{"id":65,"date":"2022-01-21T16:41:00","date_gmt":"2022-01-21T15:41:00","guid":{"rendered":""},"modified":"2023-08-04T22:46:21","modified_gmt":"2023-08-04T20:46:21","slug":"parny-alebo-neparny","status":"publish","type":"post","link":"https:\/\/www.jankafialka.sk\/?p=65","title":{"rendered":"P\u00e1rny alebo nep\u00e1rny?"},"content":{"rendered":"\n<p>V t\u00fdchto d\u0148och sa u\u010ditelia v\u0161etk\u00fdch predmetov venuj\u00fa zv\u00e4\u010d\u0161a dosk\u00fa\u0161avaniu \u017eiakov, ktor\u00ed s\u00fa &#8222;medzi&#8220;. &#8222;Keby zabezpe\u010dili, aby mali v\u0161etci \u017eiaci nep\u00e1rny po\u010det zn\u00e1mok, nemali by teraz \u017eiaden probl\u00e9m,&#8220; preblesklo mi hlavou. Ale naozaj? Tak\u00e9to tvrdenie treba dok\u00e1za\u0165. Tak po\u010fme na to:<\/p>\n\n\n\n<p>Over\u00edme, \u010di sa naozaj nem\u00f4\u017ee sta\u0165, \u017ee priemer nep\u00e1rneho po\u010dtu zn\u00e1mok bude &#8222;medzi&#8220;. Vieme, ako sa po\u010d\u00edta priemer zn\u00e1mok. Oby\u010dajne ako aritmetick\u00fd priemer, \u010di\u017ee spo\u010d\u00edtaj\u00fa sa v\u0161etky zn\u00e1mky &#8211; s\u00fa\u010det nazveme <i>s.<\/i>&nbsp;A tento s\u00fa\u010det n\u00e1sledne vydel\u00edme po\u010dtom zn\u00e1mok, ktor\u00fd m\u00e1 by\u0165 nep\u00e1rny, tak\u017ee m\u00e1me <i>s\/nep. <\/i>Toto m\u00e1 vyjs\u0165 nie\u010do medzi, tak\u017ee 1,5 alebo 2,5 a t.\u010f. To si zap\u00ed\u0161eme ako zlomok 3\/2, 5\/2, 7\/2 alebo 9\/2. Vid\u00edme, \u017ee v \u010ditateli je nep\u00e1rne \u010d\u00edslo, v menovateli dvojka.<\/p>\n\n\n\n<p>Dostaneme rovnicu <i>s\/nep.=nep.\/2<\/i>. Ktor\u00e1 po jednoduchej \u00faprave vyzer\u00e1 takto: 2<i>s= nep.\u2022nep.<\/i>&nbsp;S\u00fa\u010din dvoch nep\u00e1rnych \u010d\u00edsel ale nikdy nebude rovn\u00fd \u010d\u00edslu 2<i>s,<\/i>&nbsp;lebo 2<i>s<\/i>&nbsp;je p\u00e1rne, tak\u017ee toto sa fakt sta\u0165 nem\u00f4\u017ee.<\/p>\n\n\n\n<p>S\u00fa\u010din dvoch nep\u00e1rnych \u010d\u00edsel je nep\u00e1rny. Sta\u010d\u00ed, aby jedno z nich bolo p\u00e1rne, u\u017e bude s\u00fa\u010din p\u00e1rny. Pekne to vidie\u0165 napr\u00edklad v tabu\u013eke malej n\u00e1sobilky. Vyfarbite si v\u0161etky nep\u00e1rne v\u00fdsledky. Bude ich iba 25 zo stovky v\u0161etk\u00fdch v\u00fdsledkov.