Birden ve kendisinden ba\u015fka say\u0131ya b\u00f6l\u00fcnmeyen say\u0131lara asal say\u0131 denir1. \u00d6rne\u011fin 17 asald\u0131r, \u00e7\u00fcnk\u00fc 1 ve 17\u2019den ba\u015fka say\u0131ya (tam olarak) b\u00f6l\u00fcnmez. \u00d6te yandan 35 asal de\u011fildir, 5\u2019e ve 7\u2019ye b\u00f6l\u00fcn\u00fcr. Teknik nedenlerden 1 asal kabul edilmez.<\/p>\n
100\u2019den k\u00fc\u00e7\u00fck asallar\u0131 bulmak pek zor de\u011fildir. \u0130\u015fte o asallar: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Demek ki 100\u2019den k\u00fc\u00e7\u00fck 25 tane asal varm\u0131\u015f. Yani 100\u2019den k\u00fc\u00e7\u00fck rastgele se\u00e7ilmi\u015f bir say\u0131n\u0131n asal olma olas\u0131l\u0131\u011f\u0131 1\/4\u2019t\u00fcr.<\/p>\n
Matematiksel kan\u0131tlar aras\u0131nda bir g\u00fczellik yar\u0131\u015fmas\u0131 yap\u0131lsa, \u00d6klid\u2019in (M\u00d6. 300) \u201csonsuz tane asal say\u0131 vard\u0131r\u201d \u00f6nermesinin kan\u0131t\u0131 hi\u00e7 ku\u015fkusuz ilk on s\u0131rada yer al\u0131rd\u0131. Bu teorem \u00d6klid\u2019in \u00fcnl\u00fc \u00d6\u011feler adl\u0131 yap\u0131t\u0131n\u0131n dokuzuncu cildinde kan\u0131tlan\u0131r. \u00d6klid\u2019in teoreminin g\u00fczelli\u011finin g\u00f6klere \u00e7\u0131kar\u0131lmad\u0131\u011f\u0131 ve kan\u0131tlanmad\u0131\u011f\u0131 pop\u00fcler matematik kitab\u0131 yok gibidir. Birazdan bu g\u00fczel teoremi \u2013 ve \u00e7ok daha fazlas\u0131n\u0131 \u2013 kan\u0131tlayaca\u011f\u0131z.<\/p>\n
Bir say\u0131n\u0131n asal olup olmad\u0131\u011f\u0131n\u0131 nas\u0131l anlar\u0131z? Say\u0131m\u0131za n diyelim. n\u2019yi n\u2019den k\u00fc\u00e7\u00fck say\u0131lara b\u00f6lmeye \u00e7al\u0131\u015fal\u0131m. E\u011fer n\u2019den k\u00fc\u00e7\u00fck, 1\u2019den b\u00fcy\u00fck bir say\u0131 n\u2019yi tam b\u00f6l\u00fcyorsa, n, tan\u0131m\u0131 gere\u011fi, asal olamaz. \u00d6yle bir say\u0131 bulamazsak, n asald\u0131r.<\/p>\n
Ne var ki bu y\u00f6ntemle b\u00fcy\u00fck say\u0131lar\u0131n asall\u0131\u011f\u0131na karar vermek \u00e7ok zaman al\u0131r. Bu y\u00f6ntem ve \u00e7e\u015fitlemeleri d\u0131\u015f\u0131nda bir say\u0131n\u0131n asall\u0131\u011f\u0131na karar verebilecek genel bir y\u00f6ntem de bilinmemektedir. \u00d6rne\u011fin, \u015fu \u00e7e\u015fitleme d\u00fc\u015f\u00fcn\u00fclebilir: n\u2019yi n\u2019den k\u00fc\u00e7\u00fck her say\u0131ya b\u00f6lece\u011fimize, n\u2019yi \u221an\u2019den k\u00fc\u00e7\u00fck say\u0131lara b\u00f6lmeye \u00e7al\u0131\u015fabiliriz. \u00c7\u00fcnk\u00fc n = ab ve a \u2265 \u221an ise, b \u2264 \u221an\u2019dir. Dolay\u0131s\u0131yla n asal de\u011filse, \u221an\u2019den k\u00fc\u00e7\u00fck bir say\u0131ya b\u00f6l\u00fcn\u00fcr. B\u00f6ylece yapmam\u0131z gereken b\u00f6lme say\u0131s\u0131 azal\u0131r. Bir ba\u015fka kolayl\u0131k da \u015f\u00f6yle sa\u011flanabilir: n\u2019nin asal olup olmad\u0131\u011f\u0131na karar vermek i\u00e7in n\u2019yi \u221an\u2019den k\u00fc\u00e7\u00fck her say\u0131ya b\u00f6lmeye \u00e7al\u0131\u015faca\u011f\u0131m\u0131za, \u221an\u2019den k\u00fc\u00e7\u00fck asallara b\u00f6lmeye \u00e7al\u0131\u015fmam\u0131z yeterlidir. Bu birazdan kan\u0131tlayaca\u011f\u0131m\u0131z birinci teoremden \u00e7\u0131kar. B\u00f6ylece, n\u2019nin asall\u0131\u011f\u0131na karar vermek i\u00e7in yapmam\u0131z gereken b\u00f6lme say\u0131s\u0131 daha da azal\u0131r. \u00d6te yandan bu y\u00f6ntemi kullanabilmek i\u00e7in \u221an\u2019den k\u00fc\u00e7\u00fck asallar\u0131 bilmek gerekir. Bu asallar\u0131 bildi\u011fimizi varsaysak bile, b\u00f6lme say\u0131s\u0131 gene de b\u00fcy\u00fck say\u0131lar i\u00e7in \u00e7ok fazlad\u0131r. \u00d6rne\u011fin, n = 100.000.000.001\u2019in asal olup olmad\u0131\u011f\u0131n\u0131 anlamaya \u00e7al\u0131\u015ft\u0131\u011f\u0131m\u0131z\u0131 varsayal\u0131m bir an. E\u011fer n asal de\u011filse ve k\u00fc\u00e7\u00fck bir asala (\u00f6rne\u011fin 97\u2019ye) b\u00f6l\u00fcnebiliyorsa, n\u2019nin asal olmad\u0131\u011f\u0131na olduk\u00e7a \u00e7abuk karar veririz. Ama ya n asalsa ya da k\u00fc\u00e7\u00fck bir asala b\u00f6l\u00fcnm\u00fcyorsa? Onbinlerce b\u00f6lme i\u015flemi yapmam\u0131z gerekecek.<\/p>\n
Yukarda a\u00e7\u0131klad\u0131\u011f\u0131m\u0131z y\u00f6ntem Yunanl\u0131 matematik\u00e7i Eratosthenes taraf\u0131ndan M.\u00d6. 3. y\u00fczy\u0131lda bulunmu\u015ftur. Bu y\u00f6ntemle 50 rakaml\u0131 bir say\u0131n\u0131n en geli\u015fmi\u015f bilgisayar yard\u0131m\u0131yla asal olup olmad\u0131\u011f\u0131n\u0131 anlamak trilyonlarca y\u0131l al\u0131r. Ya\u015fam ger\u00e7ekten k\u0131sa!<\/p>\n
Baz\u0131 \u00f6zel say\u0131lar\u0131n asall\u0131\u011f\u0131na karar vermek i\u00e7in \u00f6zel y\u00f6ntemler geli\u015ftirilebilir. \u00d6rne\u011fin son rakam\u0131 \u00e7ift olan bir tek asal say\u0131 vard\u0131r, o da 2\u2019dir. \u00c7\u00fcnk\u00fc son rakam\u0131 \u00e7ift olan bir say\u0131 2\u2019ye b\u00f6l\u00fcn\u00fcr.<\/p>\n
Asal olmayan say\u0131lara bir ba\u015fka \u00f6rnek vereyim. xa \u2013 1 bi\u00e7iminde yaz\u0131lan say\u0131lar x \u20131\u2019e b\u00f6l\u00fcn\u00fcrler: <\/p>\n
xa \u2013 1 = (x \u2013 1)(xa\u20131 + xa\u20132 + … + x + 1).<\/p>\n
Dolay\u0131s\u0131yla, bir a > 1 say\u0131s\u0131 i\u00e7in, xa \u2013 1 bi\u00e7iminde yaz\u0131lan bir say\u0131n\u0131n asal olabilmesi i\u00e7in x\u2019in 2 olmas\u0131 gerekmektedir. Madem \u00f6yle, 2a \u2013 1 bi\u00e7iminde yaz\u0131lan say\u0131lara bakal\u0131m. Bu say\u0131lar asal m\u0131d\u0131r? <\/p>\n
Sav: E\u011fer a asal de\u011filse 2a \u2013 1 de asal olamaz. <\/p>\n
