Bu yaz\u0131da matematiksel bir g\u00f6steri formuna d\u00f6n\u00fc\u015ft\u00fcrd\u00fc\u011f\u00fcm\u00fcz ilgin\u00e7 bir \u00f6zde\u015fli\u011fi ve ard\u0131ndaki matemati\u011fi inceleyece\u011fiz.<\/p>\n
Kar\u015f\u0131n\u0131zdaki ki\u015fiden \u00e7ift say\u0131da ve istedi\u011fi kadar ard\u0131\u015f\u0131k tamsay\u0131y\u0131 sizden gizleyerek bir ka\u011f\u0131da yazmas\u0131n\u0131 isteyin. (\u00d6rne\u011fin kar\u015f\u0131n\u0131zdaki ki\u015fi ka\u011f\u0131da -2, -1, 0, 1, 2, 3, 4, 5 gibi sekiz (\u00e7ift say\u0131da) tane ard\u0131\u015f\u0131k say\u0131y\u0131 yazabilir.) \u201cBiraz sonra, ka\u00e7 tane ard\u0131\u015f\u0131k tamsay\u0131 yazd\u0131\u011f\u0131n\u0131 sana s\u00f6yleyece\u011fim\u201d diyebilirsiniz.<\/p>\n
Sonras\u0131nda, i\u015flemlerin sonu\u00e7lar\u0131n\u0131 yine sizden gizlemesi ko\u015fuluyla a\u015fa\u011f\u0131daki ad\u0131mlar\u0131 atmas\u0131n\u0131 isteyin.<\/p>\n
Ka\u011f\u0131ttaki t\u00fcm say\u0131lardan diledi\u011fince, eleman say\u0131lar\u0131 e\u015fit iki k\u00fcme olu\u015ftursun.<\/p>\n
K\u00fcmeleri A <\/em>ve B <\/em>olarak isimlendirirsek, A <\/em>k\u00fcmesinin en b\u00fcy\u00fck eleman\u0131ndan B <\/em>k\u00fcmesinin en k\u00fc\u00e7\u00fck eleman\u0131n\u0131 \u00e7\u0131kararak sonucun pozitif i\u015faretlisini (mutlak de\u011ferini) k\u00e2\u011f\u0131da not etsin, ayn\u0131 \u015fekilde A<\/em>\u2019daki en b\u00fcy\u00fck ikinci say\u0131dan B<\/em>\u2019deki en k\u00fc\u00e7\u00fck ikinci say\u0131y\u0131 \u00e7\u0131kar\u0131p yine pozitif i\u015faretlisini k\u00e2\u011f\u0131da ikinci say\u0131 olarak yazs\u0131n. Bu i\u015flemi bu \u015fekilde her iki k\u00fcmedeki di\u011fer say\u0131lar ( e\u011fer varsa \u00fc\u00e7\u00fcnc\u00fc, d\u00f6rd\u00fcnc\u00fc ve di\u011ferleri) i\u00e7in de uygulas\u0131n.<\/p>\n Farklar\u0131n pozitif i\u015faretlerini alarak elde etti\u011fi t\u00fcm say\u0131lar\u0131 toplay\u0131p sonucu size s\u00f6ylesin.<\/p>\n Bu toplam\u0131n karek\u00f6k\u00fcn\u00fcn iki kat\u0131 ba\u015flang\u0131\u00e7ta ka\u011f\u0131da yaz\u0131lan ard\u0131\u015f\u0131k say\u0131 adedini verir.<\/p>\n \u015eimdi, bu ad\u0131mlar\u0131 yukar\u0131da k\u00e2\u011f\u0131da yaz\u0131ld\u0131\u011f\u0131n\u0131 varsayd\u0131\u011f\u0131m\u0131z -2, -1, 0, 1, 2, 3, 4, 5 sekiz tane ard\u0131\u015f\u0131k say\u0131 i\u00e7in uygulayal\u0131m, bakal\u0131m verilen kural do\u011fru mu, sonu\u00e7 sekiz \u00e7\u0131kacak m\u0131?