5 and 5,,5, Set of all lower bound of E,,3,4,5. Set of all upper bound of E,,,...,5. E. Least Upper Bound (Supremum) Suppose S is an ordered set, E S and E is bounded above. Suppose there eists an S such that (i) is an upper bound of E. (ii) If, then is not an upper bound of E. Then is called least upper bound of E or supremum of E and written as sup E In other words is the least member of the set of upper bound of E. Eample,,3,...,5 E 5,,5,. Consider S and (i) It is clear that is upper bound of E. (ii) If then clearl is not an upper bound of E. Hence sup E..

3 Chap - 3 Greatest Lower Bound (Infimum) Suppose S is an ordered set, E S and E is bounded below. Suppose there eists a S such that (i) is a lower bound of E. (ii) If, then is not a lower bound of E. Then is called greatest lower bound of E or infimum of E and written as inf E. In other words is the greatest member of the set of lower bound of E. Eample Consider S,,3,...,5 and 5,,5, E. (i) It is clear that 5 is lower bound of E. (ii) If 5, then clearl is not lower bound of E. Hence inf E 5. Eample If is supremum of E then ma or ma not belong to E. Let E r : r r and E r : r r. Then sup Einf E but E and E. Eample Let E be the set of all numbers of the form, where n is the natural n numbers, that is, E,,, 3 4,. Then sup E which is in E, but inf E which is not in E. Eample Consider the sets where A p p p : and B p p p :, is set of rational numbers. Then the set A is bounded above. The upper bound of A are the eactl the members of B. Since B contain no smallest member therefore A has no supremum in. Similarl B is bounded below. The set of all lower bounds of B consists of A and r with r. Since A has no largest member therefore B has no infimum in.

4 Chap - 4 Least Upper Bound Propert A set S is said to have the least upper bound propert if the followings is true (i) S is non-empt and ordered. (ii) If E S and E is non-empt and bounded above then supe eists in S. Greatest lower bound propert can be defined in a similar manner. Eample Let S be set of rational numbers and E p p p : then E, E is non-empt and also bounded above but supremum of E is not in S, this implies that the set of rational number does not posses the least upper bound propert. Theorem Suppose S is an ordered set with least upper bound propert. B S, B is nonempt and is bounded below. Let L be set of all lower bound of B then sup L eists in S and also inf B. In particular infimum of B eists in S. OR An ordered set which has the least upper bound propert has also the greatest lower bound propert. Since B is bounded below therefore L is non-empt. Since L consists of eactl those S which satisf the inequalit. B We see that ever B is an upper bound of L. L is bounded above. Since S is ordered and non-empt therefore L has a supremum in S. Let us call it. If, then is not upper bound of L. B, B L. Now if then L because sup L. We have shown that L but L if. In other words, if, then is a lower bound of B, but is not, this means that inf B.

5 Chap - 5 Field A set F with two operations called addition and multiplication satisfing the following aioms is known to be field. Aioms for Addition: (i) If, F then F. Closure Law (ii),, F. Commutative Law (iii) ( z) ( ) z,, z F. Associative Law (iv) For an F, F such that Additive Identit (v) For an F, F such that ( ) ( ) +tive Inverse Aioms for Multiplication: (i) If, F then F. Closure Law (ii),, F Commutative Law (iii) ( z) ( ) z,, z F (iv) For an F, F such that Multiplicative Identit (v) For an F,, Distributive Law For an,, z F, such that F, (i) ( z) z (ii) ( ) z z z Ordered Field An ordered field is a field F which is also an ordered set such that i) z if,, z F and z. ii) if, F, and. e.g the set of rational number is an ordered field. tive Inverse. Eistance of Real Field There eists an ordered field (set of real) which has the least upper bound propert and it contain (set of rationales) as a subfield. There are man other was to construct a set of real numbers. We are not interested to do so therefore we leave it on the reader if the are interested then following page is useful:

6 Chap - 6 Theorem Let,, z. Then aioms for addition impl the following. (a) If z then z (b) If then (c) If then. (d) ( ) (Note: We have given the proofs here just to show that the things which looks simple must have valid analtical proofs under some consistence theor of mathematics) (a) Suppose z. Since z (b) Take z in (a) ( ) ( ) b Associative law ( z) b supposition ( ) z b Associative law () z (c) Take z in (a) ( ) (d) Since ( ) then (c) gives ( ) Theorem Let,, z. Then aioms of multiplication impl the following. (a) If and z then z. (b) If and then. (c) If and then (d) If, then..