Kan\u0131t: Bunu kan\u0131tlamak i\u00e7in \u00f6nce a = bc yazal\u0131m. a asal olmad\u0131\u011f\u0131ndan bu e\u015fitli\u011fi sa\u011flayan b ve c say\u0131lar\u0131 vard\u0131r. Sonra x\u2019i 2b olarak tan\u0131mlay\u0131p k\u00fc\u00e7\u00fck bir hesap yapal\u0131m: 2a \u2013 1 = 2bc \u2013 1 = (2b)c \u2013 1 = xc \u2013 1. Ama xc \u2013 1 say\u0131s\u0131n\u0131n x \u2013 1\u2019e b\u00f6l\u00fcnd\u00fc\u011f\u00fcn\u00fc yukarda g\u00f6rm\u00fc\u015ft\u00fck. Demek ki 2a \u2013 1, x \u2013 1\u2019e b\u00f6l\u00fcn\u00fcr ve asal olamaz. Dolay\u0131s\u0131yla, 2a \u2013 1\u2019in asal olmas\u0131 i\u00e7in a\u2019n\u0131n asal olmas\u0131 gerekmektedir. Kan\u0131t\u0131m\u0131z bitmi\u015ftir. <\/p>\n
Asal bir a i\u00e7in 2a \u2013 1 bi\u00e7iminde yaz\u0131lan say\u0131lara Mersenne say\u0131lar\u0131 denir2. Peki, a asalsa, <\/p>\n
Ma = 2a \u2013 1<\/p>\n
olarak tan\u0131mlanan say\u0131 da asal m\u0131d\u0131r? \u0130lk Mersenne say\u0131lar\u0131na bakal\u0131m: <\/p>\n
M2 = 3 <\/p>\n
M3 = 7 <\/p>\n
M5 = 31<\/p>\n
M7 = 127 <\/p>\n
Bu say\u0131lar\u0131n herbiri asal. Ama bundan sonraki ilk Mersenne say\u0131s\u0131, yani M11, asal de\u011fil: M11 = 23 \u00d7 89.<\/p>\n
Hangi asallar i\u00e7in Ma asald\u0131r? Yan\u0131t bilinmiyor. <\/p>\n
1972\u2019de M19937\u2019in asal oldu\u011funu Bryant Tuckerman bilgisayar yard\u0131m\u0131yla ke\u015ffetti.<\/p>\n
1975\u2019te, on be\u015f ya\u015f\u0131nda iki lise \u00f6\u011frencisi, Laura Nickel ve Curt Noll, M19937\u2019in o zamana dek bilinen en b\u00fcy\u00fck asal oldu\u011funu bir gazeteden \u00f6\u011frenince, \u00e7al\u0131\u015fmaya koyuldular ve \u00fc\u00e7 y\u0131l sonra, 1978\u2019te, bilgisayarlar\u0131n\u0131 350 saat \u00e7al\u0131\u015ft\u0131rd\u0131ktan sonra, M21701\u2019in asal oldu\u011funu buldular. Ve birdenbire \u00fcnlendiler.<\/p>\n
\u015eubat 1979\u2019da Noll, M23209\u2019un asal oldu\u011funu buldu.<\/p>\n
\u0130ki ay sonra, Slowinski M44497\u2019nin asal oldu\u011funu g\u00f6sterdi.<\/p>\n
May\u0131s 1983\u2019te Amerikal\u0131 David Slowinski, M86243\u2019\u00fcn asal oldu\u011funu, bilgisayar yard\u0131m\u0131yla tam 1 saat 3 dakika 22 saniyede kan\u0131tlad\u0131. Ama 86.243 sihirli say\u0131s\u0131n\u0131 bulmak i\u00e7in aylarca u\u011fra\u015ft\u0131. Bilinen klasik y\u00f6ntemle (yani kendisinden k\u00fc\u00e7\u00fck say\u0131lara b\u00f6lmeye \u00e7al\u0131\u015farak) bu say\u0131n\u0131n asal oldu\u011funu kan\u0131tlamak, evrenin \u00f6mr\u00fcn\u00fc a\u015fard\u0131! M86243\u2019\u00fcn tam 25.962 rakam\u0131 oldu\u011funu da ayr\u0131ca belirtelim. Bu kadar bozuk paray\u0131 \u00fcst\u00fcste y\u0131\u011fsan\u0131z, para kuleniz evrenin s\u0131n\u0131rlar\u0131n\u0131 a\u015far! [43]<\/p>\n
Yukardaki asal\u0131 bulan Slowinski, 19 Eyl\u00fcl 1983\u2019te M132049\u2019un asal oldu\u011funu bilgisayarlarla anlad\u0131. Bundan \u00e7ok daha \u00f6nce, Manfred Schroeder adl\u0131 bir matematik\u00e7i, matematiksel y\u00f6ntemlerle, sezgisinin de yard\u0131m\u0131yla, 2130.000 – 1 civarlar\u0131nda bir asal oldu\u011funu tahmin etmi\u015fti zaten.<\/p>\n
Mart 1992\u2019de M756839\u2019un asal oldu\u011fu anla\u015f\u0131ld\u0131.<\/p>\n
12 Ocak 1994\u2019te, Paul Gage ve yine David Slowinsky bilgisayar a\u011flar\u0131nda M859433\u2019\u00fcn asal oldu\u011funu kan\u0131tlad\u0131klar\u0131n\u0131 duyurdular. Hesaplar\u0131n\u0131 gene bilgisayarla yapm\u0131\u015flard\u0131 elbet. <\/p>\n
\u015eimdi, So = 4, Sk+1 = Sk2 – 2 olsun. \u00d6rne\u011fin, S1 = 42 -2 = 14\u2019t\u00fcr. Bunun gibi, S2 = S12 -2 = 142 -2 = 194\u2019t\u00fcr. Bir q asal\u0131 i\u00e7in, Mq\u2019n\u00fcn asal olmas\u0131 i\u00e7in gerekli ve yeterli ko\u015ful, Mq\u2019n\u00fcn Sq\u2019y\u00fc b\u00f6lmesidir. Bu teste Lucas testi denilir. Lucas testi sayesinde \u00e7ok b\u00fcy\u00fck asallar olduk\u00e7a kolay say\u0131lacak i\u015flemlerle bulunabilir.<\/p>\n
Bu sonu\u00e7lara bilgisayarlara g\u00fcvenebildi\u011fimiz derecede g\u00fcvenebiliriz elbet. Bilgisayarlar da hata yaparlar!<\/p>\n
B\u00fcy\u00fck say\u0131lar\u0131n asal olup olmad\u0131klar\u0131n\u0131 anlamak, \u015fifreli mesajlarda (kriptoloji) \u00e7ok \u00f6nemlidir ve geli\u015fmi\u015f \u00fclkelerin ordular\u0131 bu y\u00fczden asal say\u0131larla \u00e7ok ilgilenirler. Gizli mesaj yollamak isteyen, mesaj\u0131yla birlikte iki b\u00fcy\u00fck asal say\u0131n\u0131n \u00e7arp\u0131m\u0131n\u0131 da yollar. \u015eifreyi \u00e7\u00f6zmek i\u00e7in, \u015fifreyle birlikte yollanan say\u0131y\u0131 b\u00f6len o iki asal\u0131 bilmek gerekir, ki bu da d\u0131\u015fardan birisi i\u00e7in (say\u0131lar b\u00fcy\u00fck oldu\u011fundan) hemen hemen olanaks\u0131zd\u0131r. \u0130ki say\u0131y\u0131 \u00e7arpmak kolayd\u0131r ama bir say\u0131y\u0131 \u00e7arpanlar\u0131na ay\u0131rmak \u00e7ok daha zordur.<\/p>\n
\u015eifrelemede Mersenne say\u0131lar\u0131 kullan\u0131lmaz. \u00c7\u00fcnk\u00fc az say\u0131da (30 k\u00fcsur tane olmal\u0131) asal Mersenne say\u0131s\u0131 bilindi\u011finden, \u015fifreyle birlikte yollanan say\u0131n\u0131n asal bir Mersenne say\u0131s\u0131na b\u00f6l\u00fcn\u00fcp b\u00f6l\u00fcnmedi\u011fini anlamak kolayd\u0131r.<\/p>\n
Asal olmayan bir say\u0131y\u0131 b\u00f6lenlerine ay\u0131rman\u0131n Fermat\u2019n\u0131n buldu\u011fu \u015fu y\u00f6ntem vard\u0131r. E\u011fer n say\u0131s\u0131 iki pozitif do\u011fal say\u0131 i\u00e7in x2 – y2 bi\u00e7iminde yaz\u0131l\u0131yorsa, o zaman, <\/p>\n
n = (x – y)(x + y)<\/p>\n
e\u015fitli\u011fi do\u011frudur ve x, y +1 olmad\u0131\u011f\u0131 s\u00fcrece, n\u2019yi \u00e7arpanlar\u0131na ay\u0131rm\u0131\u015f oluruz. Bunun tersi de a\u015fa\u011f\u0131 yukar\u0131 do\u011frudur. E\u011fer n = ab ise ve n \u00e7ift de\u011filse, o zaman,<\/p>\n
x = <\/p>\n
ve<\/p>\n
y = –<\/p>\n
