{"id":72,"date":"2022-01-05T16:55:00","date_gmt":"2022-01-05T15:55:00","guid":{"rendered":""},"modified":"2023-08-04T22:53:06","modified_gmt":"2023-08-04T20:53:06","slug":"nasobenie-v-starovekom-egypte","status":"publish","type":"post","link":"https:\/\/www.jankafialka.sk\/?p=72","title":{"rendered":"N\u00e1sobenie v starovekom Egypte"},"content":{"rendered":"\n

Ke\u010f som si pred p\u00e1r rokmi pripravovala predn\u00e1\u0161ku o n\u00e1soben\u00ed, narazila som v knihe od profesora \u010ci\u017em\u00e1ra – Hist\u00f3ria matematiky na zauj\u00edmav\u00fa pas\u00e1\u017e – n\u00e1sobenie v starovekom Egypte. V tom \u010dase toti\u017e nemohli n\u00e1sobi\u0165 tak, ako to rob\u00edme dnes. Nepoznali pozi\u010dn\u00fd sp\u00f4sob z\u00e1pisu \u010d\u00edsel. \u010c\u00edsla zapisovali pomocou nieko\u013ek\u00fdch znakov. Mali znak pre jednotku, desiatku, stovku, tis\u00edcku a t.\u010f. V\u0161etky ostatn\u00e9 \u010d\u00edsla zap\u00edsali hromaden\u00edm t\u00fdchto znakov. \u010ci\u017ee dvadsa\u0165 p\u00e4\u0165 nap\u00edsali ako dva znaky desiatky a p\u00e4\u0165 znakov jednotky. Takto zap\u00edsan\u00e9 \u010d\u00edsla by sa na\u0161\u00edm zn\u00e1mym algoritmom n\u00e1sobili pomerne \u0165a\u017eko. Oni pou\u017e\u00edvali efekt\u00edvnu met\u00f3du, ktor\u00e1 sa, mimochodom, d\u00e1 rovnako dobre pou\u017e\u00edva\u0165 aj dnes.<\/p>\n\n\n\n

\"\"
Starovek\u00fd Egypt<\/figcaption><\/figure>\n\n\n\n

Ke\u010f som o n\u00e1soben\u00ed zdvojovan\u00edm \u010d\u00edtala prv\u00fd kr\u00e1t, tak ve\u013emi ma to nadchlo, \u017ee som naprogramovala aj jednoduch\u00fa aplik\u00e1ciu, ktor\u00e1 u\u017e\u00edvate\u013ea nau\u010d\u00ed, ako sa to rob\u00ed. Zverejnila som ju na google play pod n\u00e1zvom Egyptsk\u00e9 n\u00e1sobenie a delenie. Skoro nikto si ju nestiahol. Chcela som sem prida\u0165 obr\u00e1zok, ako vyzerala, ale u\u017e ju neviem n\u00e1js\u0165. Asi mi ju vymazali. \ud83d\ude15<\/p>\n\n\n\n

Princ\u00edp je zalo\u017een\u00fd na fakte, \u017ee n\u00e1sobenie je iba opakovan\u00e9 s\u010d\u00edtavanie. Unik\u00e1tna je met\u00f3da, ako toto s\u010d\u00edtavanie zr\u00fdchli\u0165 zmen\u0161en\u00edm po\u010dtu s\u010d\u00edtancov. Napr\u00edklad pri n\u00e1soben\u00ed 17tkou netreba s\u010d\u00edta\u0165 17 rovnak\u00fdch s\u010d\u00edtancov, ale ove\u013ea menej. V tomto konkr\u00e9tnom pr\u00edklade by to boli iba dva. Ako je to mo\u017en\u00e9? Odpove\u010fou je opakovan\u00e9 zdvojn\u00e1sobovanie.<\/p>\n\n\n\n

Pr\u00edklad: 17×25. 25 idem opakovane zdvojn\u00e1sobova\u0165. Dvojn\u00e1sobok 25ky je 50, dvojn\u00e1sobok 50tky je 100, dvojn\u00e1sobok stovky je 200 a t.\u010f. Vtip je v tom, \u017ee toto si p\u00ed\u0161em pod seba a do pr\u00edslu\u0161n\u00e9ho riadku v\u017edy p\u00ed\u0161em aj ko\u013eko n\u00e1sobok p\u00f4vodnej 25ky to je. Pri 50 m\u00e1m dvojku, pri 100 \u0161tvorku, pri 200 osmi\u010dku.<\/p>\n\n\n\n

