p(pi) Say\u0131s\u0131:<\/p>\n
K\u0131saca bir dairenin \u00e7evresinin \u00e7ap\u0131na oran\u0131, p say\u0131s\u0131n\u0131 verir. \u0130nsano\u011flu, asl\u0131nda \u00e7ok \u00f6nemli vazifeleri olan bu say\u0131 \u00fczerinde \u00e7ok d\u00fc\u015f\u00fcnm\u00fc\u015ft\u00fcr. Y\u0131llarca tam olarak bir de\u011fer bulamamakla beraber, ger\u00e7ek de\u011ferine en yak\u0131n sonu\u00e7lar\u0131 kullanabilmek i\u00e7in \u00e7aba sarfetmi\u015flerdir. <\/p>\n
p’ nin kronolojik geli\u015fimine bakt\u0131\u011f\u0131m\u0131zda g\u00fcn\u00fcm\u00fczde dahi tam bir sonu\u00e7 bulunamam\u0131\u015ft\u0131r. \u00c7e\u015fitli form\u00fcller \u00fcretilmesine ra\u011fmen sadece her seferinde ger\u00e7ek de\u011fere biraz daha yakla\u015f\u0131lm\u0131\u015ft\u0131r. <\/p>\n
Ar\u015fimet 3.1\/7 ile 3.10\/71 aras\u0131nda bir say\u0131 olarak hesaplad\u0131. M\u0131s\u0131rl\u0131lar 3.1605, Babilliler 3.1\/8, Batlamyus 3.14166 olarak kulland\u0131. \u0130talyan Lazzarini 3.1415929, Fibonacci ise 3.141818 ile i\u015flem yap\u0131yordu. 18.yyda 140, 19yyda 500 basama\u011fa kadar hesapland\u0131. \u0130lk bilgisayarlarla 2035 basama\u011f\u0131 hesaplan\u0131rken g\u00fcn\u00fcm\u00fczde milyonlarca basama\u011fa kadar \u00e7\u0131k\u0131l\u0131yor. \u0130\u015fin ilgin\u00e7 taraf\u0131, h\u00e2l\u00e2 tam bir sonu\u00e7 yok. Herhangi bir yerinde devir olsa i\u015f yine kolayla\u015facak. Ama hen\u00fcz \u00f6yle bir \u015feye de rastlanmad\u0131. \u015eu anda bilinen de\u011ferden birka\u00e7 basamak:<\/p>\n
p=3,1415926535897932384626433832795028841971693993 7 510582097494459230781640
\n 62862089986280348253421170679821480865132823066470 9384460955058223172535940
\n 81284811174502841027…..<\/p>\n
\u0130lgin\u00e7 Say\u0131lar(1):<\/p>\n
3\u00b2 + 4\u00b2 = 5\u00b2
\n 10\u00b2 + 11\u00b2 + 12\u00b2 = 13\u00b2 + 14\u00b2
\n 21\u00b2 + 22\u00b2 + 23\u00b2 + 24\u00b2 = 25\u00b2 + 26\u00b2 + 27\u00b2
\n 36\u00b2 + 37\u00b2 + 38\u00b2 + 39\u00b2 + 40\u00b2 = 41\u00b2 + 42\u00b2 + 43\u00b2 + 44\u00b2
\n .
\n .