<\/p>\n Bu say\u0131lar\u0131 geli\u015fi g\u00fczel bir \u015fekilde eleman say\u0131lar\u0131 e\u015fit iki k\u00fcmeye ay\u0131ral\u0131m:<\/p>\n A <\/em>= {-1, 0, 3, 4}, B <\/em>= {-2, 1, 2, 5}.<\/p>\n A <\/em>k\u00fcmesinin en b\u00fcy\u00fck eleman\u0131n\u0131 B\u2019<\/em>nin en k\u00fc\u00e7\u00fck eleman\u0131ndan \u00e7\u0131kar\u0131p sonucun pozitif i\u015faretlisini yazal\u0131m ve bu i\u015flemi s\u0131ras\u0131yla kalan di\u011fer say\u0131lar i\u00e7in de yaparak elde etti\u011fimiz say\u0131lar\u0131 toplayal\u0131m:<\/p>\n |4\u2013(\u20132)|+|3\u20131|+|0\u20132|+|\u20131\u20135| = 16.<\/p>\n 16\u2019n\u0131n karek\u00f6k\u00fcn\u00fcn 2 kat\u0131 8. Kural do\u011fruland\u0131, ama bu ilgin\u00e7 \u00f6zde\u015flik keyfi olarak se\u00e7ece\u011fimiz t\u00fcm \u00e7ift say\u0131daki ard\u0131\u015f\u0131k say\u0131lar i\u00e7in de ge\u00e7erli mi acaba? Yan\u0131t olumlu, istedi\u011finiz kadar \u00e7ift say\u0131da ard\u0131\u015f\u0131k say\u0131y\u0131 diledi\u011finiz gibi se\u00e7in, yukar\u0131daki ad\u0131mlar\u0131n ard\u0131ndan elde edece\u011finiz toplam\u0131n karek\u00f6k\u00fcn\u00fcn iki kat\u0131 ba\u015flang\u0131\u00e7ta se\u00e7ti\u011finiz ard\u0131\u015f\u0131k say\u0131lar\u0131n say\u0131s\u0131n\u0131 veriyor. Ku\u015fkusuz, b\u00f6ylesi ilgin\u00e7 bir sonu\u00e7la kar\u015f\u0131la\u015fan her matematiksever bu \u00f6zde\u015fli\u011fin ard\u0131ndaki matemati\u011fi merak ederek kan\u0131t yolculu\u011funa \u00e7\u0131kmak isteyecektir.<\/p>\n 1985 Sovyet Matematik Olimpiyatlar\u0131\u2019nda sorulan ve matematik\u00e7i Vyacheslav Proizvolov taraf\u0131ndan ke\u015ffedilen bu \u00f6nerme Proizvolov \u00d6zde\u015fli\u011fi olarak biliniyor. Proizvolov\u2019un sadece pozitif tamsay\u0131lar i\u00e7in tan\u0131mlad\u0131\u011f\u0131 \u00f6zde\u015flik, tamsay\u0131lar k\u00fcmesinin t\u00fcm\u00fcnde ge\u00e7erli, \u00e7\u00fcnk\u00fc ard\u0131\u015f\u0131k say\u0131larla \u00e7al\u0131\u015ft\u0131\u011f\u0131m\u0131zdan se\u00e7ilen say\u0131lar\u0131n negatif ya da bir b\u00f6l\u00fcm\u00fcn\u00fcn negatif bir b\u00f6l\u00fcm\u00fcn\u00fcn pozitif olmas\u0131 sonucu de\u011fi\u015ftirmiyor. Bu durumu yaz\u0131n\u0131n ikinci b\u00f6l\u00fcm\u00fcnde ele alaca\u011f\u0131z.