1, 2, 4, 8, 16, 32, … to je postupnos\u0165 mocn\u00edn dvojky. A z t\u00fdchto \u010d\u00edsel treba zlo\u017ei\u0165 druh\u00e9ho \u010dinite\u013ea, t.j. v tomto pr\u00edklade 17ku. 17=16+1 tak\u017ee si vyzna\u010d\u00edm riadky, v ktor\u00fdch je 16 a prv\u00fd, kde m\u00e1m 1 – to je ten prv\u00fd s 25kou. No a ke\u010f s\u010d\u00edtam \u0161estn\u00e1s\u0165n\u00e1sobok 25ky s jednon\u00e1sobkom, tak, pochopite\u013ene, dostanem sedemn\u00e1s\u0165n\u00e1sobok. Takto by to vyzeralo:<\/p>\n\n\n\n

1……25<\/b><\/p>\n\n\n\n

2……50<\/p>\n\n\n\n

4…..100<\/p>\n\n\n\n

8…..200<\/p>\n\n\n\n

16…400<\/b><\/p>\n\n\n\n

       400+25=425<\/p>\n\n\n\n

V poriadku, poviete si, ale pre ve\u013ek\u00e9 \u010d\u00edsla to bude ur\u010dite neskuto\u010dne zd\u013ahav\u00e9. Ale nie je to tak. Tu sa uk\u00e1\u017ee kr\u00e1sa mocn\u00edn. Rast\u00fa toti\u017e najprv nepatrne, ale nesk\u00f4r sa v\u00e1m zato\u010d\u00ed hlava z toho prudk\u00e9ho rastu. Pri sedemn\u00e1stke sme potrebovali p\u00e4\u0165 riadkov. Ko\u013eko riadkov by bolo potrebn\u00fdch pre desa\u0165miestne \u010d\u00edsla?<\/p>\n\n\n\n

2 na 33 je nie\u010do vy\u0161e 8,5 miliardy, tak\u017ee pri egyptskom n\u00e1soben\u00ed \u010d\u00edsel men\u0161\u00edch ako desa\u0165 mili\u00e1rd n\u00e1m vysta\u010d\u00ed pokojne 33 riadkov. Jedna A4.<\/p>\n\n\n\n

Pr\u00edklad<\/h2>\n\n\n\n

Tak to sa mus\u00ed vysk\u00fa\u0161a\u0165. U\u017e ste sa niekedy zam\u00fd\u0161\u013eali nad t\u00fdm, ko\u013eko je 123456789 x 987654321? Nie? Kalkula\u010dka ve\u013emi nepom\u00f4\u017ee – je to prive\u013ea, v\u00fdsledok len zaokr\u00fahli. Tak som dnes vo chv\u00ed\u013ekach, ktor\u00e9 sa mi podarilo ukradn\u00fa\u0165 si pre seba, n\u00e1sobila po egyptsky. Trvalo to asi pol hodinku. A potom e\u0161te tri hodiny, k\u00fdm som na\u0161la a opravila v\u0161etky chyby. \ud83d\ude01<\/p>\n\n\n\n

\"N\u00e1sobenie
Egyptsk\u00e9 n\u00e1sobenie <\/figcaption><\/figure>\n\n\n\n
<\/div>\n\n\n\n

Ak v\u00e1s zauj\u00edma, \u010do som to tam porobila, za chv\u00ed\u013eu vysvetl\u00edm. Ak nie, tak tu m\u00f4\u017eete skon\u010di\u0165 a pr\u00edpadne sa pobavi\u0165 pri n\u00e1soben\u00ed mali\u010dk\u00fdch \u010d\u00edsel. Napr\u00edklad 36 x 63, v\u00fdsledok si \u013eahko over\u00edte na kalkula\u010dke. Moja rada: zdvojn\u00e1sobujte to v\u00e4\u010d\u0161ie \u010d\u00edslo a to men\u0161ie potom vyskladajte z mocn\u00edn dvojky. Alebo aj nie. Niekedy to m\u00f4\u017ee by\u0165 aj dlh\u0161ie ako naopak. Napr\u00edklad 63 x 71 bude o dos\u0165 r\u00fdchlej\u0161ie, ak si vysklad\u00e1te 71tku…<\/p>\n\n\n\n

