Dördey

denklikler

${\displaystyle i^{2}=j^{2}=k^{2}=ijk=-1}$,

burada i, j, ve k H nın taban ögeleridir,i, j, ve k nın tüm olası çarpanlarını belirtir .

örneğin −1 = ijk nın sağ çarpanlarının her ikisi de k ile verilir

${\displaystyle {\begin{aligned}-k&=ijkk=ij(k^{2})=ij(-1),\\k&=ij.\end{aligned}}}$

Diğer tüm olası çarpanlar benzer yöntemlerle belirlenebilir

${\displaystyle {\begin{alignedat}{2}ij&=k,&\qquad ji&=-k,\\jk&=i,&kj&=-i,\\ki&=j,&ik&=-j,\end{alignedat}}}$

olan satır çarpanı sol faktörü teşkil eder ve bir tablo olarak ifade edilebilir,bu yazının üstünde gösterildiği gibi kendilerinin sütunlari sağ faktörü teşkil eder.

iki a₁ + b₁i + c₁j + d₁k elementler için ve a₂ + b₂i + c₂j + d₂k, burada çarpıma, Hamilton çarpımı (a₁ + b₁i + c₁j + d₁k) (a₂ + b₂i + c₂j + d₂k) denir, taban ögeler ve dağılımsal kanunun çarpımları ile tanımlanıyor.Dağılım kanunu onu çarpımın açılımı için olası yapar böylece bu taban ögelerin çarpımlarının bir toplamıdır. Bu aşağıdaki bağıntılarla veriliyor:

${\displaystyle a_{1}a_{2}+a_{1}b_{2}i+a_{1}c_{2}j+a_{1}d_{2}k}$

${\displaystyle {}+b_{1}a_{2}i+b_{1}b_{2}i^{2}+b_{1}c_{2}ij+b_{1}d_{2}ik}$

${\displaystyle {}+c_{1}a_{2}j+c_{1}b_{2}ji+c_{1}c_{2}j^{2}+c_{1}d_{2}jk}$

${\displaystyle {}+d_{1}a_{2}k+d_{1}b_{2}ki+d_{1}c_{2}kj+d_{1}d_{2}k^{2}.}$

Şimdi taban elemanları kullanılarak elde etmek için yukarıda verilen kuralları çoğaltılabilir:[1]

${\displaystyle a_{1}a_{2}-b_{1}b_{2}-c_{1}c_{2}-d_{1}d_{2}}$

${\displaystyle {}+(a_{1}b_{2}+b_{1}a_{2}+c_{1}d_{2}-d_{1}c_{2})i}$

${\displaystyle {}+(a_{1}c_{2}-b_{1}d_{2}+c_{1}a_{2}+d_{1}b_{2})j}$

${\displaystyle {}+(a_{1}d_{2}+b_{1}c_{2}-c_{1}b_{2}+d_{1}a_{2})k.}$

Hnın 1, i, j, k tabanları kullanılıyor dört katının bir kümesi olarak H yazmak için mümkün kılar:

${\displaystyle \mathbf {H} =\{(a,b,c,d)\mid a,b,c,d\in \mathbf {R} \}.}$

ise taban ögeleri:

${\displaystyle {\begin{aligned}1&=(1,0,0,0),\\i&=(0,1,0,0),\\j&=(0,0,1,0),\\k&=(0,0,0,1),\end{aligned}}}$

ve toplam ve çarpım için formüller:

${\displaystyle (a_{1},\ b_{1},\ c_{1},\ d_{1})+(a_{2},\ b_{2},\ c_{2},\ d_{2})=(a_{1}+a_{2},\ b_{1}+b_{2},\ c_{1}+c_{2},\ d_{1}+d_{2}).}$

${\displaystyle {\begin{aligned}(a_{1},\ b_{1},\ c_{1},\ d_{1})&(a_{2},\ b_{2},\ c_{2},\ d_{2})=\\&=(a_{1}a_{2}-b_{1}b_{2}-c_{1}c_{2}-d_{1}d_{2},\\&{}\qquad a_{1}b_{2}+b_{1}a_{2}+c_{1}d_{2}-d_{1}c_{2},\\&{}\qquad a_{1}c_{2}-b_{1}d_{2}+c_{1}a_{2}+d_{1}b_{2},\\&{}\qquad a_{1}d_{2}+b_{1}c_{2}-c_{1}b_{2}+d_{1}a_{2}).\end{aligned}}}$

bir dördey veriliyor,

q = a + bi + cj + dk = a + v,

üstel

${\displaystyle \exp(q)=\sum _{n=0}^{\infty }{\frac {q^{n}}{n!}}=e^{a}\left(\cos \|\mathbf {v} \|+{\frac {\mathbf {v} }{\|\mathbf {v} \|}}\sin \|\mathbf {v} \|\right)}$

olarak hesaplanıyor ve

${\displaystyle \ln(q)=\ln \|q\|+{\frac {\mathbf {v} }{\|\mathbf {v} \|}}\arccos {\frac {a}{\|q\|}}}$.[2]

bir dördeyin kutupsal çözülümünü aşağıdaki gibi yazabiliriz

${\displaystyle q=\|q\|e^{{\hat {n}}\theta }=\|q\|\left(\cos(\theta )+{\hat {n}}\sin(\theta )\right),}$

burada açı θ ve birim vektör ${\displaystyle {\hat {n}}}$ ile tanımlanıyor:

${\displaystyle a=\|q\|\cos(\theta )}$

ve

${\displaystyle \mathbf {v} ={\hat {n}}\|\mathbf {v} \|={\hat {n}}\|q\|\sin(\theta ).}$

Herhangi birim dördey ${\displaystyle e^{{\hat {n}}\theta }}$.olan kutupsal biçim içinde ifade edilebilir

bir keyfi (gerçek) üstel ${\displaystyle \alpha }$ için bir yükselen dördeyin kuvveti ile veriliyor:

${\displaystyle q^{\alpha }=\|q\|^{\alpha }e^{{\hat {n}}\alpha \theta }=\|q\|^{\alpha }\left(\cos(\alpha \theta )+{\hat {n}}\sin(\alpha \theta )\right).}$

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Şablon:Number Systems