1 / 7

ALJABAR BOOLE

ALJABAR BOOLE. Pertemuan ke-2 Oleh : Muh . Lukman Sifa , Ir. Aljabar Boole adalah suatu bentuk aljabar dimana variabel-variabel dan fungsi-fungsinya memiliki nilai 0 dan 1 .

glenna-jacobs
Download Presentation

ALJABAR BOOLE

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. ALJABAR BOOLE Pertemuan ke-2 Oleh : Muh. LukmanSifa, Ir.

  2. Aljabar Boole adalahsuatubentukaljabardimanavariabel-variabeldanfungsi-fungsinyamemilikinilai 0 dan1. • Keluaran (output) darisatuataubeberapabuahkombinasigerbangdapatdinyatakandalamsuatuteoremaAljabar Boole. • Aljabar Boole dapatdigunakanuntukmenyederhanakanpersamaanlogika. ApakahAljabar Boole itu ?

  3. Postulat (Dalil) Boolean • Postulat 1 A + 0 = A ; A + 1 = 1 ; A . 0 = 0 ; A . 1 = A • Postulat 2 A + B = B + A ; A . B = B . A • Postulat 3 A + (B + C) = (A + B) + C; A . (B.C) = (A.B) . C • Postulat 4 A + (B . C) = (A + B). (A +C) ; (A . B) +C = (A + C) . (B + C) • Postulat 5 A = 0 ; atau A = 1 • Postulat 6 A + A = 1; A . A = 0; Hukum-hukumdanTeorema- teoremaAljabar Boole sebagaiberikut :

  4. T1: Rumuskomutatif a. A + B = B + A b. A.B = B.A • T2: Rumusasosiatif a. (A + B) + C = A + (B +C) b. (A.B).C = A. (B.C) • T3: Rumusdistributif a. A.(B +C) = AB + AC b. A+(B . C) = (A+B) . (A+C) • T4: Rumusidentif a. A + A = A b. A.A = A • T5: Rumusnegatif a. (A’) = A’ b. (A)” = A • T6: Rumus redundant a. A + A.B = A b. A.(A + B) = A • T7: Rumuseliminasi a. A + A’.B = A+B b. A.(A’ + B) = A.B • T8: Rumus Van De Morgan a. A + B = A . B b. A.B = A + B TeoremaAljabar Boolean

  5. 1. A.B = A + B 2. A + B = A.B Cobaandabuktikankeduateoremadiatasdengancara menurunkantabelkebenaran Teorema De Morgan :

  6. A.(A.B + B) = A.AB + A.B = A.B + A.B = A.B 2. AC + ABC = AC(1 + B) = AC 3. ABC + AB’C + ABC’ = AC(B + B’) + ABC’ = AC + ABC’= A(C + BC’) = A(C + B) = A(B + C) 4. (A + BC) = A (B + C) = A.B + A.C ContohsoalpenyelesaiandenganAljabar Boole :

  7. Selesai…. Terimakasih.

More Related