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Panduan Pemula Belajar Aritmetika Modular

Oleh Indrawati Februari 12, 2022 Posting Komentar

belajar matematika di Hari PaskahDefenisi:
Misalkan $n$ adalah suatu bilangan bulat positif, $a$ dan $b$ adalah suatu bilangan bulat.
$a$ dikatakan kongruen $b$ modulo $n$, ditulis $a \equiv b\ (mod\ n)$ jika dan hanya jika $(a-b)$ adalah kelipatan $n$.Contoh 1:Alternatif Pembahasan: show

$3 \equiv 24\ (mod\ 7)$
Benar, karena $3 – 24 = -21$ kelipatan dari $7$

Alternatif Pembahasan: show

$-31 \equiv 11\ (mod\ 7)$
Benar, karena $–31 – 11 = -42$ kelipatan dari $7$

Alternatif Pembahasan: show

$-15 \equiv -64\ (mod\ 7)$
Benar, karena $–15 + 64 = 49$ kelipatan dari $7$

Alternatif Pembahasan: show

$13 \equiv -1\ (mod\ 7)$
Benar, karena $13 + 1 = 14$ kelipatan dari $7$

Alternatif Pembahasan: show

$23 \equiv 3\ (mod\ 7)$
Salah, karena $23 – 3 = 20$ bukan kelipatan dari $7$

Contoh 2:Alternatif Pembahasan: show

$x \equiv 1\ (mod\ 10)$ jika dan hanya jika $x-1=10k$ Untuk setiap $k$ bilangan bulat.
Jika $k = 0, 1, 2, 3, …$ maka berturut-turut $x = 1, 11, 21, 31, ...$

Begitu pula $k = -0, -1, -2, -3,...$ maka berturut-turut $x = -9, -200, 21, 31,...$
Begitu pula $k = -1, -2, -3,...$ maka berturut-turut $x = -9, -19, -29,...$

Dua barisan tersebut digabungkan sehingga himpunan penyelesaian $x \equiv 1\ (mod\ 10)$ adalah
$..., -29, -19, -9, 1, 11, 21, 31,...$

Sifat 1Alternatif Pembahasan: show

Untuk $a$ bilangan bulat sebarang dan $n$ suatu bilangan bulat positif berlaku $a – a = 0 \cdot n $.
Dengan demikian $a \equiv a\ (mod\ n)$.

Alternatif Pembahasan: show

$a \equiv b\ (mod\ n)$
Ada $k$ suatu bilangan bulat.
Akibatnya
$b-a=-(a-b)$
$b-a=-(kn)$
$b-a=(-k)n$
Karena $-k$ juga suatu bilangan bulat, $b \equiv a\ (mod\ n)$.

Alternatif Pembahasan: show

$a \equiv b\ (mod\ n)$ dan $b \equiv c\ (mod\ n)$
Ada $h$ dan $k$ suatu bilangan bulat sehingga $a-b=hn$ dan $b-c=kn$.
Akibatnya:
$a-c=(a-b)+(b-c)$
$a-c=hn+kn$
$a-c=n(h+k)$
Karena $h+k$ juga bilangan, $a \equiv c\ (mod\ n)$.

Alternatif Pembahasan: show

$a \equiv b\ (mod\ n)$ dan $c \equiv d\ (mod\ n)$
Ada $h$ dan $k$ bilangan bulat sehingga $a-b=hn$ dan $b-c=kn$.
$(a+c)-(b+d)=(a-b)+(c-d)$
$(a+c)-(b+d)=hn+kn$
$(a+c)-(b+d)=n(h+k)$
Karena $h+k$ juga bilangan bulat, $a+c \equiv b+d\ (mod\ n)$.

Alternatif Pembahasan: show

$a \equiv b\ (mod\ n)$ dan $c \equiv d\ (mod\ n)$
Pandang
$ac=(b+hn)(d+kn)
$ac=bd+(bk+dh+hkn)n$
Karena $(bk+dh+hkn)$ bilangan bulat,
Ada $h$ dan $k$ bilangan bulat, $ac \equiv bd\ (mod\ n)$.

Alternatif Pembahasan: show

$a \equiv b\ (mod\ n)$
Ada $h$ bilangan sehingga $a-b=hn$
Karena $(a+c)-(b+c)=a-b=hn$
dengan demikian $a+c \equiv b+c\ (mod\ n)$.