\n . <\/p>\n
Fermat’\u0131n Son Teoremi:<\/p>\n
Mesle\u011fi Avukatl\u0131k olan Fermat, arada bir matematikle de ilgilenirdi. Ama ne ilgilenmek. A\u015fa\u011f\u0131daki teorem, onun eseri. 1665 y\u0131l\u0131nda 64 ya\u015f\u0131nda \u00f6len Fermat’\u0131n a\u015fa\u011f\u0131daki teoremi, h\u00e2l\u00e2 ispatlanamad\u0131. Bu problem \u00fczerinde y\u0131llarca \u00e7al\u0131\u015fan \u00fcnl\u00fc alman matematik\u00e7i Wolfskehl, 1908 y\u0131l\u0131nda \u00f6ld\u00fc\u011f\u00fcnde, vasiyet olarak 100bin mark b\u0131rakt\u0131. Hem de bu problemi y\u00fczy\u0131l i\u00e7inde \u00e7\u00f6zecek ilk ki\u015fiye verilmek \u00fczere!<\/p>\n
Teorem \u015f\u00f6yle:<\/p>\n
n>2 ve a, b ve c tamsay\u0131 olmak \u00fczere<\/p>\n
an + bn= cn \u00e7\u00f6z\u00fcm\u00fc olmad\u0131\u011f\u0131n\u0131 ispatlay\u0131n.<\/p>\n
Fermat bu teoremi yazarken kulland\u0131\u011f\u0131 ka\u011f\u0131d\u0131n alt\u0131nda \u00e7ok az yer kald\u0131\u011f\u0131 i\u00e7in cevab\u0131 yazamad\u0131\u011f\u0131n\u0131, halbuki \u00e7ok g\u00fczel bir ispat\u0131 oldu\u011funu yazm\u0131\u015ft\u0131r. (Belki Fermat ta cevab\u0131 bilmiyordu)<\/p>\n
Bir hat\u0131rlatma: E\u011fer rastgele n=54179653 say\u0131s\u0131n\u0131 form\u00fcle uygulay\u0131p e\u015fitli\u011fi sa\u011flamad\u0131\u011f\u0131n\u0131 g\u00f6stermediyseniz, bu say\u0131n\u0131n h\u00e2l\u00e2 do\u011fru olma \u015fans\u0131 var demektir.<\/p>\n
\u0130lgin\u00e7 Say\u0131lar(2):<\/p>\n
\u00dc\u00e7 basamakl\u0131 herhangi bir say\u0131y\u0131 iki kere yanyana yazarak elde etti\u011fimiz yeni say\u0131, kesinlikle 7, 11, 13, 77, 91, 143, 1001 say\u0131lar\u0131na kalans\u0131z olarak b\u00f6l\u00fcn\u00fcr(neden?).<\/p>\n
\u00d6rnek: 831831<\/p>\n
831831 \/ 7 = 118833
\n 831831 \/ 11 = 75621
\n 831831 \/ 13 = 63987
\n 831831 \/ 77 = 10803
\n 831831 \/ 91 = 9141
\n 831831 \/ 143 = 5817
\n 831831 \/ 1001 = 831<\/p>\n
Sihirli Kareler:<\/p>\n
3 x 3: Birbirini yatay, dikey ve \u00e7apraz takip eden \u00fc\u00e7 karenin toplam\u0131, 15.<\/p>\n
8<\/p>\n
1<\/p>\n
6<\/p>\n
3<\/p>\n
5<\/p>\n
7<\/p>\n
4<\/p>\n
9<\/p>\n
2<\/p>\n
4 x 4: Birbirini yatay, dikey ve \u00e7apraz takip eden d\u00f6rt karenin toplam\u0131, 34.<\/p>\n
16<\/p>\n
2<\/p>\n
3<\/p>\n
13<\/p>\n
5<\/p>\n
11<\/p>\n
10<\/p>\n
8<\/p>\n
9<\/p>\n
7<\/p>\n
6<\/p>\n
12<\/p>\n
4<\/p>\n
14<\/p>\n
15<\/p>\n
1<\/p>\n
5 x 5: Birbirini yatay, dikey ve \u00e7apraz takip eden be\u015f karenin toplam\u0131, 65.<\/p>\n
3<\/p>\n
