科学の基礎を再検証するwiki

a b=abcosθ (内積)
(-1)≦cosθ≦1
(-ab)≦abcosθ≦ab
a b=abcosθ
(-ab)≦a b≦ab
a b≦ab
ab≧a b
a^2b^2≧(a b)^2
a=Δa
b=Δb
(Δa)^2(Δb)^2≧(Δa Δb)^2
Δa Δb=1/2(Δa Δb)+1/2(Δa Δb)
Δa Δb=1/2(Δa Δb)+1/2(Δa Δb)+
1/2(Δa Δb)-1/2(Δa Δb)
Δa Δb=1/2(Δa Δb)+1/2(Δb Δa)+
1/2(Δa Δb)-1/2(Δb Δa)
Δa Δb=1/2[(Δa Δb)+(Δb Δa)]+
1/2[(Δa Δb)-(Δb Δa)]
Δa Δb=1/2[(Δa Δb)+(Δb Δa)+
(Δa Δb)-(Δb Δa)]
(Δa Δb)-(Δb Δa)={(a-a')(b-b')-(b-b')(a-a')}
{(a-a')(b-b')-(b-b')(a-a')}=(a b-b a)
(Δa Δb)-(Δb Δa)=(a b-b a)
Δa Δb=1/2[(Δa Δb)+(Δb Δa)+
(Δa Δb)-(Δb Δa)]
Δa Δb=1/2[(Δa Δb)+(Δb Δa)+(a b-b a)]
Δa Δb={1/2[(Δa Δb)+(Δb Δa)]+1/2(a b-b a)}
(Δa Δb)+(Δb Δa)>(Δa Δb)-(Δb Δa)
(Δa Δb)-(Δb Δa)=(a b-b a)
(Δa Δb)+(Δb Δa)>(a b-b a)
{1/2[(Δa Δb)+(Δb Δa)]+1/2[(a b-b a)]}≧
[1/2(a b-b a)+1/2(a b-b a)]^2=
[1/4(a b-b a)]^2
{1/2[(Δa Δb)+(Δb Δa)]+1/2[(a b-b a)]}≧
[1/4(a b-b a)]^2
Δa Δb={1/2[(Δa Δb)+(Δb Δa)]+1/2(a b-b a)}
Δa Δb={1/2[(Δa Δb)+(Δb Δa)]+1/2(a b-b a)≧
[1/4(a b-b a)]^2
Δa Δb≧[1/4(a b-b a)]^2
(Δa)^2(Δb)^2≧(Δa Δb)^2
Δa Δb≧[1/4(a b-b a)]^2
(Δa Δb)^2≧[1/4(a b-b a)]^4
(Δa)^2(Δb)^2≧(Δa Δb)^2≧[1/4(a b-b a)]^4
Δa Δb≧[1/4(a b-b a)]^2
Δa Δb≧1/2(a b-b a)=1/2(h/2π)=h/4π
Δa Δb≧h/4π
a=p
b=x
Δp Δx≧h/4π
ΔpΔx≧h/4π (不確定値)
ΔpΔx=h/4π (確定値) (最小値)