✔ Panduan Pemula Berguru Aritmetika Modular

Minggu, 03 Juni 2018 Tambah Komentar Edit

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

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

Alternatif Pembahasan: show

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

Alternatif Pembahasan: show

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

Alternatif Pembahasan: show

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

Alternatif Pembahasan: show

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

Contoh 2:Alternatif Pembahasan: show

$x \equiv 1\ (mod\ 10)$ kalau dan hanya kalau $x-1=10k$ 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 bundar sebarang dan $n$ suatu bilangan bundar 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 bundar 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 bundar 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 bundar 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], kalau $a \equiv b\ (mod\ n)$ 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 sanggup dipakai untuk menemukan tanggapan atas persoalan yang diajukan. Untuk itu kita gunakan cara lain untuk menuntaskan persoalan ini.

Kita tahu bahwa suatu bilangan bundar positif dibagi oleh 5 memiliki sisa 0, 1, 2, 3, atau 4.
Penggunaan aritmetika modular akan membantu kita kalau kita sanggup menemukan bilangan bundar 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 memakai sifat [8] kita peroleh,
$(3^4)^{25} \equiv 1^{25} (mod\ 5)$, atau
$3^{100} \equiv 1 (mod\ 5)$

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

Sifat 2Alternatif Pembahasan: show

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

Kita tahu bahwa $d = FPB(c , n)$, dengan demikian, ada bilangan bundar 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 mengatakan $s \mid r(a – b)$, dan alasannya ialah $FPB (r , s) = 1$, 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 $p$ ialah bilangan prima ini menimbulkan $FPB\ (c,p)=1$

Contoh 4:

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

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

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Panduan Pemula Belajar Aritmetika Modular ✔ Panduan Pemula Belajar Aritmetika Modular

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