{"id":2183,"date":"2024-10-18T14:49:44","date_gmt":"2024-10-18T12:49:44","guid":{"rendered":"https:\/\/www.jankafialka.sk\/?p=2183"},"modified":"2024-10-21T07:31:46","modified_gmt":"2024-10-21T05:31:46","slug":"dokonale-cisla","status":"publish","type":"post","link":"https:\/\/www.jankafialka.sk\/?p=2183","title":{"rendered":"Dokonal\u00e9 \u010d\u00edsla"},"content":{"rendered":"\n

Hovor\u00ed sa, \u017ee Philo z Alexandrie raz okolo roku 0 povedal, \u017ee „Svet bol stvoren\u00fd za 6 dn\u00ed a mesiac obieha Zem raz za 28 dn\u00ed preto, \u017ee 6 a 28 s\u00fa dokonal\u00e9 \u010d\u00edsla.“<\/p>\n\n\n\n

Dokonal\u00e9 \u010d\u00edsla s\u00fa \u010d\u00edsla, ktor\u00e9 s\u00fa rovn\u00e9 s\u00fa\u010dtu v\u0161etk\u00fdch svojich delite\u013eov (okrem seba sam\u00e9ho). Najmen\u0161ie dokonal\u00e9 \u010d\u00edslo je teda 6, preto\u017ee 6 = 1 + 2 + 3. \u010eal\u0161ie tak\u00e9to \u010d\u00edslo je 28, preto\u017ee 28 = 1 + 2 + 4 + 7 + 14. To zauj\u00edmav\u00e9, naozaj zauj\u00edmav\u00e9, som sa dozvedela na jednej hodine dejepisu u deviatakov, kde som mala zastupovan\u00fa hodinu :D.<\/p>\n\n\n\n

O dokonal\u00fdch \u010d\u00edslach sa vie, \u017ee sa daj\u00fa zap\u00edsa\u0165 v tvare 2n<\/em>–1<\/sup> (2n<\/em><\/sup> — 1), ale len v pr\u00edpade, \u017ee 2n<\/em><\/sup> — 1 je prvo\u010d\u00edslo. Tak\u00e9to prvo\u010d\u00edsla sa naz\u00fdvaj\u00fa Mersennove prvo\u010d\u00edsla. V s\u00fa\u010dasnosti je zn\u00e1mych 51 Mersennovych prvo\u010d\u00edsel a teda aj 51 dokonal\u00fdch \u010d\u00edsel.<\/p>\n\n\n\n

\u010co bola pre m\u0148a ale novinka je fakt, \u017ee ke\u010f zap\u00ed\u0161eme dokonal\u00e9 \u010d\u00edslo v dvojkovej \u010d\u00edselnej s\u00fastave (o dvojkovej \u010d\u00edselnej s\u00fastave si pre\u010d\u00edtajte https:\/\/www.jankafialka.sk\/?p=59<\/a>), tak bude vyzera\u0165 nejako takto: 111110000. V\u017edy sam\u00e9 jednotky a potom len sam\u00e9 nuly. A t\u00fdch n\u00fal je v\u017edy o jednu menej ako jednotiek. N\u00e1hoda? Plat\u00ed to pre v\u0161etky dokonal\u00e9 \u010d\u00edsla? Pre\u010do? Tak o tom bude tento \u010dl\u00e1no\u010dek.<\/p>\n\n\n\n

Sk\u00fasme si nap\u00edsa\u0165 prv\u00fdch nieko\u013eko dokonal\u00fdch \u010d\u00edsel ako s\u00fa\u010det svojich delite\u013eov a z\u00e1rove\u0148 ich prep\u00ed\u0161eme do dvojkovej s\u00fastavy tak, \u017ee preusporiadame a zoskup\u00edme delite\u013eov:<\/p>\n\n\n\n

6 = 1 + 2 + 3 = (1 + 3) + 2 =
1 \u00b7 4 + 1 \u00b7 2 + 0 = 1102<\/sub><\/p>\n\n\n\n

28 = 1 + 2 + 4 + 7 + 14 =
(2 + 14) + (1 + 7) + 4 =
1 \u00b7 16 + 1 \u00b7 8 + 1 \u00b7 4 + 0 \u00b7 2 + 0 = 111002<\/sub><\/p>\n\n\n\n

496 = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 =
(8 + 248) + (4 + 124) + (2 + 62) + (1 + 31) + 16 =
1 \u00b7 256 + 1 \u00b7 128 + 1 \u00b7 64 + 1 \u00b7 32 + 1 \u00b7 16 + 0 \u00b7 8 + 0 \u00b7 4 + 0 \u00b7 2 + 0 = 1111100002<\/sub><\/p>\n\n\n\n

