Рассмотрим систему комплексных чисел , в которой операция
умножения имеет вид :
z
1
⋅
z
2
=
x
1
⋅
x
2
+
y
1
⋅
y
2
+
i
(
x
1
⋅
y
2
+
x
2
⋅
y
1
)
,
{\displaystyle z_{1}\cdot z_{2}=x_{1}\cdot x_{2}+y_{1}\cdot y_{2}+i(x_{1}\cdot y_{2}+x_{2}\cdot y_{1}),}
z
1
=
x
1
+
i
⋅
y
1
,
{\displaystyle z_{1}=x_{1}+i\cdot y_{1},}
z
2
=
x
2
+
i
⋅
y
2
.
{\displaystyle z_{2}=x_{2}+i\cdot y_{2}.}
гиперболической системой комплексных
чисел.
T
{\displaystyle T}
<
+
,
⋅
>
{\displaystyle <+,\cdot >}
T
(
+
)
{\displaystyle T(+)}
∀
a
,
b
,
c
∈
T
{\displaystyle \forall a,b,c\in T}
a
⋅
(
b
+
c
)
=
a
⋅
b
+
a
⋅
c
{\displaystyle a\cdot(b+c)=a\cdot b+a\cdot c}
(
a
+
b
)
⋅
c
=
a
⋅
c
+
b
⋅
c
{\displaystyle (a + b) \cdot c = a \cdot c + b \cdot c}
Определение . Множество G с бинарной
алгеброической операцией <<
∗
{\displaystyle *}
(
a
∗
b
)
∗
c
=
a
∗
(
b
∗
c
)
{\displaystyle (a*b)*c = a*(b*c)}
∃
e
{\displaystyle \exists e}
e
∗
a
=
a
∗
e
=
a
{\displaystyle e*a=a*e=a}
∀
a
∈
G
{\displaystyle \forall a\in G}
∀
a
∈
G
{\displaystyle \forall a\in G}
∃
{\displaystyle \exists}
a
−
1
{\displaystyle a^{-1} }
a
∗
a
−
1
=
a
−
1
∗
a
=
e
{\displaystyle a*a^{-1}=a^{-1}*a=e}
a
∗
b
=
b
∗
a
{\displaystyle a*b = b*a}
∀
{\displaystyle \forall}
a
,
b
∈
G
{\displaystyle a, b \in G}
I. Покажем, что система H гиперболических комплексных чисел является Абелевой группой по сложению
z
1
=
x
1
+
i
y
1
{\displaystyle z_{1} = x_{1} + iy_{1}}
z
2
=
x
2
+
i
y
2
{\displaystyle z_{2} = x_{2} +iy_{2}}
z
3
=
x
3
+
i
y
3
{\displaystyle z_{3}=x_{3}+iy_{3}}
x
1
,
x
2
,
x
3
,
y
1
,
y
2
,
y
3
∈
R
{\displaystyle x_{1},x_{2},x_{3},y_{1},y_{2},y_{3}\in \mathbb {R} }
⇒
)
{\displaystyle \Rightarrow )}
(
x
1
+
i
y
2
)
+
(
(
x
2
+
i
y
2
)
+
(
x
3
+
i
y
3
)
)
=
(
x
1
+
i
y
1
)
+
(
x
2
+
x
3
+
i
(
y
2
+
y
3
)
)
=
x
1
+
x
2
+
x
3
+
i
(
y
1
+
y
2
+
y
3
)
{\displaystyle (x_{1}+iy_{2})+((x_{2}+iy_{2})+(x_{3}+iy_{3}))=(x_{1}+iy_{1})+(x_{2}+x_{3}+i(y_{2}+y_{3}))=x_{1}+x_{2}+x_{3}+i(y_{1}+y_{2}+y_{3})}
⇐
)
{\displaystyle \Leftarrow )}
(
(
x
1
+
i
y
1
)
+
(
x
2
+
i
y
2
)
)
+
(
x
3
+
i
y
3
)
=
(
x
1
+
x
2
+
i
(
y
1
+
y
2
)
)
+
(
x
3
+
i
y
3
)
=
x
1
+
x
2
+
x
3
+
i
(
y
1
+
y
2
+
y
3
)
{\displaystyle ((x_{1}+iy_{1})+(x_{2}+iy_{2}))+(x_{3}+iy_{3})=(x_{1}+x_{2}+i(y_{1}+y_{2}))+(x_{3}+iy_{3})=x_{1}+x_{2}+x_{3}+i(y_{1}+y_{2}+y_{3})}
∃
{\displaystyle \exists}
z
+
e
=
e
+
z
=
z
{\displaystyle z+e=e+z=z}
z
=
x
+
i
y
{\displaystyle z = x + iy}
(
x
+
i
y
)
+
0
=
0
+
(
x
+
i
y
)
=
x
+
i
y
{\displaystyle (x+iy)+0=0+(x+iy)=x+iy}
⇒
{\displaystyle \Rightarrow}
∃
{\displaystyle \exists}
∃
{\displaystyle \exists}
∀
z
∈
H
{\displaystyle \forall z\in H}
∃
z
−
1
{\displaystyle \exists z^{-1}}
z
+
z
−
1
=
z
−
1
+
z
=
e
{\displaystyle z+z^{-1}=z^{-1}+z=e}
z
=
x
+
i
y
{\displaystyle z = x + iy}
z
−
1
{\displaystyle z^{-1}}
z
−
1
=
−
x
−
i
y
{\displaystyle z^{-1}=-x-iy}
x
+
i
y
+
(
−
x
−
i
y
)
=
(
−
x
−
i
y
)
+
(
x
+
i
y
)
=
0
{\displaystyle x+iy+(-x-iy)=(-x-iy)+(x+iy)=0}
⇒
{\displaystyle \Rightarrow}
∃
{\displaystyle \exists}
+
{\displaystyle +}
∀
z
1
,
z
2
∈
H
{\displaystyle \forall z_{1},z_{2}\in H}
⇒
{\displaystyle \Rightarrow}
z
1
+
z
2
=
z
2
+
z
1
{\displaystyle z_1+z_2=z_2+z_1}
z
1
=
x
1
+
i
y
1
{\displaystyle z_{1} = x_{1} + iy_{1}}
z
2
=
x
2
+
i
y
2
{\displaystyle z_{2} = x_{2} +iy_{2}}
x
1
+
i
y
1
+
x
2
+
i
y
2
=
x
1
+
x
2
+
i
(
y
1
+
y
2
)
=
x
2
+
x
1
+
i
(
y
2
+
y
1
)
=
x
2
+
i
y
2
+
x
1
+
i
y
1
{\displaystyle x_{1}+iy_{1}+x_{2}+iy_{2}=x_{1}+x_{2}+i(y_{1}+y_{2})=x_{2}+x_{1}+i(y_{2}+y_{1})=x_{2}+iy_{2}+x_{1}+iy_{1}}
⇒
{\displaystyle \Rightarrow}
⇒
{\displaystyle \Rightarrow}
II.Покажем что в H выполняются свойства :
(
z
1
+
z
2
)
⋅
z
3
=
z
1
⋅
z
3
+
z
2
⋅
z
3
{\displaystyle (z_{1}+z_{2})\cdot z_{3}=z_{1}\cdot z_{3}+z_{2}\cdot z_{3}}
z
1
⋅
(
z
2
+
z
3
)
=
z
1
⋅
z
2
+
z
1
⋅
z
3
{\displaystyle z_{1}\cdot (z_{2}+z_{3})=z_{1}\cdot z_{2}+z_{1}\cdot z_{3}}
∀
z
1
,
z
2
,
z
3
∈
H
{\displaystyle \forall z_{1},z_{2},z_{3}\in H}
z
1
=
x
1
+
i
y
1
{\displaystyle z_{1} = x_{1} + iy_{1}}
z
2
=
x
2
+
i
y
2