alarak, n = x2 – y2 e\u015fitli\u011fini elde ederiz. Demek ki, \u00e7ift olmayan bir n do\u011fal say\u0131s\u0131n\u0131 \u00e7arpanlar\u0131na ay\u0131rmak i\u00e7in, n = x2 – y2 e\u015fitli\u011fini sa\u011flayan x ve y bulmal\u0131y\u0131z. Bu e\u015fitlik yerine y2 = x2 – n yazal\u0131m ve x yerine teker teker say\u0131lar\u0131 koyup x2 – n say\u0131s\u0131n\u0131 hesaplayal\u0131m. Bu say\u0131 tam bir kare (y2) oldu\u011funda n = x2 – y2 e\u015fitli\u011fini bulmu\u015f oluruz. Elbette x\u2019in \u221an\u2019den b\u00fcy\u00fck olmas\u0131 gerekmektedir, yoksa x2 – n pozitif bile olamaz. Ayr\u0131ca, x2 – n say\u0131s\u0131n\u0131n tam bir kare olmas\u0131 i\u00e7in 0, 1, 4, 5, 6 ve 9\u2019la bitmesi gerekmektedir, 2, 3, 7 ve 8\u2019le biten say\u0131lar kare olamazlar.<\/p>\n
Bu y\u00f6ntemi n = 91 i\u00e7in deneyelim. x > \u221a91 olmas\u0131 gerekti\u011finden, x = 10\u2019dan ba\u015flamal\u0131y\u0131z. x = 10 ise, x2 – n = 102 – 91 = 9 = 32 dir ve y = 3 olabilir. Demek ki,<\/p>\n
91 = n = 102 – 32 = (10 – 3)(10 + 3) = 7 \u00d7 13<\/p>\n
e\u015fitli\u011fi ge\u00e7erlidir.<\/p>\n
Ayn\u0131 y\u00f6ntemi n = 143 i\u00e7in deneyecek olursan\u0131z, gene yan\u0131t\u0131 hemen bulursunuz: x = 12, y = 1. <\/p>\n
Mersenne say\u0131lar\u0131na \u00e7ok benzeyen ba\u015fka say\u0131lara bakal\u0131m. 2a + 1 bi\u00e7iminde yaz\u0131lan say\u0131lar asal m\u0131d\u0131r? Bu say\u0131lar\u0131n hangi a\u2019lar i\u00e7in asal olduklar\u0131n\u0131 bilmiyoruz ama hangi a\u2019lar i\u00e7in asal olamayacaklar\u0131n\u0131 biliyoruz: E\u011fer a, 2\u2019nin bir g\u00fcc\u00fc de\u011filse, yani 2n bi\u00e7iminde yaz\u0131lamazsa, bu say\u0131lar asal olamazlar. Bunu birazdan kan\u0131tlayaca\u011f\u0131z (Teorem 9.) Fermat, <\/p>\n
Fn = <\/p>\n
bi\u00e7iminde yaz\u0131lan b\u00fct\u00fcn say\u0131lar\u0131n asal olduklar\u0131n\u0131 san\u0131yordu. Bu y\u00fczden bu say\u0131lara Fermat say\u0131lar\u0131 denir. Ger\u00e7ekten de ilk be\u015f Fermat say\u0131s\u0131,<\/p>\n
Fo = 3<\/p>\n
F1 = 5 <\/p>\n
F2 = 17 <\/p>\n
F3 = 257 <\/p>\n
F4 = 65537<\/p>\n
asald\u0131r. Fermat, b\u00fct\u00fcn Fermat say\u0131lar\u0131n\u0131n asal olduklar\u0131n\u0131 kan\u0131tlamaya u\u011fra\u015ft\u0131 ama ba\u015faramad\u0131. Ba\u015far\u0131s\u0131zl\u0131\u011f\u0131n\u0131n nedeni vard\u0131: San\u0131s\u0131 do\u011fru de\u011fildi. F5 asal de\u011fildir. F5 on basamakl\u0131 bir say\u0131 oldu\u011fundan asall\u0131\u011f\u0131n\u0131 kan\u0131tlamak kolay de\u011fildi. Euler (1707-1783), F5\u2019in 641\u2019e b\u00f6l\u00fcnd\u00fc\u011f\u00fcn\u00fc g\u00f6sterdi: <\/p>\n
F5 = 641 \u00d7 6700417.<\/p>\n
Demek ki a = 2n bi\u00e7iminde yaz\u0131labilse bile, 2a + 1 asal olmayabiliyor.<\/p>\n
Lucas F6\u2019n\u0131n asal olmad\u0131\u011f\u0131n\u0131 kan\u0131tlad\u0131. Daha sonra, 1880\u2019de, Landry, <\/p>\n
F6 = 274177 \u00d7 67280421310721<\/p>\n
e\u015fitli\u011fini buldu. F7 ve F8 de asal de\u011filler. Bu say\u0131lar\u0131n asal olmad\u0131klar\u0131, \u00e7ok ge\u00e7 bir tarihte, 1970 ve 1981\u2019de anla\u015f\u0131ld\u0131. W. Keller, 1980\u2019de F9448\u2019in asal olmad\u0131\u011f\u0131n\u0131 g\u00f6sterdi. Bu say\u0131 19 \u00d7 29450 + 1\u2019e b\u00f6l\u00fcn\u00fcr. 1984\u2019de gene W. Keller, F23471\u2019in asal olmad\u0131\u011f\u0131n\u0131 kan\u0131tlad\u0131. Bu say\u0131n\u0131n 107000\u2019den fazla basama\u011f\u0131 vard\u0131r ve 5 \u00d7 223473 + 1\u2019e b\u00f6l\u00fcn\u00fcr. <\/p>\n
n \u2265 5 i\u00e7in, asal bir Fn\u2019nin olup olmad\u0131\u011f\u0131 \u015fimdilik bilinmiyor. Asall\u0131\u011f\u0131 bilinmeyen en k\u00fc\u00e7\u00fck Fermat say\u0131lar\u0131 \u015funlar: F22, F24, F28. <\/p>\n
Son y\u0131llarda bir say\u0131n\u0131n asall\u0131\u011f\u0131na y\u00fczde olarak olduk\u00e7a \u00e7abuk karar verebilen y\u00f6ntemler geli\u015ftirildi. \u00d6rne\u011fin, \u201c\u015eu say\u0131 y\u00fczde 99,978 olas\u0131l\u0131kla asald\u0131r,\u201d gibi \u00f6nermeler bilgisayarlar\u0131n yard\u0131m\u0131yla olduk\u00e7a k\u0131sa say\u0131labilecek zamanda kan\u0131tland\u0131. Bu konuda bilgim k\u0131s\u0131tl\u0131 oldu\u011fundan daha fazla s\u00f6z s\u00f6yleyemeyece\u011fim.<\/p>\n
11, 111, 1111, 11111 gibi her rakam\u0131 1 olan say\u0131lar asal m\u0131d\u0131r? \u0130\u00e7inde n tane 1 olan say\u0131ya Bn diyelim. E\u011fer \u00e7ift say\u0131da 1 varsa, yani n \u00e7iftse, Bn, 11\u2019e b\u00f6l\u00fcn\u00fcr ve B2 d\u0131\u015f\u0131nda bunlardan hi\u00e7biri asal olamaz. E\u011fer n \u00fc\u00e7e b\u00f6l\u00fcn\u00fcyorsa Bn de \u00fc\u00e7e b\u00f6l\u00fcn\u00fcr ve asal olamaz. <\/p>\n
Hangi n\u2019ler i\u00e7in Bn asald\u0131r? Bu asallardan ka\u00e7 tane vard\u0131r? B2, B19, B23, B317, B1031 asal say\u0131lar, bu biliniyor. Bunlardan ba\u015fka? Ben bilmiyorum. Bu say\u0131lardan daha b\u00fcy\u00fck bir asal varsa, n > 10.000 olmas\u0131 gerekti\u011fini Harvey Dubner adl\u0131 biri kan\u0131tlam\u0131\u015f, daha do\u011frusu hesaplam\u0131\u015f. [43] <\/p>\n
Asallar matematikte \u00e7ok \u00f6nemlidir elbet. Bu yaz\u0131da bu \u00f6nemli konuda bir iki teorem kan\u0131tlayaca\u011f\u0131z. \u0130lk teoremimizi okurlar\u0131n \u00e7o\u011fu biliyordur. <\/p>\n
Teorem 1. 1\u2019den b\u00fcy\u00fck her say\u01313 bir asala b\u00f6l\u00fcn\u00fcr. <\/p>\n
Kan\u0131t: Bunun kan\u0131t\u0131 olduk\u00e7a kolayd\u0131r: a > 1 bir say\u0131 olsun. a\u2019n\u0131n bir asala b\u00f6l\u00fcnd\u00fc\u011f\u00fcn\u00fc kan\u0131tlamak istiyoruz.<\/p>\n
E\u011fer a asalsa bir sorun yok: a, a\u2019y\u0131 b\u00f6ler ve teoremimiz kan\u0131tlanm\u0131\u015f olur (a bir asala (kendisine!) b\u00f6l\u00fcn\u00fcr.)<\/p>\n
E\u011fer a asal de\u011filse, a\u2019y\u0131 b\u00f6len ve 1 < b < a e\u015fitsizliklerini sa\u011flayan bir b vard\u0131r. E\u011fer b asalsa bir sorun yok: b, a\u2019y\u0131 b\u00f6ler ve teoremimiz kan\u0131tlanm\u0131\u015f olur.\n\n E\u011fer b asal de\u011filse, b\u2019yi (ve dolay\u0131s\u0131yla a\u2019y\u0131 da) b\u00f6len ve 1 < c < b e\u015fitsizliklerini sa\u011flayan bir c vard\u0131r. E\u011fer c asalsa bir sorun yok: c, a\u2019y\u0131 b\u00f6ler ve teoremimiz kan\u0131tlanm\u0131\u015f olur.\n\n E\u011fer c asal de\u011filse, c\u2019yi (ve dolay\u0131s\u0131yla a\u2019y\u0131 da) b\u00f6len ve 1 < d < c e\u015fitsizliklerini sa\u011flayan bir d vard\u0131r. E\u011fer d asalsa bir sorun yok: d, a\u2019y\u0131 b\u00f6ler ve teoremimiz kan\u0131tlanm\u0131\u015f olur.\n\n E\u011fer d asal de\u011filse.......\n\n Nereye dek gidebiliriz? Bulaca\u011f\u0131m\u0131z her say\u0131 bir \u00f6ncekinden k\u00fc\u00e7\u00fck ve 1\u2019den b\u00fcy\u00fck oldu\u011fundan sonsuza dek bunu b\u00f6yle s\u00fcrd\u00fcremeyiz. Bir zaman sonra durmal\u0131y\u0131z, yani bir zaman sonra a\u2019y\u0131 b\u00f6len bir asal buluruz. Teoremimiz kan\u0131tlanm\u0131\u015ft\u0131r. ? \n\n\n Birazdan yukarda g\u00fczelli\u011finden s\u00f6zetti\u011fimiz \u00d6klid Teoremini kan\u0131tlayaca\u011f\u0131z: Sonsuz tane asal say\u0131 vard\u0131r. Ayn\u0131 y\u00f6ntemle ba\u015fka sonu\u00e7lar da \u00e7\u0131karaca\u011f\u0131z. \u0130lk \u00f6nce biraz ilkokul aritmeti\u011fi yapal\u0131m.\n\n E\u011fer a ve b say\u0131lar\u0131 n\u2019ye b\u00f6l\u00fcn\u00fcyorsa, bu iki say\u0131n\u0131n toplam\u0131 da n\u2019ye b\u00f6l\u00fcn\u00fcr. \u00d6rne\u011fin hem 78, hem 66 \u00fc\u00e7e b\u00f6l\u00fcnd\u00fc\u011f\u00fcnden, 78 + 66 da, yani 144 de, \u00fc\u00e7e b\u00f6l\u00fcn\u00fcr.\n\n \u00d6te yandan e\u011fer a ve b say\u0131lar\u0131ndan yaln\u0131zca biri n\u2019ye b\u00f6l\u00fcn\u00fcyor, \u00f6b\u00fcr\u00fc b\u00f6l\u00fcnm\u00fcyorsa, bu iki say\u0131n\u0131n toplam\u0131 n\u2019ye b\u00f6l\u00fcnmez. \u00d6rne\u011fin 78 \u00fc\u00e7e b\u00f6l\u00fcn\u00fcr, 67 b\u00f6l\u00fcnmez. Dolay\u0131s\u0131yla 78 + 67 \u00fc\u00e7e b\u00f6l\u00fcnmez.\n\n Bir \u00fcst paragraftaki b\u2019yi 1 olarak al\u0131rsak, a\u2019y\u0131 b\u00f6len 1\u2019den b\u00fcy\u00fck bir say\u0131n\u0131n a + 1\u2019i b\u00f6lemeyece\u011fi \u00e7\u0131kar. Demek ki a ve a+1 say\u0131lar\u0131n\u0131n 1\u2019den ba\u015fka ortak b\u00f6leni yoktur.\n\n Hem ikiye, hem de \u00fc\u00e7e b\u00f6l\u00fcnen bir say\u0131ya 1 eklersek, elde etti\u011fimiz say\u0131 ne ikiye ne de \u00fc\u00e7e b\u00f6l\u00fcn\u00fcr. Bunun gibi, 2 \u00d7 3 \u00d7 4 \u00d7 5 \u00d7 6 \u00d7 7, yani 5040, 2\u2019ye, 3\u2019e, 4\u2019e, 5\u2019e, 6\u2019ya, 7\u2019ye b\u00f6l\u00fcn\u00fcr, ama bu say\u0131ya 1 ekleyerek elde etti\u011fimiz 5041, bunlardan hi\u00e7birine b\u00f6l\u00fcnmez.\n\n Ayn\u0131 \u015fey a ve a \u2013 1 say\u0131lar\u0131 i\u00e7in de ge\u00e7erlidir. \u00d6rne\u011fin, 5040\u2019\u0131 b\u00f6len 1\u2019den b\u00fcy\u00fck hi\u00e7bir say\u0131 5039\u2019u b\u00f6lemez.\n\n 5039 ve 5041 say\u0131lar\u0131n\u0131n 7 ve 7\u2019den k\u00fc\u00e7\u00fck hi\u00e7bir asala b\u00f6l\u00fcnmediklerini g\u00f6rd\u00fck. \u00d6te yandan, Teorem 1\u2019e g\u00f6re, bu say\u0131lardan herbiri bir asala b\u00f6l\u00fcnmeli. Demek ki 7\u2019den b\u00fcy\u00fck bir asal vard\u0131r. Bunun gibi 2\u2019yle 11 aras\u0131ndaki say\u0131lar\u0131n \u00e7arp\u0131m\u0131na 1 eklersek, elde edilen say\u0131 bir asala b\u00f6l\u00fcn\u00fcr ve bu asal 11\u2019den b\u00fcy\u00fck olmak zorundad\u0131r. Bu ak\u0131l y\u00fcr\u00fctmeyi genelle\u015ftirece\u011fiz: \n\n\n Teorem 2. Sonsuz tane asal say\u0131 vard\u0131r. \n\n\n Kan\u0131t: n > 1 herhangi bir say\u0131 olsun. 2\u2019den n\u2019ye kadar b\u00fct\u00fcn say\u0131lar\u0131 birbiriyle \u00e7arpal\u0131m: 2 \u00d7 3 \u00d7 … \u00d7 (n\u20132) \u00d7 (n\u20131) \u00d7 n. Kocaman bir say\u0131 elde ettik. Bu say\u0131 n! olarak simgelenir. n! say\u0131s\u0131 n + 1\u2019den k\u00fc\u00e7\u00fck b\u00fct\u00fcn say\u0131lara b\u00f6l\u00fcn\u00fcr elbet, \u00e7\u00fcnk\u00fc n! bu say\u0131lar\u0131n \u00e7arp\u0131m\u0131. Demek ki n! + 1 say\u0131s\u0131 1\u2019le n aras\u0131ndaki hi\u00e7bir say\u0131ya b\u00f6l\u00fcnemez. \u00d6te yandan, Teorem 1\u2019e g\u00f6re n! + 1 say\u0131s\u0131 bir asala b\u00f6l\u00fcnmeli. Demek ki n\u2019den b\u00fcy\u00fck bir asal vard\u0131r.<\/p>\n
Ne bulduk? Her say\u0131dan b\u00fcy\u00fck bir asal bulduk. Dolay\u0131s\u0131yla sonsuz tane asal vard\u0131r, \u00e7\u00fcnk\u00fc her asaldan b\u00fcy\u00fck bir ba\u015fka asal vard\u0131r. \u0130kinci teorem kan\u0131tlanm\u0131\u015ft\u0131r. ? <\/p>\n
Ne denli yal\u0131n bir kan\u0131t de\u011fil mi? Ve \u015fa\u015f\u0131rt\u0131c\u0131. \u015eu nedenden \u015fa\u015f\u0131rt\u0131c\u0131: Kan\u0131t, n\u2019den sonra gelen ilk asal\u0131 bulmuyor; yaln\u0131zca n\u2019den b\u00fcy\u00fck bir asal\u0131n varl\u0131\u011f\u0131 kan\u0131tlan\u0131yor. \u00d6rne\u011fin 1 milyondan b\u00fcy\u00fck bir asal vard\u0131r. Hangi asal? Yan\u0131t yok! Kan\u0131t, hangi asal\u0131n 1 milyondan b\u00fcy\u00fck oldu\u011funu g\u00f6stermiyor. \u201c\u00d6yle bir asal var\u201d demekle yetiniyor.<\/p>\n
Asl\u0131nda kan\u0131t\u0131m\u0131z n\u2019den b\u00fcy\u00fck asallar \u00fczerine hi\u00e7 de bilgi vermiyor de\u011fil. En az\u0131ndan, her n i\u00e7in, n < p \u2264 n! + 1 e\u015fitsizliklerini sa\u011flayan bir p asal\u0131n\u0131n oldu\u011funu kan\u0131tl\u0131yor. \n\n\n Teorem 3. Her n > 1 i\u00e7in, n < p \u2264 n! + 1 e\u015fitsizliklerini sa\u011flayan bir asal vard\u0131r. ? \n\n\n Hangi n asal bir say\u0131lar\u0131 i\u00e7in n! + 1 asald\u0131r? Bence bu pek ilgin\u00e7 bir soru de\u011fil ama, merakl\u0131lar b\u00f6yle sorular soruyorlar. Yan\u0131t bilinmiyor. 1987\u2019de H. Dubner, n = 13649 i\u00e7in, ki bu asal bir say\u0131d\u0131r, 5862 basamakl\u0131 n! + 1 say\u0131s\u0131n\u0131n asal oldu\u011funu g\u00f6sterdi. \n\n Yukardaki teoeremde, n! + 1 say\u0131s\u0131n\u0131 biraz daha k\u00fc\u00e7\u00fcltebiliriz. Teorem 2\u2019nin kan\u0131t\u0131n\u0131n hemen hemen ayn\u0131s\u0131, n! yerine, n\u2019den k\u00fc\u00e7\u00fck ya da e\u015fit asallar\u0131n \u00e7arp\u0131m\u0131n\u0131 alabilece\u011fimizi g\u00f6steriyor. \u00d6rne\u011fin, n = 29 ise, \n\n 29 < p \u2264 2\u00d73\u00d75\u00d77\u00d711\u00d713\u00d717\u00d719\u00d723\u00d729 + 1\n\n e\u015fitsizli\u011fini sa\u011flayan bir asal vard\u0131r. \n\n B\u00fct\u00fcn bunlar akla bir ba\u015fka soru getiriyor. Ardarda gelen, \u00f6rne\u011fin, her bin say\u0131dan en az biri asal m\u0131d\u0131r? Ba\u015fka bir deyi\u015fle, n herhangi bir say\u0131ysa, \n\n n + 1, n + 2, n + 3, ..., n + 1000\n\n say\u0131lar\u0131ndan biri asal m\u0131d\u0131r?\n\n Bu soruyu yan\u0131tlamak i\u00e7in yeterli bilgiye sahibiz. Yan\u0131t olumsuzdur. Yan\u0131t\u0131n olumsuz oldu\u011funu kan\u0131tlayal\u0131m.\n\n Bir \u00f6rnekle ba\u015flayal\u0131m. 7! = 2 \u00d7 3 \u00d7 4 \u00d7 5 \u00d7 6 \u00d7 7, yani 5040, 2\u2019ye, 3\u2019e, 4\u2019e, 5\u2019e, 6\u2019ya ve 7\u2019ye b\u00f6l\u00fcn\u00fcr. Dolay\u0131s\u0131yla \n\n 5042, 2\u2019ye \n\n 5043, 3\u2019e \n\n 5044, 4\u2019e \n\n 5045, 5\u2019e \n\n 5046, 6\u2019ya \n\n 5047, 7\u2019ye\n\n b\u00f6l\u00fcn\u00fcr ve bu say\u0131lardan hi\u00e7biri asal olamaz. Bunun gibi, a\u015fa\u011f\u0131daki bin say\u0131,\n\n 1001! + 2, 1001! + 3, ... , 1001! + 1001\n\n s\u0131ras\u0131yla 2\u2019ye, 3\u2019e, ..., 1001\u2019e b\u00f6l\u00fcn\u00fcrler ve hi\u00e7biri asal olamaz. Bu yapt\u0131\u011f\u0131m\u0131z\u0131 genelle\u015ftirmek i\u015ften bile de\u011fildir: \n\n\n Teorem 4. Ardarda gelen her n say\u0131dan birinin mutlaka asal oldu\u011fu bir n yoktur. \n\n\n Asallarla ilgili bir ba\u015fka soruya ge\u00e7elim. Say\u0131lar\u0131 \u00fc\u00e7 k\u00fcmeye ay\u0131rabiliriz: \n\n A k\u00fcmesi = {3\u2019e b\u00f6l\u00fcnen say\u0131lar } \n\n B k\u00fcmesi = {3\u2019e b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde kalan\u0131n 1 oldu\u011fu say\u0131lar} \n\n C k\u00fcmesi = {3\u2019e b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde kalan\u0131n 2 oldu\u011fu say\u0131lar}\n\n Yani, \n\n A = {3,6,9,12,15,18,...} \n\n B = {4,7,10,13,16,19,...} \n\n C = {5,8,11,14,17,20,...} \n\n B k\u00fcmesinden herhangi iki say\u0131 alal\u0131m: n1 ve n2. Bu say\u0131lar 3\u2019e b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde 1 kal\u0131yor. Dolay\u0131s\u0131yla n1 = 3q1 + 1 ve n2 = 3q2 + 1 olarak yazabiliriz. \u015eimdi n1 ve n2\u2019yi birbiriyle \u00e7arpal\u0131m: \n\n n1n2 = (3q1+1)(3q2+1) = 9q1q2 + 3q1+3q2 +1 = 3(3q1q2+q1+q2)+1\n\n Dolay\u0131s\u0131yla n1n2 say\u0131s\u0131 3\u2019e b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde 1 kal\u0131r. Ne kan\u0131tlad\u0131k? B k\u00fcmesindeki say\u0131lar\u0131n \u00e7arp\u0131mlar\u0131n\u0131n gene B k\u00fcmesinde oldu\u011funu kan\u0131tlad\u0131k. Bunu kullanarak a\u015fa\u011f\u0131daki teoremi kan\u0131tlayaca\u011f\u0131z: \n\n\n Teorem 5. C k\u00fcmesinde sonsuz tane asal vard\u0131r. \n\n\n Kan\u0131t: C k\u00fcmesindeki bir say\u0131, A k\u00fcmesindeki bir say\u0131ya b\u00f6l\u00fcnemez, \u00e7\u00fcnk\u00fc A k\u00fcmesindeki say\u0131lar 3\u2019e b\u00f6l\u00fcn\u00fcyor, oysa C k\u00fcmesindekiler 3\u2019e b\u00f6l\u00fcnm\u00fcyorlar. Demek ki C k\u00fcmesindeki bir say\u0131y\u0131 b\u00f6len say\u0131lar B ve C k\u00fcmesinde olmal\u0131d\u0131r. Ama hepsi birden B\u2019de olamaz, \u00e7\u00fcnk\u00fc B\u2019nin \u00f6\u011feleri kendileriyle \u00e7arp\u0131ld\u0131\u011f\u0131nda gene B\u2019den bir say\u0131 verir. Demek ki C k\u00fcmesinin her say\u0131s\u0131, gene C k\u00fcmesinden bir asala b\u00f6l\u00fcn\u00fcr.\n\n \u015eimdi n \u2265 3 herhangi bir say\u0131 olsun. n! \u2013 1 say\u0131s\u0131n\u0131 ele alal\u0131m. Bu say\u0131ya x diyelim. x, C\u2019dedir, \u00e7\u00fcnk\u00fc, x = (n! \u2013 3) + 2 olarak yaz\u0131labilir ve n! \u2013 3 \u00fc\u00e7e b\u00f6l\u00fcn\u00fcr. Demek ki C k\u00fcmesinde x\u2019i b\u00f6len bir asal vard\u0131r. \u00d6te yandan x\u2019i b\u00f6len say\u0131lar n\u2019den b\u00fcy\u00fckt\u00fcr elbet. Ne kan\u0131tlad\u0131k? n ka\u00e7 olursa olsun, C k\u00fcmesinde n\u2019den b\u00fcy\u00fck bir asal vard\u0131r. Yani C\u2019de sonsuz tane asal vard\u0131r. ? \n\n\n Okur buna benzer bir kan\u0131tla a\u015fa\u011f\u0131daki teoremi kan\u0131tlayabilir: \n\n\n Teorem 6. 4\u2019e b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde kalan\u0131 3 olan sonsuz tane asal vard\u0131r. \n\n\n 18. y\u00fczy\u0131l\u0131n sonlar\u0131na do\u011fru, Frans\u0131z matematik\u00e7isi Legendre (1752-1833) son iki teoremi genelle\u015ftirmek istedi. \u015eu soruyu sordu: \n\n\n Soru. a ve b, 1\u2019den ba\u015fka ortak b\u00f6leni olmayan iki say\u0131 olsun. ax+b bi\u00e7iminde yaz\u0131lan sonsuz tane asal var m\u0131d\u0131r? \n\n\n Teorem 5\u2019ten a = 3, b = 2 i\u00e7in, Teorem 6\u2019dan da a = 4, b = 3 i\u00e7in yan\u0131t\u0131n olumlu oldu\u011fu anla\u015f\u0131l\u0131yor. Legendre bu soruyu genel olarak yan\u0131tlamak istedi. \u00d6rne\u011fin 25x + 6 bi\u00e7iminde yaz\u0131lan sonsuz tane asal var m\u0131d\u0131r? E\u011fer x = 1 ise 31 buluruz ki, 31 asald\u0131r. E\u011fer x = 2, 3, 4 ise, s\u0131ras\u0131yla 56, 81, 106 buluruz ve bunlardan hi\u00e7biri asal de\u011fildir. x = 5 oldu\u011funda 131 \u00e7\u0131kar ve 131 asald\u0131r.\n\n Legendre sorunun yan\u0131t\u0131n\u0131n olumlu oldu\u011fundan hi\u00e7 ku\u015fku duymad\u0131, ancak kan\u0131tlamakta g\u00fc\u00e7l\u00fck \u00e7ekti. 1785\u2019te defterine \u201cbunu bilimsel olarak kan\u0131tlamal\u0131\u201d diye not d\u00fc\u015fm\u00fc\u015f. On d\u00f6rt y\u0131l sonra, 1798\u2019de, \u201cdo\u011frulu\u011fundan ku\u015fku duymamal\u0131y\u0131z\u201d diye yazm\u0131\u015f. Sonra da kan\u0131tlamaya \u00e7al\u0131\u015fm\u0131\u015f. Ba\u015faramadan... \u0130kinci denemesini Say\u0131lar Kuram\u0131 adl\u0131 kitab\u0131na ald\u0131\u011f\u0131n\u0131 biliyoruz [26]. Ama bu denemesi de yanl\u0131\u015f. Kan\u0131t\u0131n yanl\u0131\u015fl\u0131\u011f\u0131n\u0131n ne zaman anla\u015f\u0131ld\u0131\u011f\u0131n\u0131 bilmiyorum. 1837\u2019de, meslekta\u015f\u0131 Legendre gibi Frans\u0131z olan G. L. Dirichlet (1805-1859) teoremi do\u011fru olarak kan\u0131tlad\u0131 [8]: \n\n\n Teorem 7. a ve b ortak b\u00f6leni olmayan iki do\u011fal say\u0131ysa, ax+b bi\u00e7iminde yaz\u0131lan sonsuz tane asal say\u0131 vard\u0131r.\n\n\n\n Dirichlet\u2019nin y\u00f6nteminden bir ba\u015fka teorem daha elde edilebilir: \n\n\n Teorem 8. a, b ve c ortak b\u00f6leni olmayan \u00fc\u00e7 pozitif do\u011fal say\u0131 olsunlar. ax2 + bxy + cy2 bi\u00e7iminde yaz\u0131lan sonsuz tane asal vard\u0131r.\n\n\n\n Sadece ve sadece asal say\u0131lar\u0131 ve her asal say\u0131y\u0131 veren bir form\u00fcl var m\u0131d\u0131r? Genel olarak san\u0131lan\u0131n tersine b\u00f6yle bir form\u00fcl vard\u0131r. \u00d6yle bir form\u00fcl vard\u0131r ki, bu form\u00fclle yaln\u0131z ve yaln\u0131z asal say\u0131lar elde edilir ve her asal say\u0131 bu form\u00fclle elde edilir. Olduk\u00e7a kolay bir form\u00fcld\u00fcr bu. \u0130\u015fte form\u00fcl:\n\n n ve m herhangi iki do\u011fal say\u0131 olsun.\n\n k = m(n + 1) - (n! + 1)\n\n olarak tan\u0131mlans\u0131n. \u015eimdi,\n\n p = -|-|-- + 2\n\n her n ve m say\u0131s\u0131 i\u00e7in asald\u0131r! Ayr\u0131ca her asal say\u0131 bu bi\u00e7imde elde edilebilir.\n\n Bu form\u00fclle s\u0131k s\u0131k 2 elde ederiz, ama 2 d\u0131\u015f\u0131ndaki her asal say\u0131 bu form\u00fclle ancak bir kez, yani bir tek n ve m de\u011ferleri i\u00e7in elde edilebilir.\n\n E\u011fer k2 - 1 \u2265 0 ise, yukardaki form\u00fcl hep p = 2 verir. Ama k2 - 1 < 0 ise, yani k2 < 1 ise, yani k2 = 0 ise, yani k = 0 ise, yani m(n + 1) - (n! + 1) = 0 ise, yani,\n\n m = \n\n ise, yukardaki form\u00fcl p = n + 1 verir. Bu say\u0131 asald\u0131r, \u00e7\u00fcnk\u00fc Wilson\u2019\u0131n \u00fcnl\u00fc teoremine g\u00f6re, m\u2019nin tamsay\u0131 olabilmesi i\u00e7in, yani n + 1\u2019in n! + 1\u2019i b\u00f6lebilmesi i\u00e7in, n + 1\u2019in asal olmas\u0131 gerekmektedir.\n\n \u00d6rne\u011fin, n = 2 ve m = 1 ise, p = 3 bulunur. E\u011fer n = 4 ve m = 5 ise, p = 5 bulunur. E\u011fer, n = 6 ve m = 103 ise, p = 7 bulunur. Gelecek asal\u0131, yani 11\u2019i bulmak i\u00e7in, yani p = 11 \u00e7\u0131kmas\u0131 i\u00e7in, n\u2019nin 10 olmas\u0131, m\u2019nin de \n\n\n yani 329.891 olmas\u0131 gerekmektedir. Hangi n ve m say\u0131lar\u0131 i\u00e7in p = 13 bulunaca\u011f\u0131n\u0131 okur kolayl\u0131kla bulabilir.\n\n Hardy ve Wright, bir \u03c9 = 1,9287800\u2026 say\u0131s\u0131 i\u00e7in, \n\n f(n) = \n\n say\u0131s\u0131n\u0131n (n tane 2 var) bir asal oldu\u011funu g\u00f6sterdiler [47]4. \u00d6rne\u011fin f(1) = 3, f(2) = 13, f(3) = 16381. f(4)\u2019\u00fc hesaplamak zor, basamak say\u0131s\u0131 5000 civar\u0131nda. \u00d6te yandan, \u03c9 say\u0131s\u0131n\u0131 belirlemek i\u00e7in, asal say\u0131lar\u0131 bilmek gerekti\u011finden, bu form\u00fcl pek i\u015fe yaramaz. Gene de \u00f6yle bir \u03c9 say\u0131s\u0131n\u0131n varl\u0131\u011f\u0131 ilgin\u00e7. \n\n Her asal\u0131 veren bir form\u00fcl var ama, her asal\u0131 veren bir polinomun5 olmad\u0131\u011f\u0131 biliniyor. E\u011fer katsay\u0131lar\u0131 tamsay\u0131 olan her polinomun sonsuz tane asal olmayan say\u0131 verdi\u011fi bilinir.\n\n 1772\u2019de Euler, n2 + n + 41 polinomunun n = 0,1,2,\u2026,39 i\u00e7in asal say\u0131lar verdi\u011fini buldu. Ancak bu polinom n = 40 i\u00e7in 41\u2019e b\u00f6l\u00fcn\u00fcr ve asal de\u011fildir. \n\n\n Fermat say\u0131lar\u0131 \u00fczerine bir teorem kan\u0131tlayaca\u011f\u0131m\u0131za s\u00f6zvermi\u015ftik. S\u00f6z\u00fcm\u00fcz\u00fc tutuyoruz: \n\n\n Teorem 9. E\u011fer a = 2n bi\u00e7iminde yaz\u0131lamazsa, 2a + 1 asal olamaz. \n\n\n Kan\u0131t: \u00d6nce \u015funu belleyelim: x herhangi bir say\u0131 ve a > 1 bir tek say\u0131ysa, xa + 1 say\u0131s\u0131 asal olamaz, \u00e7\u00fcnk\u00fc x + 1\u2019e b\u00f6l\u00fcn\u00fcr. \u015e\u00f6yle b\u00f6l\u00fcn\u00fcr: <\/p>\n
xa + 1 = (x+1)(xa\u20131 \u2013 xa\u20132 + xa\u20133 \u2013 xa\u20134 + … \u2013 x + 1.)<\/p>\n
\u015eimdi a\u2019n\u0131n bir tek say\u0131ya b\u00f6l\u00fcnd\u00fc\u011f\u00fcn\u00fc varsayal\u0131m. 2a + 1\u2019in asal olamayaca\u011f\u0131n\u0131 kan\u0131tlamak istiyoruz. a\u2019y\u0131 b\u00f6len tek say\u0131ya m diyelim. Demek ki a = nm ve m bir tek say\u0131. x = 2n olsun. K\u00fc\u00e7\u00fck bir hesap yapal\u0131m: <\/p>\n
2a + 1 = 2nm + 1 = (2n)m + 1 = xm + 1. <\/p>\n
m tek oldu\u011fundan, ilk paragrafta g\u00f6rd\u00fc\u011f\u00fcm\u00fcz gibi, x + 1, xm + 1\u2019i b\u00f6ler. Yani x + 1, 2a + 1\u2019i b\u00f6ler.<\/p>\n
Demek ki a bir tek say\u0131ya b\u00f6l\u00fcn\u00fcyorsa, 2a + 1 asal olamaz. Dolay\u0131s\u0131yla a, 2\u2019nin bir kat\u0131 olmal\u0131. ? <\/p>\n
Asallar \u00fczerine bildiklerimiz bilmediklerimizin yan\u0131nda hi\u00e7 kal\u0131r. Bildiklerimiz aras\u0131ndan en \u00f6nemlilerinden biri Fermat\u2019n\u0131n K\u00fc\u00e7\u00fck Teoremi ad\u0131yla an\u0131lan \u015fu teoremdir: <\/p>\n
Teorem 10. (Fermat\u2019n\u0131n K\u00fc\u00e7\u00fck Teoremi.) n bir say\u0131ysa ve p asalsa, p, n p \u2013 n say\u0131s\u0131n\u0131 b\u00f6ler. Dolay\u0131s\u0131yla e\u011fer p, n\u2019yi b\u00f6lm\u00fcyorsa, n p\u20131\u20131\u2019i b\u00f6ler. <\/p>\n
Bu teorem, n \u00fczerine t\u00fcmevar\u0131mla kolayl\u0131kla kan\u0131tlanabilir. \u00d6rne\u011fin 23, 223\u20132 say\u0131s\u0131n\u0131 b\u00f6ler, \u00e7\u00fcnk\u00fc 23 asald\u0131r. 23, 2\u2019yi b\u00f6lmedi\u011finden, 23, 222\u20131 say\u0131s\u0131n\u0131 da b\u00f6ler. Bunun tersi do\u011fru mudur? Yani p > 1 bir say\u0131ysa ve p, 2p\u20131 \u2013 1\u2019i b\u00f6l\u00fcyorsa, p asal m\u0131d\u0131r? Eski \u00c7inliler de bu soruyu sormu\u015flar ve yapt\u0131klar\u0131 hesaplarda p hep asal \u00e7\u0131km\u0131\u015ft\u0131r. Ger\u00e7ekten de 1 < p < 300 i\u00e7in bu do\u011frudur. \u00d6te yandan p = 341 = 11 \u00d7 31 i\u00e7in do\u011fru de\u011fildir: 341 asal olmamas\u0131na kar\u015f\u0131n 2340 \u2013 1\u2019i b\u00f6ler. Demek ki \u00c7inliler yan\u0131lm\u0131\u015flar. Bir iki deney yaparak matematiksel bir ger\u00e7ek bulunmaz. Kan\u0131t gerekir. [11]\n\n E\u011fer p, 2p \u2013 2\u2019yi b\u00f6l\u00fcyorsa ve asal de\u011filse, p\u2019ye yalanc\u0131 asal ad\u0131 verilir. \u00d6rne\u011fin 341 bir yalanc\u0131 asald\u0131r6. 561, 645, 1105, 1387, 1729, 1905 de yalanc\u0131 asallard\u0131r. Ka\u00e7 tane yalanc\u0131 asal vard\u0131r? Sonsuz tane vard\u0131r, \u00e7\u00fcnk\u00fc e\u011fer p bir yalanc\u0131 asalsa, 2p\u20131 de bir yalanc\u0131 asald\u0131r. Okur bunu al\u0131\u015ft\u0131rma olarak kan\u0131tlayabilir. Demek ki 2341 \u2013 1 bir yalanc\u0131 asald\u0131r.\n\n Her p i\u00e7in, 2p \u2013 1 tek bir say\u0131d\u0131r. Dolay\u0131s\u0131yla yukardaki y\u00f6ntemle bulunan yalanc\u0131 asallar hep tektirler. Bundan da \u015fu \u201cdo\u011fal\u201d soru \u00e7\u0131kar: \u00e7ift yalanc\u0131 asal var m\u0131d\u0131r? Evet! 1950\u2019de D.H. Lehmer 161.038\u2019in bir yalanc\u0131 asal oldu\u011funu kan\u0131tlad\u0131. 161.038 say\u0131s\u0131n\u0131 bulmak kolay de\u011fil ama, bu say\u0131n\u0131n yalanc\u0131 asall\u0131\u011f\u0131n\u0131 kan\u0131tlamak olduk\u00e7a kolay. Kan\u0131tlayal\u0131m. 161.038\u2019in 2161.038 \u2013 2\u2019yi b\u00f6ld\u00fc\u011f\u00fcn\u00fc kan\u0131tlamak istiyoruz. \u00d6nce 161.038\u2019i asallar\u0131na ay\u0131ral\u0131m: 161.038 = 2 \u00d7 73 \u00d7 1103. Demek ki 73 ve 1103\u2019\u00fcn a : = 2161.037\u20131\u2019i b\u00f6ld\u00fc\u011f\u00fcn\u00fc kan\u0131tlamal\u0131y\u0131z. 161.037\u2019yi asallar\u0131na ay\u0131ral\u0131m: 161.037 = 32 \u00d7 29 \u00d7 617 = 9 \u00d7 b. Burda b = 29 \u00d7 617 olarak ald\u0131k elbet. E\u011fer c = 29 ise, bundan da \u015fu \u00e7\u0131kar: a = 2161.037 \u2013 1 = (29)b \u2013 1 = cb \u2013 1. Demek ki c \u2013 1, yani 29 \u2013 1, yani 511, yani 7 \u00d7 73, a\u2019y\u0131 b\u00f6l\u00fcyormu\u015f. Dolay\u0131s\u0131yla 73 de a\u2019y\u0131 b\u00f6l\u00fcyordur. \u015eimdi s\u0131ra 1103\u2019\u00fcn a\u2019y\u0131 b\u00f6ld\u00fc\u011f\u00fcn\u00fc kan\u0131tlamakta. Ayn\u0131 ak\u0131l y\u00fcr\u00fctmeyi yapaca\u011f\u0131z. d : = 32 \u00d7 617 ve e = 229 olsun. Hesaplayal\u0131m: a = 2161.037 \u2013 1 = (229)d \u2013 1 = ed \u2013 1. Demek ki \n\n e \u2013 1 = 1103 \u00d7 486.737, \n\n a\u2019y\u0131 b\u00f6l\u00fcyormu\u015f. Kan\u0131t\u0131m\u0131z bitmi\u015ftir.\n\n 1951\u2019de N.W.H. Beeger sonsuz tane \u00e7ift yalanc\u0131 asal oldu\u011funu kan\u0131tlad\u0131.\n\n E\u011fer p > 1, her n i\u00e7in n p \u2013 n\u2019yi b\u00f6l\u00fcyorsa ve asal de\u011filse, p\u2019ye \u00e7ok yalanc\u0131 asal ad\u0131 verilir. \u00c7ok yalanc\u0131 asal say\u0131 var m\u0131d\u0131r? Evet. En k\u00fc\u00e7\u00fck \u00e7ok yalanc\u0131 asal say\u0131 561\u2019dir. 561 = 3 \u00d7 11 \u00d7 17 oldu\u011fundan 561 asal de\u011fildir. \u00d6te yandan, 561, her n i\u00e7in a561\u2013561\u2019i b\u00f6ler. Bunu da kan\u0131tlamak olduk\u00e7a kolayd\u0131r. Kan\u0131t i\u00e7in okur [23]\u2019e bakabilir.<\/p>\n
Fermat\u2019n\u0131n K\u00fc\u00e7\u00fck Teoremi\u2019ne g\u00f6re, e\u011fer p asalsa, <\/p>\n
1p\u20131, 2p\u20131,…, (p\u20131)p\u20131 <\/p>\n
say\u0131lar\u0131 p\u2019ye b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde 1 kal\u0131r. Dolay\u0131s\u0131yla bu p\u20131 say\u0131n\u0131n toplam\u0131 olan <\/p>\n
1p\u20131 + 2p\u20131+ … + (p\u20131) p\u20131<\/p>\n
say\u0131s\u0131 p\u2019ye b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde kalan p \u2013 1\u2019dir. Bunun tersi de do\u011fru mudur? Yani n herhangi bir say\u0131ysa ve <\/p>\n
1n\u20131 + 2n\u20131 + …+ (n\u20131)n\u20131 <\/p>\n
say\u0131s\u0131 n\u2019ye b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde kalan n \u2013 1 ise, n asal m\u0131d\u0131r? 1950\u2019de Bedocchi ad\u0131nda bir matematik\u00e7i 1985\u2019de yan\u0131t\u0131n n < 101700 i\u00e7in \u201cevet\u201d oldu\u011funu g\u00f6sterdi. Genel sorunun yan\u0131t\u0131 bug\u00fcn de bilinmiyor: \n\n\n Soru: n herhangi bir say\u0131ysa ve 1n\u20131+2n\u20131+ ...