<\/p>\n Ak\u0131l dolu bir kan\u0131t<\/p>\n Se\u00e7ti\u011fimiz ard\u0131\u015f\u0131k say\u0131lar 2N <\/em>tane olsun. Bu say\u0131lar\u0131 k\u00fc\u00e7\u00fckten b\u00fcy\u00fc\u011fe do\u011fru s\u0131ralayarak a\u015fa\u011f\u0131daki gibi g\u00f6sterelim:<\/p>\n A<\/em>1, A<\/em>2, A<\/em>3, …, A<\/em>N<\/em>, A<\/em>N+<\/em>1, A<\/em>N+2<\/em>, … , A<\/em>2N<\/em>.<\/p>\n \u015eimdi, bu dizinin i\u00e7inden keyfi olarak N <\/em>tane say\u0131y\u0131 se\u00e7elim ve k\u00fc\u00e7\u00fckten b\u00fcy\u00fc\u011fe do\u011fru s\u0131ralayal\u0131m:<\/p>\n B<\/em>1 < B<\/em>2 < … < B<\/em>N<\/em>.<\/em><\/p>\n Se\u00e7medi\u011fimiz geriye kalan N <\/em>tane say\u0131y\u0131 da b\u00fcy\u00fckten k\u00fc\u00e7\u00fc\u011fe do\u011fru s\u0131ralayal\u0131m:<\/p>\n C<\/em>1 > C<\/em>2 > … > C<\/em>N<\/em>.<\/p>\n Bu durumda kan\u0131tlamam\u0131z gereken \u00f6zde\u015flik a\u015fa\u011f\u0131daki gibi yaz\u0131l\u0131r:<\/p>\n | B<\/em>1\u2013 C<\/em>1| +| B<\/em>2\u2013 C<\/em>2| +…+| B<\/em>N<\/em>\u2013 C<\/em>N<\/em>| = N<\/em>2.<\/p>\n Bu \u00f6zde\u015fli\u011fin tamsay\u0131lar k\u00fcmesinin t\u00fcm elemanlar\u0131 i\u00e7in sa\u011flan\u0131yor olmas\u0131 \u015fu \u00f6nermeye dayan\u0131yor.<\/p>\n Teorem. <\/strong>Farklar\u0131 al\u0131nan B<\/em>K<\/em>, C<\/em>K <\/em>\u00e7iftindeki say\u0131lar\u0131ndan biri X <\/em>= {A<\/em>1, A<\/em>2, A<\/em>3, …, A<\/em>N<\/em>} k\u00fcmesinin eleman\u0131, di\u011feri de Y = {A<\/em>N+<\/em>1, A<\/em>N+<\/em>2, …, A<\/em>2N<\/em>} k\u00fcmesinin eleman\u0131 olmak zorundad\u0131r.<\/p>\n Kan\u0131t. <\/strong>Teoremin olumsuzunun do\u011fru oldu\u011funu varsayal\u0131m, yani B<\/em>K <\/em>ve C<\/em>K <\/em>say\u0131lar\u0131n\u0131n her ikisi de X <\/em>ya da Y<\/em>\u2019nin eleman\u0131 olsun.<\/p>\n B<\/em>K <\/em>ve C<\/em>K <\/em>say\u0131lar\u0131n\u0131n her ikisi de X <\/em>k\u00fcmesinin elman\u0131 ise B<\/em>K <\/em>< A<\/em>N <\/em>ve C<\/em>K <\/em>< A<\/em>N <\/em>olur. Bu durumda B<\/em>K <\/em>< A<\/em>N <\/em>oldu\u011fundan en az K <\/em>tane say\u0131 N<\/em>\u2019den k\u00fc\u00e7\u00fck ya da e\u015fittir. \u00d6te yandan C<\/em>K <\/em>< A<\/em>N <\/em>oldu\u011fu i\u00e7in en az N<\/em>\u2013K<\/em>+1 tane say\u0131 yine N<\/em>\u2019den k\u00fc\u00e7\u00fck ya da e\u015fit olacakt\u0131r. B\u00f6ylece toplam olarak en az N<\/em>+1 tane say\u0131 N<\/em>\u2019den k\u00fc\u00e7\u00fck ya da e\u015fit olur ki bu da bir \u00e7eli\u015fkidir.