Tak\u017ee pre prv\u00e9 tri dokonal\u00e9 \u010d\u00edsla to plat\u00ed. Nebudeme to ale tak\u00fdmto sp\u00f4sobom overova\u0165 pre v\u0161etky zn\u00e1me, ke\u010f\u017ee najv\u00e4\u010d\u0161ie z nich m\u00e1 viac ako 49 mili\u00f3nov cifier. Namiesto toho urob\u00edme nie\u010do ove\u013ea lep\u0161ie. Sprav\u00edme d\u00f4kaz, \u017ee tento tvar maj\u00fa v dvojkovej s\u00fastave v\u0161etky dokonal\u00e9 \u010d\u00edsla. Teda aj tie, ktor\u00e9 e\u0161te neboli n\u00e1jden\u00e9. Vych\u00e1dza\u0165 budeme zo zn\u00e1meho tvaru, ktor\u00fd som u\u017e vy\u0161\u0161ie spom\u00ednala. Dokonal\u00e9 \u010d\u00edsla maj\u00fa tvar 2n<\/em>–1<\/sup> (2n<\/em><\/sup> — 1), pri\u010dom 2n<\/em><\/sup> — 1 je Mersennove prvo\u010d\u00edslo. Ozna\u010dme ho p\u00edsmenkom p<\/em>.<\/p>\n\n\n\n

Tak\u017ee dokonal\u00e9 \u010d\u00edslo m\u00e1 tvar 2 \u00b7 2 \u00b7 2 \u00b7 2 \u00b7 … \u00b7 2 \u00b7 p<\/em>.<\/p>\n\n\n\n

Ak\u00e9 delitele m\u00e1 tak\u00e9to \u010d\u00edslo? No, pravda\u017ee, delitele bud\u00fa v\u0161etky mocniny dvojky a\u017e do 2n<\/em>–1<\/sup> , potom tam bude to prvo\u010d\u00edslo p<\/em> a \u010falej v\u0161etky n\u00e1sobky tohto prvo\u010d\u00edsla s t\u00fdmi mocninami dvojky, ale bez poslednej, preto\u017ee to u\u017e by bolo to samo dokonal\u00e9 \u010d\u00edslo, ktor\u00e9 medzi delite\u013eov teraz po\u010d\u00edta\u0165 nebudeme. \u010ci\u017ee p<\/em>, 2p<\/em>, 4p<\/em>, …, 2n<\/em>–2<\/sup>p<\/em>.<\/p>\n\n\n\n

Teda s\u00fa\u010det v\u0161etk\u00fdch delite\u013eov je
S<\/em> = 1 + 2 + 4 + … + 2n<\/em>–2<\/sup> + 2n<\/em>–1<\/sup> +
+ p<\/em> + 2p<\/em> + 4p<\/em> + … + 2n<\/em>–2<\/sup> p<\/em><\/p>\n\n\n\n

V hornom s\u00fa\u010dte nie s\u00fa \u010d\u00edsla pod sebou podp\u00edsan\u00e9 n\u00e1hodne. Tieto budeme toti\u017e zoskupova\u0165 a uvid\u00edme, ako n\u00e1m spolu vytv\u00e1raj\u00fa mocniny dvojky. <\/p>\n\n\n\n

S<\/em> = (p<\/em> + 1) + (2p<\/em> + 2) + (4p<\/em> + 4) + … + 2n<\/em>–2<\/sup>(p<\/em> + 1) + 2n<\/em>–1<\/sup><\/p>\n\n\n\n

Ke\u010f\u017ee \u010d\u00edslo p <\/em>m\u00e1 tvar p<\/em> = 2n<\/em> <\/sup>— 1, tak p<\/em> + 1 = 2n<\/em><\/sup>.
A teda n\u00e1\u0161 s\u00fa\u010det bude vyzera\u0165 takto:
S<\/em> = 2n<\/sup><\/em> + 2 \u00b7 2n<\/em><\/sup> + 4 \u00b7 2n<\/em><\/sup> + … + 2n<\/em>–2<\/sup> \u00b7 2n<\/em><\/sup> + 2n<\/em>–1<\/sup><\/p>\n\n\n\n

Po drobn\u00fdch \u00faprav\u00e1ch a usporiadan\u00ed od najv\u00e4\u010d\u0161ej mocniny dost\u00e1vame S<\/em> = 22n–2<\/sup><\/em> + … + 2n+2<\/em><\/sup> + 2n+1<\/em><\/sup> + 2n<\/em><\/sup> + 2n<\/em>–1<\/sup><\/p>\n\n\n\n

A to je v dvojkovej s\u00fastave presne to, \u010do sme chceli dosta\u0165 – najprv sam\u00e9 jednotky, konkr\u00e9tne n<\/em> kusov a potom len nuly, ktor\u00fdch je o jednu menej.<\/p>\n\n\n\n

