N
{\displaystyle \mathbb {N} }
natūrinių skaičių aibė :
1
,
2
,
3
,
…
{\displaystyle {1,2,3,\ldots }}
Z
{\displaystyle \mathbb {Z} }
…
,
−
2
,
−
1
,
0
,
1
,
2
,
…
{\displaystyle {\ldots ,-2,-1,0,1,2,\ldots }}
Q
{\displaystyle \mathbb {Q} }
I
{\displaystyle \mathbb {I} }
(
π
,
e
,
…
)
{\displaystyle (\pi ,e,\ldots )}
(
2
,
cos
3
,
…
)
{\displaystyle ({\sqrt {2}},\cos {3},\ldots )}
R
{\displaystyle \mathbb {R} }
C
{\displaystyle \mathbb {C} }
a
+
i
b
{\displaystyle a+ib}
a
,
b
{\displaystyle a,b}
i
=
−
1
{\displaystyle i={\sqrt {-1}}}
∞
{\displaystyle \infty }
Aibes galima išdėstyti taip:
N
⊂
Z
⊂
Q
⊂
R
{\displaystyle \mathbb {N} \subset \mathbb {Z} \subset \mathbb {Q} \subset \mathbb {R} }
Teisinga, jog
Q
∪
I
=
R
{\displaystyle \mathbb {Q} \cup \mathbb {I} =\mathbb {R} }
Q
∩
I
=
∅
{\displaystyle \mathbb {Q} \cap \mathbb {I} =\emptyset }
Tarkime, jog
a
<
b
{\displaystyle a<b}
a
,
b
∈
R
{\displaystyle a,b\in \mathbb {R} }
[
a
,
b
]
=
{
x
|
a
≤
x
≤
b
}
{\displaystyle [a,b]=\left\{x|\,a\leq x\leq b\right\}}
uždaras intervalas arba atkarpa
(
a
,
b
)
=
{
x
|
a
<
x
<
b
}
{\displaystyle (a,b)=\left\{x|\,a<x<b\right\}}
atviras intervalas
[
a
,
b
)
=
{
x
|
a
≤
x
<
b
}
{\displaystyle [a,b)=\left\{x|\,a\leq x<b\right\}}
pusiau atviras arba pusiau uždaras intervalas
(
a
,
b
]
=
{
x
|
a
<
x
≤
b
}
{\displaystyle (a,b]=\left\{x|\,a<x\leq b\right\}}
pusiau atviras arba pusiau uždaras intervalas
(
a
,
∞
)
=
{
x
|
a
<
x
<
∞
}
{\displaystyle (a,\infty )=\left\{x|\,a<x<\infty \right\}}
atviras intervalas arba atvirasis spindulys
[
a
,
∞
)
=
{
x
|
a
≤
x
<
∞
}
{\displaystyle [a,\infty )=\left\{x|\,a\leq x<\infty \right\}}
pusiau atviras arba spindulys
(
−
∞
,
∞
)
=
R
{\displaystyle (-\infty ,\infty )=\mathbb {R} }
visa realiųjų skaičių tiesė
keisti
Bet kuriems realies skaiciams
a
,
b
,
c
{\displaystyle a,b,c}
Sudėtiės aksiomos:
a
+
b
=
b
+
a
{\displaystyle a+b=b+a}
a
+
(
b
+
c
)
=
(
a
+
b
)
+
c
{\displaystyle a+(b+c)=(a+b)+c}
a
+
0
=
a
{\displaystyle a+0=a}
a
+
(
−
a
)
=
0
{\displaystyle a+(-a)=0}
Daugybos aksiomos:
a
⋅
b
=
b
⋅
a
{\displaystyle a\cdot b=b\cdot a}
a
⋅
(
b
⋅
c
)
=
(
a
⋅
b
)
⋅
c
{\displaystyle a\cdot (b\cdot c)=(a\cdot b)\cdot c}
a
⋅
1
=
a
{\displaystyle a\cdot 1=a}
a
⋅
(
b
+
c
)
=
a
⋅
b
+
a
⋅
c
{\displaystyle a\cdot (b+c)=a\cdot b+a\cdot c}
keisti
