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# Fndns Of Analysis I MATH 3210

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This 3 page Class Notes was uploaded by Miss Noel Mertz on Monday October 26, 2015. The Class Notes belongs to MATH 3210 at University of Utah taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/229940/math-3210-university-of-utah in Mathematics (M) at University of Utah.

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Date Created: 10/26/15

Rational numbers De nitions and problems Let m and n be integers with n f 0 We de ne a the symbol to be the set of all ordered pairs of integers a7b with I f 0 such that mb nal Here is the same de nition of using mathematical symbols 3 abab e m x 0 and mb nu n We call a rational Humbert The set of all rational numbers is denoted by 1 Show that for any integer k f 0 the ordered pair kn7 km is in 2 If gcdnm 1 show that every ordered pair in is of the form kn7 km where k is a nonzero integeri 3 Show that if the ordered pair a7 12 is in then is equal to as sets We de ne the addition of two rational numbers as follows De ne EEE n q 7 We de ne multiplication by m 17 mp i X 7 it n q 7 These de nitions are very easy to deal with since they are exactly the ones that we are used to However to be rigorous we need to make sure that they are well de ned In particular by 3 we know that and i are the same setsl We need to check that the above rules for addition and multiplication give the same answer wether we use 5 or It Here is an example By the de nition of addition 1 2 7 l X 3 2 X 2 7 7 2 3 2x3 6 and g 272gtltS2gtlt47E 4 3 4x3 712 Since 7 X 12 6 X 14 the order pair 1412 is in By 3 we have so in this example we get the same answer using 5 and It 4 Show that if then m a EE n 4 b q for all rational numbers 5 Show that if then m a XEXE n 4 b q for all rational numbers 3 De ne a function f from Z to Q by We will use this function in the next two problems 6 Show that 7 Show that X m gtlt Solutions n 1 We need to show that the ordered pair kn7 km is in the set of ordered pairs El First we observe that kn and km are integers since the product of two integers is an integer Second km 0 since k f 0 and m f 0 Finally nkm mkn since multiplication is commutative By de nition an ordered pair that satis es these three properties is in 2 Let a7 12 be an ordered pair in Then a and b are integers7 I f 0 and no mbl By the last property we see that m is a factor of the integer nal Since gcdnm 17 the only common factor of n and m is 1 Therefore m must be a factor of or That is there is a nonzero integer k such that a kml If we replace a in the equation na ml with km we get the equation nkm mbl Since m f 0 this implies that kn 12 3 Let 07d be an ordered pair in We will show that ad is also in f Since ad is in we know that c and d are integers7 d f 0 and me n i To show that ad is in we are only left to show that be adi Since a7b is in we also have ma nbi Since mc nd and ma nb we have me gtlt n12 nd gtlt ma which implies that be ad as desired Therefore ad is in f We have shown that if a7 12 is in i then is a subset of ii To nish the proof we note that the ordered pair n7 m is contained in and is therefore also contained in lf mm is an element of we have just shown that g is a subset of a Since 9 is a subset of 1 and E is a subset of 9 we must have that 1 a b m m b m b 4 By the de nition of addition m 10 mg n i 7 7 n q 7 and b a a 7 g 7 u b q 7 b4 By 3 if the ordered pair mg np mg is in the set 1ng then 1ng mag We now check that mg np nq satis es the three de ning properties of with First we note that mg np and m are integers since they are products and sums of integers and that m f 0 since neither n nor 4 are 0 This is the rst two properties Using this fact we have 124 X mg np 1272142 bnpq 7Lan bnpq m X W b where we are using the fact that mb no in the second inequalityl This is the third property so mg np mg is in 1121 Combining the equalities we have pmqnpaqbp Elf EE n 4 m M b q as desired 5 By the de nition of multiplication mp XE7 4 m1 and a I7 up 5 q 7 W As in 4 we need to show that the ordered pair mp7 mg is the set gr Checking the rst two properties we see that mp and m are integers since they are the pro uct of integers and that m f 0 since neither n nor 4 are zero Finally we see that 124 X mp bmqp anqp up gtlt nq where the second inequality uses the fact that mb nai We have shown that mp7nq is in H bq so by 3 we have 77 7 Exactly as in 4 combining the equalities gives m 7 X g E X E n 4 b q as desired 6 By the de ntion of f7 By the de nition of addition7 m mi Again using the de nition off we have m Combining the equalities gives m 7 By the de ntion of f7 gtlt gtlt By the de nition of multiplication X m mm nxm We again use the de nition off to see that X m As in 6 combining the1eualilies gives X m gtlt

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