16<\/p>\n
9<\/p>\n
22<\/p>\n
15<\/p>\n
20<\/p>\n
8<\/p>\n
21<\/p>\n
14<\/p>\n
2<\/p>\n
7<\/p>\n
25<\/p>\n
13<\/p>\n
1<\/p>\n
19<\/p>\n
24<\/p>\n
12<\/p>\n
5<\/p>\n
18<\/p>\n
6<\/p>\n
11<\/p>\n
4<\/p>\n
17<\/p>\n
10<\/p>\n
23<\/p>\n
\u0130lgin\u00e7 Say\u0131lar(3):<\/p>\n
1 x 8 + 1 = 9
\n 12 x 8 + 2 = 98
\n 123 x 8 + 3 = 987
\n 1234 x 8 + 4 = 9876
\n 12345 x 8 + 5 = 98765
\n 123456 x 8 + 6 = 987654
\n 1234567 x 8 + 7 = 9876543
\n 12345678 x 8 + 8 = 98765432
\n 123456789 x 8 + 9 = 987654321<\/p>\n
Teorem:<\/p>\n
B\u00fct\u00fcn kare say\u0131lar, 1’den ba\u015flamak \u00fczere s\u0131ras\u0131yla tek tamsay\u0131lar\u0131n toplam\u0131 olarak yaz\u0131labilir.<\/p>\n
\u00d6rnekler:<\/p>\n
5\u00b2=25
\n 1 + 3 + 5 + 7 + 9 = 25<\/p>\n
11\u00b2 = 121
\n 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 = 121<\/p>\n
\u00dc\u00e7gen Say\u0131lar:<\/p>\n
1’den ba\u015flamak \u00fczere kendisinden \u00f6nceki t\u00fcm say\u0131lar\u0131n toplam\u0131na kar\u015f\u0131l\u0131k gelen say\u0131lar\u0131n dizisidir. <\/p>\n
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, … pozitif do\u011fal say\u0131lar ise, \u00fc\u00e7gen say\u0131lar:<\/p>\n
1, 3(1+2), 6(1+2+3), 10(1+2+3+4), 15(1+2+3+4+5),… \u00fc\u00e7gen say\u0131lard\u0131r. Yani:<\/p>\n
1, 3, 6, 10, 15, 21, 28, 36, 45, 55…<\/p>\n
Pascal \u00dc\u00e7geni:<\/p>\n
Pascal \u00fc\u00e7geni, \u015fekilde de g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi kenarlarda “1” olmak \u00fczere her say\u0131, \u00fcst\u00fcndeki iki say\u0131n\u0131n toplam\u0131 olarak yaz\u0131lacak \u015fekilde olu\u015fturulur. <\/p>\n
Pascal \u00fc\u00e7geninin baz\u0131 \u00f6zellikleri:
\n Kenarlar “1”den olu\u015fur
\n ikinci(k\u0131rm\u0131z\u0131) s\u0131ra, pozitif tamsay\u0131lar serisidir.
\n \u00dc\u00e7\u00fcnc\u00fc(mavi) s\u0131ra, \u00fc\u00e7gen say\u0131lard\u0131r. (1, 3, 6, 10 15,…)
\n Ayn\u0131 y\u00f6ndeki say\u0131lar\u0131n(sar\u0131) toplam\u0131, se\u00e7ti\u011fimiz son say\u0131n\u0131n ters y\u00f6n\u00fcndeki say\u0131ya e\u015fittir.
\n (\u00d6rnek: 1+2+3+4+5+6+7=28, 1+4+10+20+35=70 gibi)
\n Her s\u0131radaki say\u0131lar\u0131n toplam\u0131, ‘s\u0131f\u0131r’dan ba\u015flamak \u00fczere “2”nin \u00fcslerini verir. 20, 21, 22, 23 ,24 ,…
\n (\u00d6rnek: 5. s\u0131radaki say\u0131lar\u0131n toplam\u0131, 1+4+6+4+1=16=24 )
\n Her s\u0131ra, yine ‘s\u0131f\u0131r’dan ba\u015flamak \u00fczere kendi derecesinden bir polinomun katsay\u0131lar\u0131n\u0131 verir.