<\/p>\n B<\/em>K <\/em>ve C<\/em>K <\/em>say\u0131lar\u0131n\u0131n her ikisinin de Y <\/em>k\u00fcmesinin elman\u0131 oldu\u011funu varsayarsak yukar\u0131dakine benzer bir yolla yine \u00e7eli\u015fkili bir sonuca ula\u015f\u0131r\u0131z.<\/p>\n O halde B<\/em>K <\/em>ve C<\/em>K<\/em>\u2019dan biri X<\/em>\u2019in di\u011feri de Y<\/em>\u2019nin eleman\u0131 olmak zorundad\u0131r. Teorem kan\u0131tlanm\u0131\u015ft\u0131r.<\/p>\n \u015eimdi art\u0131k, | B<\/em>1\u2013 C<\/em>1| +| B<\/em>2\u2013 C<\/em>2| +…+| B<\/em>N<\/em>\u2013 C<\/em>N<\/em>| toplam\u0131n\u0131n N<\/em>2\u2019ye e\u015fit oldu\u011funu g\u00f6sterebiliriz.<\/p>\n Hemen belirtelim, (daha \u00f6nce s\u00f6z\u00fcn\u00fc etmi\u015ftik) ba\u015flang\u0131\u00e7ta se\u00e7ilen A<\/em>1, A<\/em>2, A<\/em>3, …, A<\/em>N<\/em>, A<\/em>N+<\/em>1, A<\/em>N+2<\/em>, … , A<\/em>2N<\/em>. say\u0131lar\u0131n\u0131n (2N <\/em>tane) hangi aral\u0131kta oldu\u011funun hi\u00e7 bir \u00f6nemi yok, \u00e7\u00fcnk\u00fc se\u00e7ilen say\u0131lar ard\u0131\u015f\u0131k oldu\u011fundan farklar\u0131n toplam\u0131 de\u011fi\u015fmiyor. \u00d6rne\u011fin, -2, -1, 0, 1, 2, 3, 4, 5 say\u0131lar\u0131n\u0131 se\u00e7mekle 1, 2, 3, 4, 5, 6, 7, 8 say\u0131lar\u0131n\u0131 se\u00e7mek aras\u0131nda fark yok.Farklar\u0131 al\u0131nan say\u0131lardanbiri ilk d\u00f6rt say\u0131 aras\u0131ndan al\u0131n\u0131rken, di\u011feri kalan d\u00f6rt say\u0131 aras\u0131ndan al\u0131nd\u0131\u011f\u0131ndan farklar\u0131n birindeki eksilme di\u011ferinde art\u0131\u015fa neden oluyor. \u00d6rne\u011fin, 1, 2, 3, 4, 5, 6, 7, 8 say\u0131lar\u0131n\u0131 keyfi olarak K <\/em>= {2, 3, 6, 8}, L <\/em>= {1, 4, 5, 7}. gibi iki k\u00fcmeye ay\u0131ral\u0131m ve Proizvolov \u00d6zde\u015fli\u011fi\u2019ni yazal\u0131m:<\/p>\n |8\u20131|+|6\u20134|+|3\u20135|+|2\u20137| = 16.<\/p>\n Burada farklar\u0131n mutlak de\u011ferini alarak elde etti\u011fimiz 7, 2, 2, 5 say\u0131lar\u0131n\u0131, yaz\u0131n\u0131n ba\u015f\u0131nda se\u00e7ti\u011fimiz say\u0131lardan olu\u015fan (A <\/em>ve B <\/em>k\u00fcmelerinden se\u00e7ilen) farklar\u0131n mutlak de\u011feri olan 6, 2, 2, 6 say\u0131lar\u0131yla kar\u015f\u0131la\u015ft\u0131r\u0131rsak, 7 ile 6 aras\u0131ndaki 1 kadar eksilme 5 ile 6 aras\u0131ndaki 1 art\u0131\u015fa neden oluyor. Bu y\u00fczden ard\u0131\u015f\u0131k tamsay\u0131lar\u0131n se\u00e7ildi\u011fi aral\u0131\u011f\u0131n \u00f6nemi yok. \u00d6rnekle g\u00f6sterdi\u011fimiz bu sonucu merakl\u0131 okur de\u011fi\u015fken kullanarak kolayca kan\u0131tlayabilir.