{\displaystyle z_{2} = x_{2} +iy_{2}}
z
3
=
x
3
+
i
y
3
{\displaystyle z_{3}=x_{3}+iy_{3}}
(
z
1
+
z
2
)
⋅
z
3
=
(
x
1
+
i
y
1
+
x
2
+
i
y
2
)
(
x
3
+
i
y
3
)
=
(
x
1
+
x
2
+
i
(
y
1
+
y
2
)
)
(
x
3
+
i
y
3
)
=
(
(
x
1
+
x
2
)
x
3
+
(
y
1
+
y
2
)
y
3
+
i
(
(
x
1
+
x
2
)
y
3
+
x
3
(
y
1
+
y
2
)
)
=
x
1
x
3
+
x
2
x
3
+
y
1
y
3
+
y
2
y
3
+
i
(
(
x
1
+
x
2
)
y
3
+
x
3
(
y
1
+
y
2
)
)
)
{\displaystyle (z_{1}+z_{2})\cdot z_{3}=(x_{1}+iy_{1}+x_{2}+iy_{2})(x_{3}+iy_{3})=(x_{1}+x_{2}+i(y_{1}+y_{2}))(x_{3}+iy_{3})=((x_{1}+x_{2})x_{3}+(y_{1}+y_{2})y_{3}+i((x_{1}+x_{2})y_{3}+x_{3}(y_{1}+y_{2}))=x_{1}x_{3}+x_{2}x_{3}+y_{1}y_{3}+y_{2}y_{3}+i((x_{1}+x_{2})y_{3}+x_{3}(y_{1}+y_{2})))}
z
1
⋅
z
3
+
z
2
⋅
z
3
=
x
1
x
3
+
y
1
y
3
+
i
(
x
1
y
3
+
x
3
y
1
)
+
x
2
x
3
+
y
2
y
3
+
i
(
x
2
y
3
+
x
3
y
2
)
=
x
1
x
3
+
x
2
x
3
+
y
1
y
3
+
y
2
y
3
+
i
(
x
1
y
3
+
x
3
y
1
+
x
2
y
3
+
x
3
y
2
)
{\displaystyle z_{1}\cdot z_{3}+z_{2}\cdot z_{3}=x_{1}x_{3}+y_{1}y_{3}+i(x_{1}y_{3}+x_{3}y_{1})+x_{2}x_{3}+y_{2}y_{3}+i(x_{2}y_{3}+x_{3}y_{2})=x_{1}x_{3}+x_{2}x_{3}+y_{1}y_{3}+y_{2}y_{3}+i(x_{1}y_{3}+x_{3}y_{1}+x_{2}y_{3}+x_{3}y_{2})}
⇒
{\displaystyle \Rightarrow}
z
1
⋅
(
z
2
+
z
3
)
=
(
x
1
+
i
y
1
)
(
(
x
2
+
i
y
2
)
+
(
x
3
+
i
y
3
)
)
=
(
x
1
+
i
y
1
)
(
x
2
+
x
3
+
i
(
y
2
+
y
3
)
)
=
x
1
⋅
(
x
2
+
x
3
)
+
y
1
⋅
(
y
2
+
y
3
)
+
i
(
x
1
⋅
(
y
2
+
y
3
)
+
(
x
2
+
x
3
)
⋅
y
1
)
=
x
1
x
2
+
x
1
x
3
+
y
1
y
2
+
y
1
y
3
+
i
(
x
1
y
2
+
x
1
y
3
+
x
2
y
1
+
x
3
y
1
)
{\displaystyle z_{1}\cdot (z_{2}+z_{3})=(x_{1}+iy_{1})((x_{2}+iy_{2})+(x_{3}+iy_{3}))=(x_{1}+iy_{1})(x_{2}+x_{3}+i(y_{2}+y_{3}))=x_{1}\cdot (x_{2}+x_{3})+y_{1}\cdot (y_{2}+y_{3})+i(x_{1}\cdot (y_{2}+y_{3})+(x_{2}+x_{3})\cdot y_{1})=x_{1}x_{2}+x_{1}x_{3}+y_{1}y_{2}+y_{1}y_{3}+i(x_{1}y_{2}+x_{1}y_{3}+x_{2}y_{1}+x_{3}y_{1})}
z
1
⋅
z
2
+
z
1
⋅
z
3
=
(
x
1
+
i
y
1
)
(
x
2
+
i
y
2
)
+
(
x
1
+
i
y
1
)
(
x
3
+
i
y
3
)
=
x
1
x
2
+
y
1
y
2
+
i
(
x
1
y
2
+
x
2
y
1
)
+
x
1
x
3
+
y
1
y
3
+
i
(
x
1
y
3
+
x
3
y
1
)
=
x
1
x
2
+
x
1
x
3
+
y
1