+(n\u20131)n\u20131 say\u0131s\u0131 n\u2019ye b\u00f6l\u00fcnd\u00fc\u011f\u00fcnde kalan n \u2013 1 ise, n asal m\u0131d\u0131r? \n\n\n Ger\u00e7ek asallara geri d\u00f6nelim. Wilson Teoremi, hemen hemen Fermat\u2019n\u0131n K\u00fc\u00e7\u00fck Teoremi kadar \u00f6nemlidir: \n\n\n Teorem 11. E\u011fer p asalsa, p, (p \u2013 1)! + 1\u2019i b\u00f6ler. \n\n\n Asallar \u00fczerine yan\u0131t\u0131 bilinmeyen bir ba\u015fka soru ge\u00e7eyim. Goldbach, bir mektubunda a\u015fa\u011f\u0131daki soruyu Euler\u2019e sordu (1972): \n\n\n Goldbach San\u0131s\u0131 (1): 5\u2019ten b\u00fcy\u00fck her say\u0131 \u00fc\u00e7 asal\u0131n toplam\u0131na e\u015fittir. \n\n\n Euler, Goldbach\u2019a sorunun yan\u0131t\u0131n\u0131 bilmedi\u011fini, ama sorunun a\u015fa\u011f\u0131daki soruyla e\u015fde\u011fer oldu\u011funu yazd\u0131: \n\n\n Goldbach San\u0131s\u0131 (2): 4\u2019ten b\u00fcy\u00fck her \u00e7ift say\u0131 iki asal\u0131n toplam\u0131d\u0131r. \n\n\n \u00d6rne\u011fin,\n\n 4 = 2+2\n\n 6 = 3+3\n\n 8 = 3+5\n\n 10 = 3+7 = 5+5\n\n 12 = 5+7\n\n 14 = 3+11 = 7+7\n\n 16 = 3+13 = 5+11\n\n 18 = 5+13 = 7+11\n\n 20 = 3+17 = 7+13\n\n 22 = 3+19 = 5+17 = 11+11 \n\n 24 = 5+19 = 7+17 = 11+13 \n\n 26 = 3+23 = 7+19 = 13+13\n\n Y\u00fcz milyondan k\u00fc\u00e7\u00fck say\u0131lar i\u00e7in Goldbach san\u0131s\u0131n\u0131n do\u011fru oldu\u011fu biliniyor. \u00d6nermenin her say\u0131 i\u00e7in do\u011fru oldu\u011fu bilinmiyor, ancak do\u011fru oldu\u011fu san\u0131l\u0131yor. Bu san\u0131y\u0131 kan\u0131tlayabilirseniz \u00f6l\u00fcms\u00fczler aras\u0131nda yerinizi al\u0131rs\u0131n\u0131z.\n\n Asal say\u0131lar \u00fczerine dahaca \u00e7\u00f6z\u00fclememi\u015f bir ba\u015fka \u00fcnl\u00fc san\u0131 vard\u0131r: \n\n\n \u0130kiz Asallar San\u0131s\u0131: Sonsuz tane ikiz asal say\u0131 vard\u0131r. \n\n\n E\u011fer iki asal say\u0131n\u0131n aras\u0131ndaki fark 2 ise, bu iki asal say\u0131ya ikiz denir. \u00d6rne\u011fin, (3,5), (5,7), (11,13), (17,19), (29,31), (41,43) ikiz asal say\u0131lard\u0131r. Sonsuz tane ikiz asal\u0131n olup olmad\u0131\u011f\u0131 bilinmiyor. \u201cBilinse ne olur, bilinmese ne olur?\u201d demeyin. Yan\u0131t\u0131 bilinmeyen her soru ilgin\u00e7tir, \u00fczerinde d\u00fc\u015f\u00fcnmeye de\u011fer. \u0130nsan yaln\u0131zca \u201cd\u00fc\u015f\u00fcnen hayvan\u201d de\u011fildir, nedenli nedensiz d\u00fc\u015f\u00fcnen hayvand\u0131r.\n\n 1966\u2019da, sonsuz tane asal p say\u0131s\u0131 i\u00e7in, p + 2 say\u0131s\u0131n\u0131n ya asal ya da iki asal\u0131n \u00e7arp\u0131m\u0131 oldu\u011fu kan\u0131tland\u0131.\n\n Bilinen en b\u00fcy\u00fck ikiz asallar 1.706.595 \u00d7 211235 \u00b1 1 asallar\u0131d\u0131r, 1990\u2019da Parady, Smith ve Zarantonello bulmu\u015ftur.\n\n \u00dc\u00e7\u00fcz asal var m\u0131d\u0131r? (3,5,7)\u2019den ba\u015fka yoktur. Okur bunu kolayl\u0131kla kan\u0131tlayabilir. Bir ipucu verelim: e\u011fer n bir tamsay\u0131ysa, n, n+2, n+4 say\u0131lar\u0131ndan biri 3\u2019e b\u00f6l\u00fcn\u00fcr.\n\n Yukarda sonsuz tane asal say\u0131n\u0131n oldu\u011funu g\u00f6rd\u00fck. Gene de o kadar fazla asal say\u0131 yoktur. \u00d6rne\u011fin, \u00e7ift say\u0131lar (2 d\u0131\u015f\u0131nda) asal olamayacaklar\u0131ndan, say\u0131lar\u0131n \u201cyar\u0131s\u0131ndan fazlas\u0131\u201d asal de\u011fildir. 1\u2019le n aras\u0131ndan rastgele bir say\u0131 se\u00e7sek, bu say\u0131n\u0131n asal olma olas\u0131l\u0131\u011f\u0131 ka\u00e7t\u0131r? Bu olas\u0131l\u0131k n\u2019ye g\u00f6re de\u011fi\u015fir elbet. E\u011fer n = 100 ise, bu olas\u0131l\u0131\u011f\u0131n 1\/4 oldu\u011funu yaz\u0131n\u0131n en ba\u015f\u0131nda g\u00f6rm\u00fc\u015ft\u00fck.\n\n E\u011fer n bir tamsay\u0131ysa, \u03c0(n), n\u2019den k\u00fc\u00e7\u00fck asallar\u0131n say\u0131s\u0131 olsun. \u03c0(n)\/n, n\u2019den k\u00fc\u00e7\u00fck rastgele se\u00e7ilmi\u015f bir say\u0131n\u0131n asal olma olas\u0131l\u0131\u011f\u0131d\u0131r. n sonsuza gitti\u011finde, bu olas\u0131l\u0131\u011f\u0131n de\u011feri ka\u00e7t\u0131r? Okur, n b\u00fcy\u00fcd\u00fck\u00e7e, asal se\u00e7me olas\u0131l\u0131\u011f\u0131n\u0131n da k\u00fc\u00e7\u00fclece\u011fini ve n sonsuza gitti\u011finde bu olas\u0131l\u0131\u011f\u0131n 0\u2019a yak\u0131nsayaca\u011f\u0131n\u0131 tahmin edebilir. Bu tahmin do\u011frudur7:\n\n limn\u2192\u221e \u03c0(n)\/n = 0. \n\n Bundan \u00e7ok daha iyi bir sonu\u00e7 bilinmektedir. \u03c0(n)\/n ve 1\/log(n), n b\u00fcy\u00fcd\u00fck\u00e7e birbirlerine \u00e7ok yak\u0131nsamaktad\u0131rlar8. Ba\u015fka bir deyi\u015fle, e\u011fer n b\u00fcy\u00fckse, \u03c0(n) a\u015fa\u011f\u0131 yukar\u0131 n\/log(n) dur, yani \u03c0(n) \u2248 n\/log(n). Bu sonuca Asal Say\u0131lar Teoremi ad\u0131 verilir.\n\n Asal say\u0131lar son derece ilgin\u00e7 bir konudur. Asal say\u0131lar konusunda bilgilenmek isteyen okur [33] ve [40]\u2019a bakabilir. Hele Euler\u2019in sonsuz tane asal say\u0131n\u0131n oldu\u011funu (bir kez daha) kan\u0131tlayan bir kan\u0131t\u0131 vard\u0131r ki...\n<\/p>\n","protected":false},"excerpt":{"rendered":"
Birden ve kendisinden ba\u015fka say\u0131ya b\u00f6l\u00fcnmeyen say\u0131lara asal say\u0131 denir1. \u00d6rne\u011fin 17 asald\u0131r, \u00e7\u00fcnk\u00fc 1 ve 17\u2019den ba\u015fka say\u0131ya (tam olarak) b\u00f6l\u00fcnmez. \u00d6te yandan 35 asal de\u011fildir, 5\u2019e ve 7\u2019ye b\u00f6l\u00fcn\u00fcr. Teknik nedenlerden 1 asal kabul edilmez. 100\u2019den k\u00fc\u00e7\u00fck asallar\u0131 bulmak pek zor de\u011fildir. \u0130\u015fte o asallar: 2, 3, 5, 7, 11, 13, 17, 19, …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1404,1403],"tags":[7208,3424,7243],"class_list":["post-3193","post","type-post","status-publish","format-standard","hentry","category-matematik-odevleri","category-odevler","tag-asal-sayilar","tag-matematik","tag-oklid"],"_links":{"self":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3193","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/comments?post=3193"}],"version-history":[{"count":0,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3193\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/media?parent=3193"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/categories?post=3193"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/tags?post=3193"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}