<\/p>\n \u015eimdi art\u0131k, A<\/em>1, A<\/em>2, A<\/em>3, …, A<\/em>N<\/em>, A<\/em>N+<\/em>1, A<\/em>N+2<\/em>, … , A<\/em>2N<\/em>. say\u0131lar\u0131 yerine 1, 2, 3, … N<\/em>, N<\/em>+1, N<\/em>+2, … 2N <\/em>say\u0131lar\u0131n\u0131 alabiliriz.<\/p>\n Kan\u0131tlad\u0131\u011f\u0131m\u0131z teorem gere\u011fi B<\/em>K<\/em>, C<\/em>K <\/em>\u00e7iftindeki say\u0131lardan birinin ilk N <\/em>tamsay\u0131 aras\u0131ndan, di\u011ferinin de sonraki ilk N <\/em>tamsay\u0131 aras\u0131ndan oldu\u011funu bildi\u011fimizden | B<\/em>1\u2013 C<\/em>1| +| B<\/em>2\u2013 C<\/em>2| +…+| B<\/em>N<\/em>\u2013 C<\/em>N<\/em>| toplam\u0131n\u0131 hesaplayabiliriz.<\/p>\n Bu toplam, (N<\/em>+1)+(N<\/em>+2)+…+2N <\/em>ile 1+2+3+…+N <\/em>aras\u0131ndaki farka e\u015fittir. Buradan N <\/em>tane N<\/em>\u2019nin toplam\u0131 bulunur ki, arad\u0131\u011f\u0131m\u0131z sonu\u00e7 N<\/em>2\u2019dir. B\u00f6ylece Proizvolov \u00d6zde\u015fli\u011fi\u2019nin do\u011fru oldu\u011funu \u00e7ift say\u0131daki ard\u0131\u015f\u0131k t\u00fcm tam say\u0131lar i\u00e7in kan\u0131tlam\u0131\u015f olduk.<\/p>\n","protected":false},"excerpt":{"rendered":" Bu yaz\u0131da matematiksel bir g\u00f6steri formuna d\u00f6n\u00fc\u015ft\u00fcrd\u00fc\u011f\u00fcm\u00fcz ilgin\u00e7 bir \u00f6zde\u015fli\u011fi ve ard\u0131ndaki matemati\u011fi inceleyece\u011fiz. Kar\u015f\u0131n\u0131zdaki ki\u015fiden \u00e7ift say\u0131da ve istedi\u011fi kadar ard\u0131\u015f\u0131k tamsay\u0131y\u0131 sizden gizleyerek bir ka\u011f\u0131da yazmas\u0131n\u0131 isteyin. (\u00d6rne\u011fin kar\u015f\u0131n\u0131zdaki ki\u015fi ka\u011f\u0131da -2, -1, 0, 1, 2, 3, 4, 5 gibi sekiz (\u00e7ift say\u0131da) tane ard\u0131\u015f\u0131k say\u0131y\u0131 yazabilir.) \u201cBiraz sonra, ka\u00e7 tane ard\u0131\u015f\u0131k tamsay\u0131 […]<\/p>\n","protected":false},"author":375,"featured_media":23827,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[167,514],"tags":[2934,208],"class_list":["post-23826","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-130-sayi","category-matematik-sohbetleri","tag-farklarin-toplami","tag-matematik"],"acf":[],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t