Alternatif Pembahasan: show

$a \equiv b\ (mod\ n)$
Ada $h$ bilangan bulat sehingga $a – b = hn $.
$ac-bc=(a-b)c=hnc=(hc)n$
Karena $hc$ bilangan bulat,
dengan demikian $ac \equiv bc\ (mod\ n)$.

Alternatif Pembahasan: show

Untuk bukti ini ita gunakan Induksi Matematika.
Untuk $k=1$, berlaku $a \equiv a\ (mod\ n)$.
Kita asumsikan $a^k \equiv b^k\ (mod\ n)$ berlaku,
Akan ditunjukkan $a^{k+1} \equiv b^{k+1}\ (mod\ n)$ juga berlaku.

Dari sifat [5], jika $a \equiv b\ (mod\ n)$ dan $c \equiv d\ (mod\ n)$ maka $ac \equiv bd\ (mod\ n)$
kita ganti $c$ oleh $a^k$ dan $d$ oleh $b^k$ diperoleh;
$aa^k \equiv bb^k (mod\ n)$ atau $a^{k+1} \equiv b^{k+1} (mod\ n)$

Contoh 3:Alternatif Pembahasan: show

Tampaknya kalkulator tidak dapat digunakan untuk menemukan jawaban atas masalah yang diajukan. Untuk itu kita gunakan cara lain untuk menyelesaikan masalah ini.

Kita tahu bahwa suatu bilangan bulat positif dibagi oleh 5 mempunyai sisa 0, 1, 2, 3, atau 4.
Penggunaan aritmetika modular akan membantu kita jika kita dapat menemukan bilangan bulat terkecil yang ekuivalen dengan pangkat dari 3, dan penggunaan sifat [7] dan [8] untuk membangun $3^{100}$ dan menemukan ekuivalensi mod 5.

Kita tahu bahwa $3^{2} \equiv 4\ (mod\ 5)$
Dengan demikian,
$3^3 \equiv 3 \cdot 4 \equiv 2\ (mod\ 5)$
$3^4 \equiv 3 \cdot 2 \equiv 1\ (mod\ 5)$

Dengan menggunakan sifat [8] kita peroleh,
$(3^4)^{25} \equiv 1^{25} (mod\ 5)$, atau
$3^{100} \equiv 1 (mod\ 5)$

Jadi: $3^{100}$ dibagi oleh $5$ mempunyai sisa $1$.

Sifat 2Alternatif Pembahasan: show

Karena $ca \equiv cb\ (mod\ n)$,
$c(a – b) = ca – cb = kn$ untuk suatu bilangan bulat $k$.

Kita tahu bahwa $d = FPB(c , n)$, dengan demikian, ada bilangan bulat saling prima [relative prime] $r$ dan $s$ yang memenuhi $c = dr$, $n = ds$.

Jika hasil ini kita substitusikan ke persamaan $c(a – b) = ca – cb = kn$ maka kita peroleh $r(a – b) = ks$.

Hasil ini menunjukkan $s \mid r(a – b)$, dan karena $FPB (r , s) = 1$, diperoleh $s (a – b)$. Dengan kata lain,
$ca \equiv cb\ (mod\ n)$ maka $a \equiv b\ (mod\ n/d )$

Sifat 3Sifat 3sifat 2Sifat 4Alternatif Pembahasan: show

Kondisi $p \nmid c$ dan $p$ adalah bilangan prima ini mengakibatkan $FPB\ (c,p)=1$

Contoh 4:

Perhatikan $33 \equiv 15\ (mod\ 9)$, atau $3 \cdot 11 \equiv 3 \cdot 5\ (mod\ 9)$.
Karena $FPB(3,9)=1$ mengakibatkan $11 \equiv 5\ (mod\ 9)$

Perhatikan $-35 \equiv 45\ (mod\ 8)$, atau $5 \cdot (-7) \equiv 5 \cdot 9\ (mod\ 8)$.
Karena $5$ dan $8$ bilangan bulat saling prima mengakibatkan $-7 \equiv 9\ (mod\ 8)$

silahkan download dimari

Panduan Pemula Belajar Aritmetika Modular Panduan Pemula Belajar Aritmetika Modular

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