7 Chap - 7 (Note: We have given the proofs here just to show that the things which looks simple must have valid analtical proofs under some consistence theor of mathematics) (a) Suppose z Since b associative law z z z b associative law z z (b) Take z in (a) (c) Take z in (a) i.e. (d) Since then (c) give Theorem Let,, z. Then field aioms impl the following. (i) (ii) if, then. (iii) ( ) ( ) ( ) (iv) ( )( )

8 Chap - 8 (i) Since ( ) (ii) Suppose, but Since () from (i) a contradiction, thus (ii) is true. (iii) Since ( ) ( ).. () Also ( ) ( ) () Also ( ). (3) Combining () and () ( ) ( ) ( ) ( ) (4) Combining () and (3) ( ) ( ) ( ). (5) From (4) and (5) ( ) ( ) (iv) ( )( ) ( ) using (iii) Theorem Let,, z. Then the following statements are true in ever ordered field. i) If then and vice versa. ii) If and z then z. iii) If and z then z. iv) If then v) If then in particular..

9 Chap - 9 i) If then so that. If then so that. ii) Since z we have z which means that z also ( z ) z z z z iii) Since z z z Also Therefore ( z ) z z z z z iv) If then If then ( )( ) i.e. if then, since ( ) then. v) If and v then v, But Likewise as If we multipl both sides of the inequalit we obtain i.e. finall. b the positive quantit

10 Chap - Theorem (Archimedean Propert) If, and then there eists a positive integer n such that n. Let A n : n, Suppose the given statement is false i.e. n. is an upper bound of A. Since we are dealing with a set of real therefore it has the least upper bound propert. Let sup A is not an upper bound of A. m where m A for some positive integer m. ( m ) where m + is integer, therefore ( m ) A. This is impossible because is least upper bound of A i.e. sup A. Hence we conclude that the given statement is true i.e. n. The Densit Theorem If, and then there eists p such that p. i.e. between an two real numbers there is a rational number or is dense in. Since, therefore there eists a positive integer n such that n( ) (b Archimedean Propert) n n (i) We appl (a) part of the theorem again to obtain two +ive integers m and m such that m n and m n m n m, then there eists and integers m( m m m) such that m n m n m and m n n m n n m n from (i) m n

11 Chap - p where m p is a rational. n Theorem Given two real numbers and, there is an irrational number u such that u. Take, Then a rational number q such that q where is an irrational. q where u u, q is an irrational as product of rational and irrational is irrational. Theorem For ever real number there is a set E of rational number such that Take E { q : q } where is a real. sup E. Then E is bounded above. Since E therefore supremum of E eists in. Suppose sup E. It is clear that. If then there is nothing to prove. If then q such that q, which can not happened hence we conclude that real is supe. Question Let E be a non-empt subset of an ordered set, suppose is a lower bound of E and is an upper bound then prove that. Since E is a subset of an ordered set S i.e. E S. Also is a lower bound of E therefore b definition of lower bound E (i) Since is an upper bound of E therefore b the definition of upper bound E (ii) Combining (i) and (ii) as required.

12 Chap - The Etended Real Numbers The etended real number sstem consists of real field and, We preserve the original order in. and define and two smbols The etended real number sstem does not form a field. Mostl we write. We make following conventions: i) If is real the,,. ii) If then ( ), ( ). iii) If then ( ), ( ). Euclidean Space For each positive integer k, let k be the set of all ordered k-tuples (,,..., k ) where,,..., k are real numbers, called the coordinates of. The elements of k are called points, or vectors, especiall when k. (,,..., n ) and is a real number, put If (,,..., ) and (,,..., k ) So that k and k k k. These operations make over the real field. The inner product or scalar product of and is defined as k. (... ) i And the norm of is defined b The vector space Euclidean k-space. i i k k k ( ) i k with the above inner product and norm is called k into a vector space

13 Chap - 3 Theorem n Let, then i) i) Since ii) (Cauch-Schwarz s inequalit) ( ) therefore ii) If or, then Cauch-Schwarz s inequalit holds with equalit. If and, then for we have ( ) ( ) ( ) ( ) ( ) Now put (certain real number) 4. Which hold if and onl if Question i.e.. n Suppose,, z the prove that a) b) z z a) Consider a a a,

14 Chap - 4 a a a.. (i) b) We have z z z from (i) Relativel Prime Let ab,. Then a and b are said to be relativel prime or co-prime if a and b don t have common factor other than. If a and b are relativel prime then we write ( ab, ). Question If r is non-zero rational and is irrational then prove that r and r are irrational. Let r be rational. a r b where ab,, b such that ab,, a r b c Since r is rational therefore r where cd, d, d such that cd,, a c ad bc. b d bd Which is rational, which can not happened because is given to be irrational. Similarl let us suppose that r is rational then a r for some ab,, b a b r ab, b such that

15 Chap - 5 c Since r is rational therefore r where cd,, d such that cd, d a a d ad b c b c bc d Which shows that is rational, which is again contradiction; hence we conclude that r and r are irrational. References: () Principles of Mathematical Analsis Walter Rudin (McGraw-Hill, Inc.) () Introduction to Real Analsis R.G.Bartle, and D.R. Sherbert (John Wile & Sons, Inc.) (3) Mathematical Analsis, Tom M. Apostol, (Pearson; nd edition.) A password protected zip archive of above resources can be downloaded from the following URL: