827, 016 ẳ 8105ỵ2104ỵ7103ỵ0102ỵ110ỵ6
64 SOME PROPERTIES OF INTEGERS [CHAP. 5
What is not so well known is that this representation is an application of the congruence properties of integers. For, supposeais a positive integer. By the division algorithm,aẳ10q0ỵr0, 0r0<10.
If q0ẳ0, we write aẳr0; if q0>0, then q0 ẳ10q1ỵr1, 0r1<10. Now if q1ẳ0, then aẳ10r1ỵr2 and we write aẳr1r0; if q1>0, then q1 ẳ10q2ỵr2, 0r2<10. Again, if q2ẳ0, then aẳ102r2ỵ10r1ỵr0 and we write aẳr2r1r0; if q2>0, we repeat the process. That it must end eventually and we have
a ẳ 10srsỵ10s1rs1 ỵ ỵ10r1 ỵr0 ẳ rsrs1 r1r0
follows from the fact that the qs constitute a set of decreasing non-negative integers. Note that in this representation the symbols ri used are from the set f0, 1, 2, 3,. . ., 9g of remainders modulo 10.
(Why is this representation unique?)
In the paragraph above, we chose the particular integer 10, called the base, since this led to our system of representation. However, the process is independent of the base and any other positive integer may be used. Thus, if 4 is taken as base, any positive integer will be represented by a sequence of the symbols 0, 1, 2, 3. For example, the integer (base 10) 155ẳ432ỵ421ỵ42ỵ3ẳ2123 (base 4).
Now addition and multiplication are carried out in much the same fashion, regardless of the base;
however, new tables for each operation must be memorized. These tables for base 4 are:
Table 5-3 Table 5-4
þ 0 1 2 3
0 0 1 2 3
1 1 2 3 10
2 2 3 10 11
3 3 10 11 12
and
0 1 2 3
0 0 0 0 0
1 0 1 2 3
2 0 2 10 12 3 0 3 12 21
See Problem 5.15.
Solved Problems
5.1. Prove: Ifajbandajc, then ajðbxỵcyịwherex,y2Z.
Since ajb and ajc, there exist integers s,t such that bẳas and cẳat. Then bxỵcyẳasxỵatyẳ aðsxỵtyịandajðbxỵcyị.
5.2. Prove: Ifajbandb6ẳ0, thenjbj jaj.
Since ajb, we havebẳacfor some c2Z. Then jbj ẳ jaj jcj withjcj 1. Sincejcj 1, it follows that jaj jcj jaj, that is,jbj jaj.
5.3. Prove: Ifajbandbja, thenbẳaorbẳ a.
Since ajb implies a6ẳ0 and bja implies b6ẳ0, write bẳac and aẳbd, where c,d2Z. Now abẳ ðbdịðacị ẳabcd and, by the Cancellation Law, 1ẳcd. Then by Problem 5.16, Chapter 4,cẳ1 or 1 andbẳacẳaora.
5.4. Prove: The number of positive primes is infinite.
Suppose the contrary, i.e., suppose there are exactlynpositive primesp1,p2,p3,. . .,pn, written in order of magnitude. Now form aẳp1p2p3. . . pn and consider the integeraỵ1. Since no one of the ps is a divisor ofaþ1, it follows thataþ1 is either a prime>pnor has a prime>pnas a factor, contrary to the assumption thatpnis the largest. Thus, there is no largest positive prime and their number is infinite.
CHAP. 5] SOME PROPERTIES OF INTEGERS 65
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5.5. Prove the Division Algorithm: For any two non-zero integersaandb, there exist unique integers q andrsuch that
aẳbqỵr, 0r<jbj
DefineSẳ fabx:x2Zg. Ifb<0, i.e.,b 1, thenb jaj jaj aandab jaj 0. Ifb>0, i.e., b1, thenb ðjajị jaj aandabðjajị 0. Thus,Scontains non-negative integers; denote byr the smallest of these (r0) and suppose rẳabq. Now if r jbj, then r jbj 0 and r jbj ẳ abq jbj ẳa ðqỵ1ịb<rora ðq1ịb<r, contrary to the choice ofras the smallest non-negative integer2S. Hence,r<jbj.
Suppose we should find another pairq0 andr0 such that aẳbq0ỵr0, 0r0<jbj
Now bq0ỵr0ẳbqỵr or bðq0qị ẳrr0 implies bjðrr0ị and, since jrr0j<jbj, then rr0ẳ0; alsoq0qẳ0 sinceb6ẳ0. Thus,r0ẳr,q0ẳq, andq andrare unique.
5.6. Find ð389, 167ịand express it in the form 389m ỵ167n.
From We find
389ẳ1672ỵ55 1ẳ55227 167ẳ553ỵ2 ẳ558216727
55ẳ227ỵ1 ẳ38982167191 2ẳ12
Thus,ð389, 167ị ẳ1ẳ82389ỵ ð191ịð167ị.
5.7. Prove: Ifcjaband ifða,cị ẳ1, thencjb.
From 1ẳmaỵnc, we havebẳmabỵncb. Sincecis a divisor ofmabỵncb, it is a divisor ofbandcjbas required.
5.8. Prove: Ifða,sị ẳ ðb,sị ẳ1, thenðab,sị ẳ1.
Suppose the contrary, i.e., suppose ðab,sị ẳd>1 and letdẳ ðab,sị ẳmabỵns. Now djab and djs.
Since ða,sị ẳ1, it follows that d6 j a; hence, by Problem 7, djb. But this contradicts ðb,sị ẳ1; thus, ðab,sị ẳ1.
5.9. Prove: Ifpis a prime and if pjab, wherea,b2Z, thenpja orpjb.
Ifpjawe have the theorem. Suppose then thatp6 ja. By definition, the only divisors ofpare1 andp;
thenðp,aị ẳ1ẳmpỵnafor somem,n2Zby Theorem II. Nowbẳmpbỵnaband, sincepjðmpbỵnabị, pjbas required.
5.10. Prove: Every integera>1 has a unique factorization (except for order) aẳp1p2p3 . . . pn
into a product of positive primes.
Ifa is a prime, the representation in accordance with the theorem is immediate. Suppose then that a is composite and consider the set Sẳ fx:x>1,xjag. The least element s of S has no positive factors except 1 ands; hencesis a prime, sayp1, and
aẳp1b1, b1>1
Now eitherb1is a prime, sayp2, andaẳp1p2orb1, being composite, has a prime factorp2and aẳp1p2b2, b2>1
66 SOME PROPERTIES OF INTEGERS [CHAP. 5
A repetition of the argument leads toaẳp1p2p3or
aẳp1p2p3b3, b3>1 and so on.
Now the elements of the setBẳ fb1,b2,b3,. . .ghave the propertyb1>b2>b3> ; hence,Bhas a least element, saybn, which is a primepnand
aẳp1p2p3. . . pn
as required.
To prove uniqueness, suppose we have two representations
aẳp1p2p3. . .pnẳq1q2q3. . . qm
Nowq1is a divisor ofp1p2p3. . . pn; hence, by Theorem V0,q1is a divisor of some one of theps, say p1. Thenq1ẳp1, since both are positive primes, and byM4of Chapter 4,
p2p3. . . pnẳq2q3. . . qm
After repeating the argument a sufficient number of times, we find thatmẳnand the factorization is unique.
5.11. Find the least positive integers modulo 5 to which 19, 288, 19288 and 1932882are congruent.
We find
19ẳ53ỵ4; hence 194ðmod 5ị:
288ẳ557ỵ3; hence 2883ðmod 5ị:
19288ẳ5ð ị ỵ12; hence 192882ðmod 5ị:
1932882ẳ5ð ị ỵ4332ẳ5ð ị ỵ576; hence 19328821ðmod 5ị:
5.12. Prove: Letðc,mị ẳdand writemẳm1d. Ifcacbðmodmị, thenabðmodm1ịand conversely.
Write cẳc1d so that ðc1,m1ị ẳ1. If mjcðabị, that is, if m1djc1dðabị, then m1jc1dðabị and, sinceðc1,m1ị ẳ1,m1jðabịandabðmodm1ị.
For the converse, suppose abðmod m1ị. Since m1jðabị, it follows that m1jc1ðabị and m1djc1dðabị. Thus,mjcðabịandcacbðmodmị.
5.13. Show that, whena,b,p>02Z, ðaỵbịpapỵbpðmodpị.
By the binomial theorem,ðaỵbịpẳapỵpð ị ỵbpand the theorem is immediate.
5.14. Find the least positive incongruent solutions of:
ðaị13x9ðmod 25ị ðcị259x5ðmod 11ị ðeị222x12ðmod 18ị ðbị207x6ðmod 18ị ðdị7x5ðmod 256ị
(a) Sinceð13, 25ị ẳ1, the congruence has, by Theorem XI, a single incongruent solution.
Solution I. Ifx1is the solution, then it is clear thatx1is an integer whose unit’s digit is either 3 or 8;
thusx12 f3, 8, 13, 18, 23g. Testing each of these in turn, we findx1ẳ18.
Solution II. By the greatest common divisor process we find ð13, 25ị ẳ1ẳ 125ỵ213. Then 9ẳ 925ỵ1813 and 18 is the required solution.
CHAP. 5] SOME PROPERTIES OF INTEGERS 67
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(b) Since 207ẳ1811ỵ9, 2079ðmod 18ị, 207x9xðmod 18ịand, by transitivity, the given congruence is equivalent to 9x6ðmod 18ị. By Theorem IX this congruence may be reduced to 3x2ðmod 6ị.
Nowð3, 6ị ẳ3 and 36 j2; hence, there is no solution.
(c) Since 259ẳ1123ỵ6, 2596ðmod 11ị and the given congruence is equivalent to 6x5ðmod 11ị.
This congruence has a single incongruent solution which by inspection is found to be 10.
(d) Using the greatest common divisor process, we find ð256, 7ị ẳ1ẳ2256ỵ7ð73ị; thus, 5ẳ10256ỵ7ð365ị. Now365147ðmod 256ịand the required solution is 147.
(e) Since 222ẳ1812ỵ6, the given congruence is equivalent to 6x12ðmod 18ị. Sinceð6, 18ị ẳ6 and 6j12, there are exactly 6 incongruent solutions. As shown in the proof of Theorem XI, these 6 solutions are the first 6 positive integers in the set of all solutions ofx2ðmod 3ị, that is, the first 6 positive integers inẵ2 2Zmod 3. They are then 2, 5, 8, 11, 14, 17.
5.15. Write 141 and 152 with base 4. Form their sum and product, and check each result.
141ẳ432ỵ 420ỵ 43ỵ 1; the representation is 2031 152ẳ432ỵ 421ỵ 42ỵ 0; the representation is 2120 Sum.
1 ỵ0 ẳ 1; 3ỵ2 ẳ 11, we write 1 and carry 1; 1 ỵ 1 ỵ0 ẳ 2; 2 ỵ2 ẳ 10 Thus, the sum is 10211, base 4, and 293, base 10.
Product.
Multiply by 0: 0000
Multiply by 2:21ẳ2; 23ẳ12, write 2 and carry 1; etc: 10122
Multiply by 1: 2031
Multiply by 2: 10122
11032320 The product is 11032320, base 4, and 21432, base 10.
Supplementary Problems
5.16. Show that the relation (j) is reflexive and transitive but not symmetric.
5.17. Prove: Ifajb, thenajb,aj b, andaj b.
5.18. List all the positive primes (a)<50, (b)<200.
Ans. (a) 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
5.19. Prove: If aẳbqỵr, where a,b,q,r2Z, then any common divisor of aandb also divides r while any common divisor ofbandralso dividesa.
5.20. Find the greatest common divisor of each pair of integers and express it in the form of Theorem II:
ðaị237, 81 Ans: 3ẳ13237ỵ ð38ị 81 ðbị616, 427 Ans: 7ẳ 9616ỵ13427 ðcị936, 666 Ans: 18ẳ5936ỵ ð7ị 666 ðdị1137, 419 Ans: 1ẳ2061137ỵ ð559ị 419
68 SOME PROPERTIES OF INTEGERS [CHAP. 5
5.21. Prove: Ifs6ẳ0, thenðsa,sbị ẳ jsj ða,bị.
5.22. Prove:
(a) Ifajs,bjsandða,bị ẳ1, thenabjs.