Tak\u017ee to nie je n\u00e1hoda a skuto\u010dne v\u0161etky dokonal\u00e9 \u010d\u00edsla s\u00fa tohto tvaru. Dokonca m\u00f4\u017eeme poveda\u0165, \u017ee v\u0161etky \u010d\u00edsla (nielen dokonal\u00e9) tvaru 2n<\/em>–1<\/sup> (2n<\/em><\/sup> — 1) maj\u00fa v dvojkovej s\u00fastave tvar 11111…10000…0 (n<\/em> jednotiek a n<\/em> — 1 n\u00fal). Preto\u017ee
22n–2<\/sup><\/em> + … + 2n+2<\/em><\/sup> + 2n+1<\/em><\/sup> + 2n<\/em><\/sup> + 2n<\/em>–1<\/sup> = 2n<\/em>–1<\/sup>(1 + 2 + 4 + … + 2n–1<\/em><\/sup> )
a to v z\u00e1tvorke je geometrick\u00e1 postupnos\u0165, ktorej s\u00fa\u010det je 2n<\/em><\/sup> — 1. (Ak si spom\u00ednate na vzorec sk<\/sub><\/em> = (q<\/em>k<\/em>+1<\/sup> — 1)\/(q<\/em> — 1), pri\u010dom na\u0161e q<\/em> = 2.)<\/p>\n\n\n\n

Alebo inak povedan\u00e9 to, \u017ee \u010d\u00edslo 2n<\/em>–1<\/sup> (2n<\/em><\/sup> — 1) m\u00e1 v dvojkovej s\u00fastave tvar 11111…10000…0 je analogick\u00e9 tvrdeniu, \u017ee v desiatkovej s\u00fastave je \u010d\u00edslo napr\u00edklad 100 \u00b7 (1000 — 1) = 99900. A toto je n\u00e1m vcelku jasn\u00e9. \ud83d\ude42<\/p>\n\n\n\n

\u010c\u00edsel, ktor\u00e9 sa daj\u00fa zap\u00edsa\u0165 sp\u00f4sobom 1111…1000…0 je teda naozaj ve\u013ea. Dokonal\u00e9 s\u00fa ale iba tie z nich, pre ktor\u00e9 je 2n<\/em><\/sup> — 1 prvo\u010d\u00edslo. A t\u00fdch je prekvapivo m\u00e1lo, aj ke\u010f je ich nekone\u010dne ve\u013ea \ud83d\ude01. Vysvetl\u00edm. Naozaj je ich nekone\u010dne ve\u013ea, ale doteraz pozn\u00e1me iba prv\u00fdch 51 z nich. To posledn\u00e9 bolo n\u00e1jden\u00e9 v decembri 2018 a je tak ve\u013ek\u00e9, \u017ee keby sme ho chceli zap\u00edsa\u0165 do knihy, t\u00e1to kniha by mala nieko\u013eko tis\u00edc str\u00e1n husto pop\u00edsan\u00fdch iba \u010d\u00edslicami tvoriacimi toto \u010d\u00edslo. A keby sme ho chceli pre\u010d\u00edta\u0165 a hovorili by sme r\u00fdchlos\u0165ou 5 \u010d\u00edslic za sekundu, tak by sme museli bez prest\u00e1vky rozpr\u00e1va\u0165 viac ako tri mesiace. A to je iba p\u00e4\u0165desiateprv\u00e9 dokonal\u00e9 \u010d\u00edslo. Ak\u00e9 ve\u013ek\u00e9 bud\u00fa \u010fal\u0161ie? Uvid\u00edme…<\/p>\n\n\n\n

\u010co ale vieme poveda\u0165 s absol\u00fatnou istotou je fakt, \u017ee v dvojkovej \u010d\u00edselnej s\u00fastave bud\u00fa ma\u0165 tvar 1111…1000…0, pri\u010dom jednotiek bude presne n<\/em>. To n<\/em>, ktor\u00e9 zodpoved\u00e1 Mersennovmu prvo\u010d\u00edslu 2n<\/em><\/sup> — 1.<\/p>\n\n\n\n

Kompletn\u00fd zoznam doteraz n\u00e1jden\u00fdch dokonal\u00fdch \u010d\u00edsel n\u00e1jdete na https:\/\/en.m.wikipedia.org\/wiki\/List_of_Mersenne_primes_and_perfect_numbers<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"

Hovor\u00ed sa, \u017ee Philo z Alexandrie raz okolo roku 0 povedal, \u017ee „Svet bol stvoren\u00fd za 6 dn\u00ed a mesiac obieha Zem raz za 28 dn\u00ed preto, \u017ee 6 a 28 s\u00fa dokonal\u00e9 \u010d\u00edsla.“ Dokonal\u00e9 \u010d\u00edsla s\u00fa \u010d\u00edsla, ktor\u00e9 s\u00fa rovn\u00e9 s\u00fa\u010dtu v\u0161etk\u00fdch svojich delite\u013eov (okrem seba sam\u00e9ho). Najmen\u0161ie dokonal\u00e9 \u010d\u00edslo je teda 6, preto\u017ee […]<\/p>\n","protected":false},"author":1,"featured_media":2216,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-2183","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-nezaradene"],"yoast_head":"\nDokonal\u00e9 \u010d\u00edsla - jankafialka<\/title>\n<meta name=\"description\" content=\"Dokonal\u00e9 \u010d\u00edsla a d\u00f4kaz toho, \u017ee maj\u00fa v dvojkovej s\u00fastave tvar 111110000 (najprv sam\u00e9 jednotky, potom iba nuly).\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.jankafialka.sk\/?p=2183\" \/>\n<meta property=\"og:locale\" content=\"sk_SK\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Dokonal\u00e9 \u010d\u00edsla - 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