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Pytagorejci<\/h2>\n\n\n\n<p>Tento jednoduch\u00fd fakt stoj\u00ed aj za jedn\u00fdm z mojich ob\u013e\u00faben\u00fdch d\u00f4kazov. Rozpr\u00e1vanie by som za\u010dala d\u00e1vno prad\u00e1vno. V \u010dasoch, ke\u010f matematika bola z\u00e1bavou vyvolen\u00fdch. Mu\u017ei, o ktor\u00fdch budem hovori\u0165, sa dnes naz\u00fdvaj\u00fa Pytagorejci, n\u00e1stupcovia sl\u00e1vneho Pytagora. Ako to tak b\u00fdvalo, nezaoberali sa len matematikou, ale aj filozofiou a kade\u010d\u00edm in\u00fdm, o \u010dom sa dalo h\u00fata\u0165 pri \u010d\u013eapkan\u00ed sa v baz\u00e9ne. No, a oni verili, \u017ee v\u0161etko na svete sa d\u00e1 pop\u00edsa\u0165 \u010d\u00edslami. (\u00c1no, odtia\u013eto poch\u00e1dza numerol\u00f3gia.) A to nie hocijak\u00fdmi, ale \u010d\u00edslami, ktor\u00e9 sa daj\u00fa nap\u00edsa\u0165 v tvare zlomku v z\u00e1kladnom tvare. (T.j. zlomkom, kde \u010ditate\u013e aj menovate\u013e s\u00fa prirodzen\u00e9 \u010d\u00edsla, ktor\u00e9 nemaj\u00fa \u017eiadneho spolo\u010dn\u00e9ho delite\u013ea- nedaj\u00fa sa u\u017e vykr\u00e1ti\u0165.)<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"768\" height=\"768\" src=\"https:\/\/www.jankafialka.sk\/wp-content\/uploads\/2022\/01\/95209167.jpg\" alt=\"\" class=\"wp-image-1943\" srcset=\"https:\/\/www.jankafialka.sk\/wp-content\/uploads\/2022\/01\/95209167.jpg 768w, https:\/\/www.jankafialka.sk\/wp-content\/uploads\/2022\/01\/95209167-300x300.jpg 300w, https:\/\/www.jankafialka.sk\/wp-content\/uploads\/2022\/01\/95209167-150x150.jpg 150w\" sizes=\"auto, (max-width: 768px) 100vw, 768px\" \/><figcaption class=\"wp-element-caption\">Pytagorejci pri \u00favahach o matematike<\/figcaption><\/figure>\n\n\n\n<p>Je celkom ironick\u00e9, \u017ee jedna z najsl\u00e1vnej\u0161\u00edch viet, ktor\u00e1 je po Pytagorovi pomenovan\u00e1, v\u00f4bec nie je jeho objavom a navy\u0161e pr\u00e1ve t\u00e1to veta stoj\u00ed za devast\u00e1ciou hlavn\u00e9ho sveton\u00e1zorov\u00e9ho presved\u010denia Pytagorejcov.<\/p>\n\n\n\n<p>Ist\u00fd Pytagoreho \u017eiak sa toti\u017e podujal n\u00e1js\u0165 ten zlomok, ktor\u00fd vyjadruje uhloprie\u010dku \u0161tvorca so stranou d\u013a\u017eky jeden. Pod\u013ea Pytagoreho vety je to \u010d\u00edslo, ktor\u00e9 ke\u010f vyn\u00e1sob\u00edme sam\u00e9 sebou, vyjde dvojka. H\u013eadan\u00fd zlomok bude teda <i>p\/q<\/i>, pri\u010dom minim\u00e1lne jedno &#8211; \u010ditate\u013e alebo menovate\u013e mus\u00ed by\u0165 nep\u00e1rne \u010d\u00edslo. (Zlomok m\u00e1 by\u0165 v z\u00e1kladnom tvare.)<\/p>\n\n\n\n<p>M\u00e1me teda rovnicu <i>p\/q\u2022p\/q = 2<\/i>.<\/p>\n\n\n\n<p>Uprav\u00edme: <i>p\u2022p = 2\u2022q\u2022q.<\/i><\/p>\n\n\n\n<p>Teraz je jasn\u00e9, \u017ee <i>p<\/i>&nbsp;je p\u00e1rne, preto\u017ee vpravo m\u00e1me p\u00e1rne \u010d\u00edslo. Z toho ale vypl\u00fdva, \u017ee <i>q<\/i>&nbsp;je nep\u00e1rne. Ke\u010f\u017ee&nbsp;<i>p <\/i>je n\u00e1sobkom dvojky, m\u00f4\u017eeme ho k\u013eudne nap\u00edsa\u0165 ako 2<i>k. <\/i>Dosa\u010fme:<\/p>\n\n\n\n<p>2<i>k\u2022<\/i>2<i>k = 2q\u2022q.<\/i>&nbsp;Cel\u00fa rovnicu vydel\u00edme dvojkou a dostaneme:<\/p>\n\n\n\n<p>2<i>k\u2022k = q\u2022q<\/i>.<\/p>\n\n\n\n<p>A tu sme narazili na neprekonate\u013en\u00fd probl\u00e9m. Vpravo m\u00e1me s\u00fa\u010din dvoch nep\u00e1rnych \u010d\u00edsel a v\u013eavo p\u00e1rne \u010d\u00edslo. Koniec. Cel\u00e1 viera pad\u00e1. Nie v\u0161etky \u010d\u00edsla sa daj\u00fa zap\u00edsa\u0165 ako zlomok.<\/p>\n\n\n\n<p>Dnes sa u\u017e v\u00f4bec ne\u010dudujeme. Vieme, \u017ee \u010d\u00edsla, ktor\u00e9 sa daj\u00fa zap\u00edsa\u0165 ako zlomok, sa naz\u00fdvaj\u00fa racion\u00e1lne. V\u0161etky ostatn\u00e9 s\u00fa iracion\u00e1lne. A spolu tvoria mno\u017einu re\u00e1lnych \u010d\u00edsel. Keby Pytagoras vedel, \u017ee iracion\u00e1lnych \u010d\u00edsel je dokonca viac ako racion\u00e1lnych&#8230;(Pozri rozpr\u00e1vku o nekone\u010dnom hoteli &#8211; racion\u00e1lne by sme do neho ubytovali, iracion\u00e1lne nie. Je to v\u00e4\u010d\u0161ie nekone\u010dno.)<\/p>\n","protected":false},"excerpt":{"rendered":"<p>V t\u00fdchto d\u0148och sa u\u010ditelia v\u0161etk\u00fdch predmetov venuj\u00fa zv\u00e4\u010d\u0161a dosk\u00fa\u0161avaniu \u017eiakov, ktor\u00ed s\u00fa &#8222;medzi&#8220;. &#8222;Keby zabezpe\u010dili, aby mali v\u0161etci \u017eiaci nep\u00e1rny po\u010det zn\u00e1mok, nemali by teraz \u017eiaden probl\u00e9m,&#8220; preblesklo mi hlavou. Ale naozaj? Tak\u00e9to tvrdenie treba dok\u00e1za\u0165. Tak po\u010fme na to: Over\u00edme, \u010di sa naozaj nem\u00f4\u017ee sta\u0165, \u017ee priemer nep\u00e1rneho po\u010dtu zn\u00e1mok bude &#8222;medzi&#8220;. Vieme, [&hellip;]<\/p>\n","protected":false},"author":2,"featured_media":1943,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[20,21,22,23],"class_list":["post-65","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-nezaradene","tag-aritmeticky-priemer","tag-iracionalne-cisla","tag-parita","tag-pytagoras"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>P\u00e1rny alebo nep\u00e1rny? - jankafialka<\/title>\n<meta name=\"description\" content=\"P\u00e1rny alebo nep\u00e1rny? Ot\u00e1zka, ktor\u00e1 pom\u00e1ha rie\u0161i\u0165 mnoho matematick\u00fdch ot\u00e1zok. 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