y
2
+
y
1
y
3
+
i
(
x
1
y
2
+
x
2
y
1
+
x
1
y
3
+
x
3
y
1
)
{\displaystyle z_{1}\cdot z_{2}+z_{1}\cdot z_{3}=(x_{1}+iy_{1})(x_{2}+iy_{2})+(x_{1}+iy_{1})(x_{3}+iy_{3})=x_{1}x_{2}+y_{1}y_{2}+i(x_{1}y_{2}+x_{2}y_{1})+x_{1}x_{3}+y_{1}y_{3}+i(x_{1}y_{3}+x_{3}y_{1})=x_{1}x_{2}+x_{1}x_{3}+y_{1}y_{2}+y_{1}y_{3}+i(x_{1}y_{2}+x_{2}y_{1}+x_{1}y_{3}+x_{3}y_{1})}
⇒
{\displaystyle \Rightarrow}
III.Покажем ,что H - коммутотивное кольцо , т.е
∀
z
1
,
z
2
∈
H
{\displaystyle \forall z_{1},z_{2}\in H}
⇒
{\displaystyle \Rightarrow}
z
1
⋅
z
2
=
z
2
⋅
z
1
{\displaystyle z_{1}\cdot z_{2}=z_{2}\cdot z_{1}}
(
x
1
+
i
y
1
)
⋅
(
x
2
+
i
y
2
)
=
x
1
x
2
+
y
1
y
2
+
i
(
x
1
y
2
+
x
2
y
1
)
{\displaystyle (x_{1}+iy_{1})\cdot (x_{2}+iy_{2})=x_{1}x_{2}+y_{1}y_{2}+i(x_{1}y_{2}+x_{2}y_{1})}
(
x
2
+
i
y
2
)
⋅
(
x
1
+
i
y
1
)
=
x
2
x
1
+
y
2
y
1
+
i
(
x
2
y
1
+
x
1
y
2
)
{\displaystyle (x_{2}+iy_{2})\cdot (x_{1}+iy_{1})=x_{2}x_{1}+y_{2}y_{1}+i(x_{2}y_{1}+x_{1}y_{2})}
⇒
{\displaystyle \Rightarrow}
⇒
{\displaystyle \Rightarrow}
∃
{\displaystyle \exists}
Определение. Пара ненулевых элементов кольца,
произведение которых равно 0 называется делителями нуля.
z
1
=
x
1
+
i
y
1
{\displaystyle z_{1} = x_{1} + iy_{1}}
z
2
=
x
2
+
i
y
2
{\displaystyle z_{2} = x_{2} +iy_{2}}
{
x
1
≠
0
u
y
1
≠
0
x
2
≠
0
u
y
2
≠
0
{\displaystyle \left\{{\begin{matrix}x_{1}\neq 0\qquad u\qquad y_{1}\neq 0\\x_{2}\neq 0\qquad u\qquad y_{2}\neq 0\end{matrix}}\right.}
z
1
⋅
z
2
=
x
1
x
2
+
y
1
y
2
+
i
(
x
1
y
2
+
x
2
y
1
)
=
0
{\displaystyle z_{1}\cdot z_{2}=x_{1}x_{2}+y_{1}y_{2}+i(x_{1}y_{2}+x_{2}y_{1})=0}
{
x
1
⋅
x
2
+
y
1
⋅
y
2
=
0
x
1
⋅
y
2
+
x
2
⋅
y
1
=
0
{\displaystyle \left\{{\begin{matrix}x_{1}\cdot x_{2}+y_{1}\cdot y_{2}=0\\x_{1}\cdot y_{2}+x_{2}\cdot y_{1}=0\end{matrix}}\right.}
⇒
{\displaystyle \Rightarrow}
{
x
1
⋅
x
2
=
−
y
1
⋅
y
2
x
1
⋅
y
2
=
−
x
2
⋅
y
1
{\displaystyle \left\{{\begin{matrix}x_{1}\cdot x_{2}=-y_{1}\cdot y_{2}\\x_{1}\cdot y_{2}=-x_{2}\cdot y_{1}\end{matrix}}\right.}
x
1
{\displaystyle x_1}