Pre t\u00fdch, ktor\u00ed maj\u00fa podobn\u00fa predstavu o z\u00e1bave ako ja, priklad\u00e1m p\u00e1r postrehov. St\u013apec \u00faplne v\u013eavo obsahuje mocniny dvojky. \u00daplne vpravo zasa dvojn\u00e1sobky \u010d\u00edsla 987654321. V strede mi zostalo dos\u0165 miesta, tak som tam umiestnila v\u00fdpo\u010det, ktor\u00fdm som ur\u010dila, ktor\u00e9 riadky budem potrebova\u0165. Mohla som to, samozrejme, spravi\u0165 aj klasicky – pozriem najvy\u0161\u0161iu mocninu men\u0161iu ako 123456789, ozna\u010d\u00edm ten riadok a zist\u00edm rozdiel. K nemu zasa n\u00e1jdem najvy\u0161\u0161iu mocninu men\u0161iu od neho a t.\u010f. To sa mi ale nechcelo, tak som pou\u017eila algoritmus, ktor\u00fdm sa prev\u00e1dza \u010d\u00edslo do dvojkovej s\u00fastavy. Je dos\u0165 \u013eahk\u00fd: iba del\u00edte dvojkou a zapisujete zvy\u0161ky. Ja som si to tak pekne rozmiestnila, \u017ee tam, kde m\u00e1m zvy\u0161ok jeden, to je pr\u00e1ve ten riadok, ktor\u00fd potrebujem ozna\u010di\u0165. Tak\u017ee v riadkoch s jednotkou som vpravo obtiahla \u010d\u00edsla a tie som n\u00e1sledne s\u010d\u00edtala. To je v\u0161etko. Brnka\u010dka.<\/p>\n\n\n\n

Sk\u00fa\u0161ka spr\u00e1vnosti<\/h2>\n\n\n\n

A\u017e na to, \u017ee som, ako v\u017edy, porobila p\u00e1r numerick\u00fdch ch\u00fdb, ktor\u00e9 treba n\u00e1js\u0165 a opravi\u0165. Ako som vedela, \u017ee m\u00e1m zl\u00fd v\u00fdsledok? Tak po prv\u00e9, u\u017e sa pozn\u00e1m. A po druh\u00e9 nesed\u00ed deviatkov\u00e1 sk\u00fa\u0161ka.<\/p>\n\n\n\n

Deviatkov\u00e1 sk\u00fa\u0161ka<\/b> je \u00faplne n\u00e1dhern\u00e1 vec. Nech\u00e1pem, \u017ee sa o nej skoro nikde neu\u010d\u00ed. Ja som na \u0148u narazila v podstate n\u00e1hodou. Ide o to, \u017ee ke\u010f pou\u017e\u00edvate iba s\u00fa\u010det, rozdiel a s\u00fa\u010din a m\u00e1te ve\u013ek\u00e9 \u010d\u00edsla, zredukujete cel\u00e9 zadanie na zvy\u0161ky po delen\u00ed deviatkou. Zvy\u0161ok po delen\u00ed deviatkou sa toti\u017e n\u00e1jde ve\u013emi \u013eahko – sta\u010d\u00ed ur\u010di\u0165 zvy\u0161ok cifern\u00e9ho s\u00fa\u010dtu. V mojom pr\u00edpade 1+2+3+4+5+6+7+8+9. Druh\u00fd \u010dinite\u013e m\u00e1 cifern\u00fd s\u00fa\u010det \u00faplne rovnak\u00fd, v\u0161ak\u017ee. Ja som tak leniv\u00e1, \u017ee ani toto nes\u010d\u00edtavam. \u010c\u00edsla zoskupujem do deviatok a tie vyhadzujem, a\u017e mi nie\u010do zostane. Tu m\u00e1m 9, potom 1+8, 2+7, 3+6, 4+5. Ni\u010d nezostalo, \u010d\u00edslo je teda delite\u013en\u00e9 deviatkou bezo zvy\u0161ku. No a ke\u010f vyn\u00e1sob\u00edme dve \u010d\u00edsla delite\u013en\u00e9 deviatkou, tak aj v\u00fdsledok mus\u00ed byt delite\u013en\u00fd deviatkou. Ak nie je, niekde je zaru\u010dene chyba. Ak je, st\u00e1le tam niekde m\u00f4\u017ee by\u0165 chyba, ale je to m\u00e1lo pravdepodobn\u00e9.<\/p>\n\n\n\n