Sakysime, jog
a
,
b
,
c
∈
R
{\displaystyle a,b,c\in \mathbb {R} }
Jei
a
>
b
{\displaystyle a>b}
b
<
a
{\displaystyle b<a}
Jei
a
>
b
{\displaystyle a>b}
b
>
c
{\displaystyle b>c}
a
>
c
{\displaystyle a>c}
Jei
a
>
b
{\displaystyle a>b}
a
+
c
>
b
+
c
{\displaystyle a+c>b+c}
Jei
a
>
b
{\displaystyle a>b}
c
>
0
{\displaystyle c>0}
a
c
>
b
c
{\displaystyle ac>bc}
Jei
a
>
b
{\displaystyle a>b}
c
<
0
{\displaystyle c<0}
a
c
<
b
c
{\displaystyle ac<bc}
Jei
a
>
1
{\displaystyle a>1}
a
n
>
a
m
{\displaystyle a^{n}>a^{m}}
n
>
m
,
m
,
n
∈
N
{\displaystyle n>m,m,n\in \mathbb {N} }
Jei
0
<
a
<
1
{\displaystyle 0<a<1}
a
n
<
a
m
{\displaystyle a^{n}<a^{m}}
n
>
m
,
n
,
n
∈
M
{\displaystyle n>m,n,n\in \mathbb {M} }
keisti
Modulio apibrėžimas:
|
a
|
=
{
a
,
k
a
i
a
≥
0
,
−
a
,
k
a
i
a
<
0
{\displaystyle |a|={\begin{cases}a,\mathrm {kai} \,a\geq 0,\\-a,\mathrm {kai} \,a<0\,\end{cases}}}
Modulio savybes:
|
a
|
≥
0
{\displaystyle |a|\geq 0}
|
a
|
=
|
−
a
|
{\displaystyle |a|=|-a|}
|
a
|
2
=
a
2
{\displaystyle |a|^{2}=a^{2}}
|
a
b
|
=
|
a
|
|
b
|
{\displaystyle \left|{\frac {a}{b}}\right|={\frac {|a|}{|b|}}}
b
≠
0
{\displaystyle b\neq 0}
|
a
⋅
b
|
=
|
a
|
⋅
|
b
|
{\displaystyle |a\cdot b|=|a|\cdot |b|}
|
a
+
b
|
≤
|
a
|
+
|
b
|
{\displaystyle |a+b|\leq |a|+|b|}
|
a
−
b
|
≥
|
a
|
−
|
b
|
{\displaystyle |a-b|\geq |a|-|b|}
a
b
=
{
|
a
|
⋅
|
b
|
,
k
a
i
(
a
>
0
i
r
b
>
0
)
a
r
b
a
(
a
<
0
i
r
b
<
0
)
−
|
a
|
⋅
|
b
|
,
k
a
i
(
a
>
0
i
r
b
<
0
)
a
r
b
a
(
a
<
0
i
r
b
>
0
)
{\displaystyle ab={\begin{cases}|a|\cdot |b|,\mathrm {kai} \,(a>0\,\mathrm {ir} \,b>0)\,\mathrm {arba} \,(a<0\,\mathrm {ir} \,b<0)\\-|a|\cdot |b|,\mathrm {kai} \,(a>0\,\mathrm {ir} \,b<0)\,\mathrm {arba} \,(a<0\,\mathrm {ir} \,b>0)\end{cases}}}
keisti
Sumos dalumo teorema: jeigu
a
c
∈
Z
{\displaystyle {\frac {a}{c}}\in \mathbb {Z} }
b
c
∈
Z
{\displaystyle {\frac {b}{c}}\in \mathbb {Z} }
a
⋅
b
c
∈
Z
{\displaystyle {\frac {a\cdot b}{c}}\in \mathbb {Z} }
Sandaugos dalumo teorema: jeigu
a
c
∈
Z
{\displaystyle {\frac {a}{c}}\in \mathbb {Z} }
b
d
∈
Z
{\displaystyle {\frac {b}{d}}\in \mathbb {Z} }
a
⋅
b
c
∈
Z
{\displaystyle {\frac {a\cdot b}{c}}\in \mathbb {Z} }
a
⋅
b
d
∈
Z
{\displaystyle {\frac {a\cdot b}{d}}\in \mathbb {Z} }
Sveikasis skaičius dalijasi iš 2, kai jo paskutinis skaitmuo yra 0, 2, 4, 6, 8, t.y. lyginis.
