Program matematika (Bahasa Indonesia)
Mathematics program (English)
Sumber asli / Authoritative original
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Daftar Isi Indeks
Cari di Buku
Hasil Pencarian:
Tidak ada hasil.
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Subbab B.8 Perkalian sebagai Operasi Biner
Kita mendefinisikan operasi biner \(\times\text{,}\) yang disebut perkalian , pada himpunan bilangan asli. Jika \(m\) dan \(n\) merupakan bilangan asli, \(m\times n\) juga disebut hasil kali dari \(m\) dan \(n\text{,}\) serta kadang-kadang dinotasikan dengan \(m*n\text{,}\) atau secara lebih ringkas dengan \(mn\text{.}\) Dalam materi berikutnya, kita menggunakan konvensi terakhir ini. Misalkan \(n\in \nonnegints\text{.}\) Kita definisikan
\(n0=0\text{,}\) dan
\(n(k+1)=nk +n\text{.}\)
Perhatikan bahwa
\(10=0\) dan
\(01=00+0=0\text{.}\) Perhatikan pula bahwa
\(11=10+1=0+1=1\text{.}\) Secara lebih umum, dari (ii) dan
Lema B.19 , kita menyimpulkan bahwa jika
\(m,n\neq0\text{,}\) maka
\(mn\neq0\text{.}\)
Teorema B.21 . Hukum Distributif Kiri.
\(m(n+p)=mn + mp\) untuk setiap
\(m,n,p\in \nonnegints\text{.}\)
Bukti.
Misalkan \(m,n\in \nonnegints\text{.}\) Maka
\begin{equation*}
m(n+0)=mn = mn +0 = mn+ m0.
\end{equation*}
Sekarang andaikan \(m(n+k) = mn + mk\text{.}\) Maka
\begin{align*}
m[n+(k+1)] \amp = m[(n+k)+1]=m(n+k)+m\\
\amp =(mn+mk)+m=mn+(mk+m)= mn+m(k+1).
\end{align*}
Teorema B.22 . Hukum Distributif Kanan.
\((m+n)p=mp + np\) untuk setiap
\(m,n,p\in \nonnegints\text{.}\)
Bukti.
Misalkan \(m,n\in \nonnegints\text{.}\) Maka
\begin{equation*}
(m+n)0 =0 = 0+0 = m0 + n0.
\end{equation*}
Sekarang andaikan \((m+n)k = mk + nk\text{.}\) Maka
\begin{align*}
(m+n)(k+1)\amp =(m+n)k+(m+n)= (mk+nk) +(m+n)\\
\amp =(mk+m)+(nk+n)=m(k+1)+n(k+1).
\end{align*}
Teorema B.23 . Hukum Asosiatif Perkalian.
\(m(np) = (mn)p\) untuk setiap
\(m,n,p\in \nonnegints\text{.}\)
Bukti.
Misalkan \(m,n\in \nonnegints\text{.}\) Maka
\begin{equation*}
m(n0)= m0 = 0 = (mn)0.
\end{equation*}
Sekarang andaikan \(m(nk)=(mn)k\text{.}\) Maka
\begin{equation*}
m[n(k+1)]= m(nk + n)= m(nk) + mn =(mn)k + mn = (mn)(k+1).
\end{equation*}
Hukum komutatif memerlukan beberapa hasil pendahuluan.
Lema B.24 .
\(n0= 0n=0\) untuk setiap
\(n\in \nonnegints\text{.}\)
Bukti.
Lemma ini langsung berlaku ketika \(n=0\text{.}\) Andaikan \(k0=
0k=0\text{.}\) Maka
\begin{equation*}
(k+1)0 =0 = 0+0= 0k+0=0(k+1).
\end{equation*}
Lema B.25 .
\(n1 =1n=n\) untuk setiap
\(n\in \nonnegints\text{.}\)
Bukti.
\(01=00+0=0 =10\text{.}\) Andaikan \(k1=1k=k\text{.}\) Maka
\begin{equation*}
(k+1)1=k1+11=1k+1=1(k+1).
\end{equation*}
Teorema B.26 . Hukum Komutatif Perkalian.
\(mn=nm\) untuk setiap
\(m,n\in \nonnegints\text{.}\)
Bukti.
Misalkan \(m\in \nonnegints\text{.}\) Maka \(m0=0m\text{.}\) Andaikan \(mk=km\text{.}\) Maka
\begin{equation*}
m (k+1) = mk +m = km+m= km +1m=(k+1)m.
\end{equation*}