E\u0161te pozn\u00e1mka k deviatkovej sk\u00fa\u0161ke – ak by tam nejak\u00e9 zvy\u0161ky vy\u0161li, treba urobi\u0165 v\u0161etky oper\u00e1cie s t\u00fdmito zvy\u0161kami a ak\u00fd zvy\u0161ok bude ma\u0165 v\u00fdsledok, tak\u00fd zvy\u0161ok mus\u00ed ma\u0165 aj cel\u00fd vypo\u010d\u00edtan\u00fd v\u00fdsledok. \u010ci\u017ee, ak by sme n\u00e1sobili \u010d\u00edsla, z ktor\u00fdch jedno d\u00e1va zvy\u0161ok 4 a druh\u00e9 7, ich s\u00fa\u010din je 28, \u010do je zvy\u0161ok 1. A teda celkov\u00fd v\u00fdsledok mus\u00ed ma\u0165 zvy\u0161ok 1. Elegantn\u00e9 a r\u00fdchle. \ud83d\ude0d<\/p>\n\n\n\n

No, ale mne deviatkov\u00e1 sk\u00fa\u0161ka nevy\u0161la, tak\u017ee som musela p\u00e1tra\u0165. A tu prin\u00e1\u0161am e\u0161te dva sp\u00f4soby overenia:<\/p>\n\n\n\n

Kontrola spredu<\/b> spo\u010d\u00edva v tom, \u017ee \u010d\u00edsla norm\u00e1lne vyn\u00e1sob\u00edm na kalkula\u010dke. Ona sice v\u00fdsledok zaokr\u00fahli, ale prv\u00fdch nieko\u013eko cifier bude spr\u00e1vnych. Sta\u010d\u00ed porovna\u0165.<\/p>\n\n\n\n

Kontrola odzadu<\/b> je tak\u00e1 menej zn\u00e1ma finta. Ak toti\u017e vyn\u00e1sob\u00edte len nieko\u013eko posledn\u00fdch cifier, to\u013eko isto cifier odzadu bude spr\u00e1vnych. Mne kalkula\u010dka zvl\u00e1dla n\u00e1sobi\u0165 bez zaokr\u00fah\u013eovania p\u00e4\u0165cifern\u00e9 \u010d\u00edsla. Tak som n\u00e1sobila 56789 x 54321. Posledn\u00fdch p\u00e4\u0165 cifier v\u00fdsledku je spr\u00e1vnych. T\u00fato kontrolu pou\u017e\u00edvam na posledn\u00fa jednu cifru dos\u0165 \u010dasto. \ud83d\ude09<\/p>\n\n\n\n

Po t\u00fdchto kontrol\u00e1ch zostali len tri a\u017e \u0161tyri miesta, kde e\u0161te mohla zosta\u0165 chyba. Z\u00e1kon schv\u00e1lnosti op\u00e4\u0165 funguje, preto\u017ee pr\u00e1ve tam bola. \ud83d\ude0f<\/p>\n\n\n\n

Tak\u017ee 123.456.789 x 987.654.321 = 121.932.631.112.635.269, tak\u00e9 pekn\u00e9 zadanie a tak nezauj\u00edmav\u00fd v\u00fdsledok. \ud83e\udd23<\/p>\n","protected":false},"excerpt":{"rendered":"

Ke\u010f som si pred p\u00e1r rokmi pripravovala predn\u00e1\u0161ku o n\u00e1soben\u00ed, narazila som v knihe od profesora \u010ci\u017em\u00e1ra – Hist\u00f3ria matematiky na zauj\u00edmav\u00fa pas\u00e1\u017e – n\u00e1sobenie v starovekom Egypte. V tom \u010dase toti\u017e nemohli n\u00e1sobi\u0165 tak, ako to rob\u00edme dnes. Nepoznali pozi\u010dn\u00fd sp\u00f4sob z\u00e1pisu \u010d\u00edsel. \u010c\u00edsla zapisovali pomocou nieko\u013ek\u00fdch znakov. Mali znak pre jednotku, desiatku, stovku, […]<\/p>\n","protected":false},"author":2,"featured_media":1967,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[33,34,17,35],"class_list":["post-72","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-nezaradene","tag-aritmetika","tag-deviatkova-skuska","tag-historia","tag-nasobenie"],"yoast_head":"\nN\u00e1sobenie v starovekom Egypte - jankafialka<\/title>\n<meta name=\"description\" content=\"N\u00e1sobenie v starovekom Egypte met\u00f3dou zdvojovania. Podrobn\u00fd postup, nieko\u013eko mo\u017enost\u00ed pre kontrolu spravnosti. 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