(b) Ifmẳdm1and ifmjam1, thendja.
5.23. Prove: Ifp, a prime, is a divisor ofabc, thenpjaorpjborpjc.
5.24. The integereẳ ẵa,bis called theleast common multipleof the positive integersaandbwhen (1)ajeandbje, (2) ifajxandbjxthenejx.
5.25. Find: (a)ẵ3, 7, (b)ẵ3, 12, (c)ẵ22, 715.
Ans. (a) 21, (b) 12, (c) 1430
5.26. (a) Write the integersaẳ19, 500 andbẳ54, 450 as products of positive primes.
(b) Finddẳ ða,bịandeẳ ẵa,b.
(c) Verifydeẳab.
(d) Prove the relation in (c) whenaandbare any positive integers.
Ans. (b) 2352; 22325311213
5.27. Prove: Ifm>1,m6 ja,m6 jb, thenmjðabịimpliesamq1ẳrẳbmq2, 0<r<m, and conversely.
5.28. Find all solutions of:
ðaị4x3ðmod 7ị ðeị153x6ðmod 12ị ðbị9x11ðmod 26ị ðfịxỵ13ðmod 7ị ðcị3xỵ14ðmod 5ị ðgị8x6ðmod 422ị ðdị8x6ðmod 14ị ðhị363x345ðmod 624ị
Ans. (a)ẵ6, (b)ẵ7, (c)ẵ1, (d)ẵ6,ẵ13, (e)ẵ2,ẵ6,ẵ10, (f)ẵ2, (g)ẵ159,ẵ370, (h)ẵ123,ẵ331,ẵ539
5.29. Prove Theorems V, VI, VII, VIII.
5.30. Prove: Ifabðmodmịandcbðmodmị, thenacðmodmị. See Examples 10(a), (b), (c).
5.31. (a) Prove: Ifaỵxbỵxðmodmị, thenabðmodmị.
(b) Give a single numerical example to disprove: Ifaxbxðmodmị, thenaxbðmodmị.
(c) Modify the false statement in (b) to obtain a true one.
5.32. (a) Interpretabðmod 0ị.
(b) Show that everyx2Zis a solution ofaxbðmod 1ị.
5.33. (a) Construct addition and multiplication tables forZ5.
(b) Use the multiplication table to obtain 324ðmod 5ị, 341ðmod 5ị, 381ðmod 5ị.
(c) Obtain 32561ðmod 5ị, 35144ðmod 5ị, 310241ðmod 5ị.
5.34. Construct addition and multiplication tables forZ2,Z6,Z7,Z9.
CHAP. 5] SOME PROPERTIES OF INTEGERS 69
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5.35. Prove: Ifẵs 2Zm and ifa,b2 ẵs, thenabðmodmị.
5.36. Prove: Ifẵs,ẵt 2Zmand ifa2 ẵsandb2 ẵt, thenabðmodmịif and only ifẵs ẳ ẵt.
5.37. Express 212 using in turn the base (a) 2, (b) 3, (c) 4, (d) 7, and (e) 9.
Ans. (a) 11010100, (b) 21212, (c) 3110, (d) 422, (e) 255 5.38. Express 89 and 111 with various bases, form the sum and product, and check.
5.39. Prove the first part of the Unique Factorization Theorem using the induction principle stated in Problem 3.27, Chapter 3.
70 SOME PROPERTIES OF INTEGERS [CHAP. 5
The Rational Numbers
INTRODUCTION
The system of integers has an obvious defect in that, given integers m6ẳ0 and s, the equationmxẳs may or may not have a solution. For example, 3xẳ6 has the solutionxẳ2 but 4xẳ6 has no solution.
This defect is remedied by adjoining to the integers additional numbers (common fractions) to form the systemQofrational numbers. The construction here is, in the main, that used in Chapter 4.