x
1
=
−
y
1
⋅
y
2
x
2
{\displaystyle x_{1}=-{\frac {y_{1}\cdot y_{2}}{x_{2}}}}
−
y
1
⋅
y
2
2
x
2
=
−
x
2
⋅
y
1
{\displaystyle -{\frac {y_{1}\cdot y_{2}^{2}}{x_{2}}}=-x_{2}\cdot y_{1}}
⇒
{\displaystyle \Rightarrow}
−
y
2
2
x
2
=
−
x
2
{\displaystyle -{\frac {y_{2}^{2}}{x_{2}}}=-x_{2}}
⇒
{\displaystyle \Rightarrow}
y
2
2
=
x
2
2
{\displaystyle y_{2}^{2}=x_{2}^{2}}
⇒
{\displaystyle \Rightarrow}
y
2
=
±
x
2
{\displaystyle y_{2}=\pm x_{2}}
z
2
{\displaystyle z_{2}}
y
=
x
{\displaystyle y = x}
y
=
−
x
{\displaystyle y=-x}
(
0
;
0
)
{\displaystyle (0;0)}
y
2
{\displaystyle y_2}
y
2
=
−
x
2
⋅
y
1
x
1
{\displaystyle y_{2}=-{\frac {x_{2}\cdot y_{1}}{x_{1}}}}
x
1
⋅
x
2
=
y
1
⋅
x
2
⋅
y
1
x
1
{\displaystyle x_{1}\cdot x_{2}=y_{1}\cdot {\frac {x_{2}\cdot y_{1}}{x_{1}}}}
⇒
{\displaystyle \Rightarrow}
x
1
=
y
1
2
x
1
{\displaystyle x_{1}={\frac {y_{1}^{2}}{x_{1}}}}
⇒
{\displaystyle \Rightarrow}
y
1
2
=
x
1
2
{\displaystyle y_{1}^{2}=x_{1}^{2}}
⇒
{\displaystyle \Rightarrow}
y
1
=
±
x
1
{\displaystyle y_{1}=\pm x_{1}}
z
1
{\displaystyle z_1}
y
=
x
{\displaystyle y = x}
y
=
−
x
{\displaystyle y=-x}
(
0
;
0
)
{\displaystyle (0;0)}
y
=
x
{\displaystyle y = x}
y
=
x
{\displaystyle y = x}
y
=
−
x
{\displaystyle y=-x}
Доказательство.
x
1
=
y
1
{\displaystyle x_{1}=y_{1}}
x
2
=
y
2
{\displaystyle x_{2}=y_{2}}
⇒
{\displaystyle \Rightarrow}
{
y
1
⋅
y
2
=
−
y
1
⋅
y
2
y
1
⋅
y
2
=
−
y
1
⋅
y
2
{\displaystyle \left\{{\begin{matrix}y_{1}\cdot y_{2}=-y_{1}\cdot y_{2}\\y_{1}\cdot y_{2}=-y_{1}\cdot y_{2}\end{matrix}}\right.}
⇒
{\displaystyle \Rightarrow}
x
1
=
y
1
{\displaystyle x_{1}=y_{1}}
x
2
=
−
y
2
{\displaystyle x_{2}=-y_{2}}
⇒
{\displaystyle \Rightarrow}
{
−
y
1
⋅
y
2
=
−
y
1
⋅
y
2
y
1
⋅
y
2
=
y
2
⋅
y
1
{\displaystyle \left\{{\begin{matrix}-y_{1}\cdot y_{2}=-y_{1}\cdot y_{2}\\y_{1}\cdot y_{2}=y_{2}\cdot y_{1}\end{matrix}}\right.}
⇒
{\displaystyle \Rightarrow}
∃
{\displaystyle \exists}
y
=
x
{\displaystyle y = x}
y
=
−
x
{\displaystyle y=-x}
(
0
;
0
)
{\displaystyle (0;0)}