\n ( \u00d6rnek: (a+b)3=1a3+3ab2+3a2b+1b3) <\/p>\n
Teorem:<\/p>\n
B\u00fct\u00fcn say\u0131lar 2’nin \u00fcsleri toplam\u0131 (tekrars\u0131z) olarak yaz\u0131labilir.<\/p>\n
\u00d6rnekler:<\/p>\n
12 = 23 + 22
\n 12 = 8 + 4<\/p>\n
45 = 25 + 23 + 22 + 20
\n 45 = 32 + 8 + 4 + 1<\/p>\n
\u0130lgin\u00e7 Say\u0131lar(4):<\/p>\n
12 x 42 = 21 x 24
\n 23 x 96 = 32 x 69
\n 24 x 84 = 42 x 48
\n 13 x 62 = 31 x 26
\n 46 x 96 = 64 x 69<\/p>\n
Fibonacci Dizisi:<\/p>\n
1’den ba\u015flamak \u00fczere kendisinden \u00f6nceki iki say\u0131n\u0131n toplam\u0131na kar\u015f\u0131l\u0131k gelen say\u0131lar\u0131n dizisidir. <\/p>\n
1, 2, 3, 4, 5, 6, 7, 8, 9, …ise, fibonacci dizisi:<\/p>\n
1, 1(0+1), 2(1+1), 3(1+2), 5(2+3), 8(3+5), 13(5+8),… yani:<\/p>\n
1, 2, 3, 5, 8, 13, 21, 34, 55…<\/p>\n
Fibonacci dizisinin kullan\u0131ld\u0131\u011f\u0131 pek\u00e7ok yerden biri de “[Linkleri Sadece uyelerimiz gorebilirler.Uye Olmak icin Tiklayiniz]”ndaki \u00fc\u00e7genli ve kareli sorulard\u0131r.<\/p>\n
\u0130lgin\u00e7 Say\u0131lar(5):<\/p>\n
3 x 37 = 111
\n 6 x 37 = 222
\n 9 x 37 = 333
\n 12 x 37= 444
\n 15 x 37 = 555
\n 18 x 37 = 666
\n 21 x 37 = 777
\n 24 x 37 = 888
\n 27 x 37 = 999<\/p>\n
e Say\u0131s\u0131:<\/p>\n
1 + (1\/1!) + (1\/2!) + (1\/3!) + (1\/4!) + … + (1\/n!) serisinin toplam\u0131 “e” say\u0131s\u0131n\u0131 verir. Yakla\u015f\u0131k de\u011feri:<\/p>\n
e = 2.71828182…dir. (e sabit say\u0131s\u0131n\u0131n kullan\u0131ld\u0131\u011f\u0131 yerler ayr\u0131ca anlat\u0131lacakt\u0131r)<\/p>\n
(Sonsuz):<\/p>\n
\u00a5, sadece matematik\u00e7ilerin de\u011fil, d\u00fc\u015f\u00fcnen herkesin ilgisini ve merak\u0131n\u0131 \u00e7ekmi\u015ftir. \u00a5’u say\u0131 olarak d\u00fc\u015f\u00fcn\u00fcrsek; akl\u0131m\u0131z\u0131 zorlay\u0131p “en b\u00fcy\u00fck say\u0131”ya ula\u015ft\u0131\u011f\u0131m\u0131z\u0131 kabul edelim. O say\u0131n\u0131n mutlaka 1 fazlas\u0131 olaca\u011f\u0131ndan yeni say\u0131lar elde ederiz. <\/p>\n
Mesel\u00e2 say\u0131 do\u011frusunda 0 ile 1 aras\u0131nda sonsuz adet reel say\u0131 vard\u0131r. 0 ile 10 aras\u0131nda da sonsuz adet say\u0131 oldu\u011funa g\u00f6re bu iki sonsuz da birbirine e\u015fit olamaz. Bu y\u00fczden matematikte “\u00a5\/\u00a5” ifadesi tan\u0131ms\u0131zd\u0131r. Ayn\u0131 \u015fekilde 1\u00a5 ifadesi de hen\u00fcz tan\u0131mlanamam\u0131\u015ft\u0131r. H\u00e2lbuki 1’in t\u00fcm \u00fcsleri 1’ e\u015fit olmal\u0131d\u0131r. <\/p>\n
K\u00e2inatta ka\u00e7 adet “atom” oldu\u011fu sorulsa ka\u00e7 derdiniz? Herhalde akl\u0131n\u0131za gelebilecek en b\u00fcy\u00fck say\u0131y\u0131 s\u00f6ylersiniz. Sizce 1073 nas\u0131l bir say\u0131? B\u00fcy\u00fck bir ihtimalle sizin tahmininizden k\u00fc\u00e7\u00fck. Ama t\u00fcm k\u00e2inattaki gezegenlerin, y\u0131ld\u0131zlar\u0131n, asteroidlerin … atom say\u0131s\u0131 i\u015fte bu kadar. (Ara\u015ft\u0131rmalar sonucundaki tahmini say\u0131). <\/p>\n
