Kenangan dalam Diskusi di thescienceforum.com (Matematika)

[trans]

Om santi santi om.

You forgot one more condition: multiplication must be distributive over addition. The two operations in a field are not independent of each other: they are related by the distributive law.

I’m sorry ... .  I had forgotten to add this requirement for building a field ... .
For [tex]a,b,c\in{F}[/tex], where [tex]F[/tex] is a field, then [tex]a(b+c)=ab+ac[/tex] and [tex](a+b)c=ac+bc[/tex] ... .

When defining a field, mathematicians usually insist that [tex]0\ne1[/tex], that is, the additive identity and the multiplicative identity must be distinct. Hence [tex]\{0\}[/tex] is not normally considered a field since a field must contain at least two elements.

Why do the additive identity and the multiplicative identity must be distinct ... ?  Is the distinction one of all conditions of a field ... ?

Isn’t [tex]0+0=0[/tex], [tex]0\cdot0=0[/tex], [tex]0\cdot(0+0)=0\cdot0+0\cdot0[/tex], and [tex](0+0)\cdot0=0\cdot0+0\cdot0[/tex] ... ?


Gloria in excelsis Deo.

[trans]

Dalam Nama Bapa dan Putera dan Roh Kudus. Amin.

Excuse me ... .



[tex]\displaystyle\tan\theta=\frac{h}{x}[/tex] ... .

[tex]\displaystyle\sec^2\theta\frac{d\theta}{dt}=-\frac{h}{x^2}\frac{dx}{dt}[/tex] ... .

[tex]\displaystyle\frac{d\theta}{dt}=-\frac{h}{x^2}\frac{dx}{dt}\cos^2\theta=-\frac{h}{x^2}\frac{dx}{dt}\frac{x^2}{x^2+h^2}=-\frac{h}{x^2+h^2}\frac{dx}{dt}[/tex] ... .


Sampai jumpa lagi.

[trans]

Salam damai Kristus.

I try to answer ... .  I’m sorry if I wrong ... .

[tex]n(S)=10^3\cdot(10\cdot9\cdot8)=72\cdot10^4[/tex] ... .

[tex]n(A)=6^3\cdot(6\cdot4\cdot3)=15552[/tex] ... .

[tex]\displaystyle{P(A)}=\frac{n(A)}{n(S)}=\frac{15552}{72\cdot10^4}\approx0,0216[/tex] ... .


Allahu Akbar.