Sveikasis skaičius dalijasi iš 3, kai jo visų skaitmenų suma dalijasi iš 3.
Sveikasis skaičius dalijasi iš 4, kai iš 4 dalijasi dviženklis skaičius, sudarytas iš paskutinių dviejų skaičiaus skaitmenų arba paskutinai skaitmenys yra nuliai.
Sveikasis skaičius dalijasi iš 5, kai jo paskutinis skaitmuo yra 5 arba 0.
Sveikasis skaičius dalijasi iš 11, kai lyginėse ir nelyginėse vietose esančių skaitmenų sumos sutampa arba skiriasi skaičiumi, kuris yra 11 kartotinis.
keisti
...
...
aš noriu sužinoti apie daugybos skirstymo dėsnį
Pagrindinės logaritmų savybės . Su kiekvienu
a
>
0
,
a
≠
1
{\displaystyle a>0,\;\;a\neq 1\;}
log
a
1
=
0
;
{\displaystyle \log _{a}1=0;}
log
a
a
=
1
;
{\displaystyle \log _{a}a=1;}
log
a
(
x
y
)
=
log
a
x
+
log
a
y
,
{\displaystyle \log _{a}(xy)=\log _{a}x+\log _{a}y,}
x >0 ir y >0;
log
a
x
y
=
log
a
x
−
log
a
y
,
{\displaystyle \log _{a}{\frac {x}{y}}=\log _{a}x-\log _{a}y,}
x >0 ir y >0;
log
a
x
p
=
p
log
a
x
,
{\displaystyle \log _{a}x^{p}=p\log _{a}x,}
x >0, p - realusis skaičius;
log
a
x
=
log
b
x
log
b
a
,
{\displaystyle \log _{a}x={\frac {\log _{b}x}{\log _{b}a}},}
x >0, b >0,
b
≠
1
;
{\displaystyle b\neq 1;}
a
log
a
x
=
x
,
{\displaystyle a^{\log _{a}x}=x,}
x >0 (pagrindinė logaritmų tapatybė).
log
a
b
=
1
log
b
a
.
{\displaystyle \log _{a}b={\frac {1}{\log _{b}a}}.}
log
a
b
=
log
a
r
b
r
.
{\displaystyle \log _{a}b=\log _{a^{r}}b^{r}.}
a
2
−
b
2
=
(
a
−
b
)
(
a
+
b
)
.
{\displaystyle a^{2}-b^{2}=(a-b)(a+b).}
a
3
−
b
3
=
(
a
−
b
)
(
a
2
+
a
b
+
b
2
)
.
{\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2}).}
a
3
+
b
3
=
(
a
+
b
)
(
a
2
−
a
b
+
b
2
)
.
{\displaystyle a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2}).}
(
a
+
b
)
2
=
a
2
+
2
a
b
+
b
2
.
{\displaystyle (a+b)^{2}=a^{2}+2ab+b^{2}.}
(
a
−
b
)
2
=
a
2
−
2
a
b
+
b
2
.
{\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2}.}
(
a
n
)
m
=
a
n
⋅
m
.
{\displaystyle (a^{n})^{m}=a^{n\cdot m}.}
(
a
b
)
n
=
a
n
b
n
{\displaystyle \left({\frac {a}{b}}\right)^{n}={\frac {a^{n}}{b^{n}}}}
b
≠
0
{\displaystyle b\neq 0}
Sakykime,
r
=
m
1
n
1
,
s
=
m
2
n
2
.