K\u00e2inat\u0131n sonu neresi? Herhalde k\u00e2inat da bir yerde bulunuyor. Ayr\u0131ca geni\u015fledi\u011fi (\u015fi\u015fen bir balon gibi) ilm\u00ee bir ger\u00e7ek. Nerede, neyin i\u00e7inde, nereleri kaplayarak geni\u015fliyor? Bundan sonras\u0131 ancak tahmin edilebilir. \u015eimdilik bunlar s\u0131r.<\/p>\n
\u015eimdi \u00a5’un ne kadar b\u00fcy\u00fck oldu\u011fu daha iyi anla\u015f\u0131l\u0131yor (veya anla\u015f\u0131lam\u0131yor) de\u011fil mi?<\/p>\n
\u0130lgin\u00e7 Say\u0131lar(6):<\/p>\n
(0 x 9) + 8 = 8
\n (9 x 9) + 7 = 88
\n (98 x 9) + 6 = 888
\n (987 x 9) + 5 = 8888
\n (9876 x 9) + 4 = 88888
\n (98765 x 9) + 3 = 888888
\n (987654 x 9) + 2 = 8888888
\n (9876543 x 9) + 1 = 88888888
\n (98765432 x 9) + 0 = 888888888
\n (987654321 x 9) – 1 = 8888888888<\/p>\n","protected":false},"excerpt":{"rendered":"
p(pi) Say\u0131s\u0131: K\u0131saca bir dairenin \u00e7evresinin \u00e7ap\u0131na oran\u0131, p say\u0131s\u0131n\u0131 verir. \u0130nsano\u011flu, asl\u0131nda \u00e7ok \u00f6nemli vazifeleri olan bu say\u0131 \u00fczerinde \u00e7ok d\u00fc\u015f\u00fcnm\u00fc\u015ft\u00fcr. Y\u0131llarca tam olarak bir de\u011fer bulamamakla beraber, ger\u00e7ek de\u011ferine en yak\u0131n sonu\u00e7lar\u0131 kullanabilmek i\u00e7in \u00e7aba sarfetmi\u015flerdir. p’ nin kronolojik geli\u015fimine bakt\u0131\u011f\u0131m\u0131zda g\u00fcn\u00fcm\u00fczde dahi tam bir sonu\u00e7 bulunamam\u0131\u015ft\u0131r. \u00c7e\u015fitli form\u00fcller \u00fcretilmesine ra\u011fmen sadece her …<\/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":[7497,7501,7496,7494,7495,7210,7500,7498,7499],"class_list":["post-3248","post","type-post","status-publish","format-standard","hentry","category-matematik-odevleri","category-odevler","tag-fermatin-son-teoremi","tag-fibonacci-dizisi","tag-ilginc-sayilar","tag-matematigin-sirlari","tag-ppi-sayisi","tag-pascal-ucgeni","tag-pascal-ucgeninin-bazi-ozellikleri","tag-sihirli-kareler","tag-ucgen-sayilar"],"_links":{"self":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3248","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=3248"}],"version-history":[{"count":0,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/3248\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/media?parent=3248"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/categories?post=3248"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/tags?post=3248"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}