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PERTEMUAN 1

PERTEMUAN 1. BAB I. SISTEM BILANGAN. 1.1 SISTEM BILANGAN RIL. 1.1.1 BILANGAN RIL. RIL ( R ). IRRASIONAL ( I ). RASIONAL ( Q ). Bulat ( J ). Pecahan. Negatif. Desimal Berulang. Cacah ( W ). Desimal Terbatas. Asli ( N ). Nol. Himpunan Bilangan cacah ( W ) W = { 0, 1, 2, 3, … }.

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PERTEMUAN 1

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  1. PERTEMUAN 1

  2. BAB I SISTEM BILANGAN

  3. 1.1 SISTEM BILANGAN RIL 1.1.1 BILANGAN RIL RIL (R) IRRASIONAL (I) RASIONAL (Q) Bulat (J) Pecahan Negatif DesimalBerulang Cacah (W ) DesimalTerbatas Asli (N) Nol

  4. HimpunanBilangancacah (W) W = { 0, 1, 2, 3, … } HimpunanBilanganAsli (N) N = { 1, 2, 3, … } HimpunanBilanganBulat (J) J = { … , -3, -2, -1, 0, 1, 2, 3, … } Himpunanbilanganrasional (Q) Himpunanbilanganrasionaladalahhimpunanbilangan yang mempunyaibentukp/qataubilangan yang dapatditulisdalam bentukp/q, dimanapdanqadalahanggotabilanganbulat danq 0 P q | pdanq J, q  0 Q =

  5. Contoh 1.1 Buktikanbahwabilangan-bilangan 3, (4,7) dan 2,5858…) adalahbilangan-bilanganrasional! Bukti: • a) Bilangan 3 dapatditulisdalambentukp/qyaitu 3/1 atau 6/2 danseterusnya. b) Bilangan 4,7 dapatditulisdalambentuk 47/10 • c) Bilangan2,5858… dapatditulisdalambentukp/q • dengancara, x = 2,5858… 100 x = 258,5858… 99 x = 256 x = 256/99

  6. Contoh 1.2 Buktikanbahwabilangan 2,342121212121… adalah bilanganrasional! Penyelesaian x = 2,342121212121 100 x = 234,2121212121 10000 x = 23421,21212121 10000 x = 23421,2121212121 … 100 x = 234,2121212121 … 9900 x = 23187 x = 23187/9900 Jadibilangan 2,3421212121212121 … = 23187/9900

  7. Latihan Tulisbilanganberikutdalambentukpecahanp/q a) 3,41222… b) 17,282828…

  8. 1.1.2 GARIS BILANGAN RIL Garisbilanganriladalahtempatkedudukantitik-titik. Setiaptitikmenunjukkansatubilanganriltertentu yang tersusunsecaraterurut.       -3 1,5 2,5 0 -1 -2 1.1.3 HUKUM-HUKUM BILANGAN RIL Jikaadanbadalahduabilanganrilmakaberlaku: (i) a + badalahbilanganril (ii) a . badalahbilanganril (iii) a + b = b + aHukumKomutatifPenjumlahan (iv) a . b = b . aHukumkomutatifPerkalian

  9. Jikaa, b, danc adalahtigabilanganrilmakaberlaku: (v) (a + b) + c = a + (b + c) adalahbilanganril (vi) (ab)c = a (bc) adalahbilanganril (vii) a(b + c) = ab + acHukumKomutatifPenjumlan (viii) a + 0 = 0 + aHukumPenjumlahanNol (ix) a . 1 = 1 . a = aHukumPerkalianSatu (x) a.0 = 0.a = 0 HukumPerkalianNol (xi) a + (-a) = -a + aHukumInversPenjumlahan (xii) a (1/a) = 1 , a 1 HukumInversPerkalian

  10. 1.2 BILANGAN KOMPLEKS Bentukumumz = a + ib adanbadalahbilanganril amerupakanbagianrildaribilangankompleks, ditulisRe(z) bmerupakanbagianimajinerdaribilangan kompleks , ditulisIm(z) imerupakanbilanganimajiner = Dari keterangandiatasdidapat

  11. 1.2.1 SIFAT-SIFAT BILANGAN KOMPLEKSS Misalz1 = x1 + iy1danz2= x2+ iy2,makaberlaku: a) z1 = z2 x1 = x2 dany1 = y2sifatkesamaan b) z1 + z2 = (x1 + x2) +i(y1 + y2) sifatpenjumlahan c) z1 - z2 = (x1 + x2) +i(y1 - y2) sifatpengurangan d) z1 . z2 = (x1x2 - y1y2) +i(x1 y2 – x2y1) sifatperkalian 1.2.2 KONJUGAT Jikaz = x+ iy, makakonjugatdariz (ditulis z ) adalahx – iy Jikaz = x - iy, Maka

  12. 1.2.3 PERKALIAN BILANGAN KOMPLEKS DENGAN KONJUGATNYA (x + iy) (x – iy) = z.z z.z – i2y2 – ixy + ixy x2 = (–1)y2 – = x2 y2 x2 + =

  13. 1.2.4 PEMBAGIAN DUA BUAH BILANGAN KOMPLEKS = = x1x2 – ix1y2 + ix2y1 – i2x1y2 = x12– i2y22 x1x2 – ix1y2 + ix2y1 + x1y2 = x12 + y22 x1x2 + x1y2 + ix2y1 – ix1y2 = x12 + y22 z1 z1 x1x2 + x1y2 + i( x2y1 – x1y2) z2 z2 z2 = x12 + y22 z2 z1 x1x2 + x1y2 x2y1 – x1y2 z2 i = + x12 + y22 x12 + y22

  14. Contoh 1.2 Diketahui z1 = – 5 + 7i z2 = 3 – 2i Tentukan a) z1 + z2 b) z1 – z2 c) z1 . z2 d) z1/z2 e) z1. f ) z2. z2 z1 Penyelesaian –2 + 5i (–5 + 3) (7i –2i) (3 – 2i) + (– 5 + 7i) a) z1 + z2 = + = = –8 9i (–5 – 3) (3 – 2i) 7i– (–2i) – + (– 5 + 7i) = + = b) z1 – z2 = – 14i 2 c) z1 . z –15 + 10i + 21i (– 5+7i ) (3 – 2i ) = = – 14(–1) –15 –1 –15 + 31i + 31i + 31i +14 = = =

  15. d) = –29 11 z1 . z2 + i 13 13 = (–5 + 7i)(3 + 2i) =– 5(3) – 5(2i) + 7i(3) +7i(2i) = –15 – 10i + 21i – 14 –10i + 21i + 14i2 –15 = e) z1 . z2 –29 + 11i = f ) = (-5 - 7i)(3 - 2i) = – 15 + 10i – 21i + 14i2 = –29 – 11i = –15 – 11i –14

  16. Latihan 1. Selesaikan a) (3 + 5i) + (4 – 7i) b) (–2 – 4i) – (– 5 –8i) c) (2 – i)(5 +8i) d) (3/4 – 2/5 i) – (2/3 + 5/6 i) e) (3/7 – 3i)(2/3 + 3/8i) a) z1. b ) z2. 2. Jikaz1 = –7 – 2idanz2= 5 + 4i Tentukan z2 z1

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