{\displaystyle r={\frac {m_{1}}{n_{1}}},\;\;s={\frac {m_{2}}{n_{2}}}.}
m
1
{\displaystyle m_{1}}
m
2
{\displaystyle m_{2}}
n
1
{\displaystyle n_{1}}
n
2
{\displaystyle n_{2}}
a
r
⋅
a
s
=
a
m
1
n
1
⋅
a
m
2
n
2
=
a
m
1
n
2
n
1
n
2
⋅
a
n
1
m
2
n
1
n
2
=
a
m
1
n
2
n
1
n
2
+
n
1
m
2
n
1
n
2
=
a
m
1
n
2
+
m
2
n
1
n
1
n
2
=
a
r
+
s
.
{\displaystyle a^{r}\cdot a^{s}=a^{\frac {m_{1}}{n_{1}}}\cdot a^{\frac {m_{2}}{n_{2}}}=a^{\frac {m_{1}n_{2}}{n_{1}n_{2}}}\cdot a^{\frac {n_{1}m_{2}}{n_{1}n_{2}}}=a^{{\frac {m_{1}n_{2}}{n_{1}n_{2}}}+{\frac {n_{1}m_{2}}{n_{1}n_{2}}}}=a^{\frac {m_{1}n_{2}+m_{2}n_{1}}{n_{1}n_{2}}}=a^{r+s}.}
Čia a gali būti bet koks teigiamas (realusis) skaičius. Jeigu sandauga
n
1
⋅
n
2
{\displaystyle n_{1}\cdot n_{2}}
a gali būti bet koks realusis skaičius (tame tarpe ir neigiamas), išskyrus 0 (nes nulio negalima pakelti neigiamu laipsniu).
a
4
−
b
4
=
(
a
−
b
)
(
a
3
+
a
2
b
+
a
b
2
+
b
3
)
.
{\displaystyle a^{4}-b^{4}=(a-b)(a^{3}+a^{2}b+ab^{2}+b^{3}).}
a
5
−
b
5
=
(
a
−
b
)
(
a
4
+
a
3
b
+
a
2
b
2
+
a
b
3
+
b
4
)
.
{\displaystyle a^{5}-b^{5}=(a-b)(a^{4}+a^{3}b+a^{2}b^{2}+ab^{3}+b^{4}).}
a
4
−
b
4
=
(
a
+
b
)
(
a
3
−
a
2
b
+
a
b
2
−
b
3
)
.
{\displaystyle a^{4}-b^{4}=(a+b)(a^{3}-a^{2}b+ab^{2}-b^{3}).}
a
5
+
b
5
=
(
a
+
b
)
(
a
4
−
a
3
b
+
a
2
b
2
−
a
b
3
+
b
4
)
.
{\displaystyle a^{5}+b^{5}=(a+b)(a^{4}-a^{3}b+a^{2}b^{2}-ab^{3}+b^{4}).}
x
n
−
y
n
=
(
x
−
y
)
(
x
n
−
1
+
x
n
−
2
y
+
x
n
−
3
y
2
+
.
.
.
+
x
y
n
−
2
+
y
n
−
1
)
.
{\displaystyle x^{n}-y^{n}=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^{2}+...+xy^{n-2}+y^{n-1}).}
x
n
+
y
n
=
(
x
+
y
)
(
x
n
−
1
−
x
n
−
2
y
+
x
n
−
3
y
2
−
.
.
.
−
x
y
n
−
2
+
y
n
−
1
)
{\displaystyle x^{n}+y^{n}=(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^{2}-...-xy^{n-2}+y^{n-1})\;}
n nelyginis!).
x
n
−
y
n
=
(
x
+
y
)
(
x
n
−
1
−
x
n
−
2
y
+
x
n
−
3
y
2
−
.
.
.
+
x
y
n
−
2
−
y
n
−
1
)
{\displaystyle x^{n}-y^{n}=(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^{2}-...+xy^{n-2}-y^{n-1})\;}
n lyginis!).
![{\displaystyle [a,b]=\left\{x|\,a\leq x\leq b\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0be700279b9e206ae56c7995a5b7d2e38a61b55)
![{\displaystyle (a,b]=\left\{x|\,a<x\leq b\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fcd79e7bd9c53331adb4c7f1527428a5f9021e19)
