it is mainly for passing through laws. if the mainparty only has 51% its tough, because individual members can disagree
so in the USA, if the ruling party is for something but not all of them ,say 51% of the party or effectivly only 25% of all politicians a law can be passed through anyway?
Guy i got a question I need to show that K[X] is K-Vektorroom. I know the definition of a polynom and a vektorroom, but my question is can i just choose 3 random polynoms from my vektorroom to show for example associativity under addition?
Just consider three arbitrary polynomials $\sum_{i=0}^n a_ix^i$, $\sum_{i=0}^m b_ix^i$, $\sum_{i=0}^p c_ix^i$, and apply the definition of adding polynomials, then use associativity of $K$
You can't just pick three particular random polynomials--it has to hold in general. Unless what I said is what you meant, in which case my bad. ;)
if G is any group and $a\in G$ for which $a\circ a=a$ holds, then a is the identity element of G. (i asked that previously) So is the inverse-axiom here to be used?
Put more precisely, if $ac = bc$, then $a = b$, and similarly if $ab = ac$, then $b = c$.
In the first case, you right-multiply by $c^{-1}$ ("right-divide by c", heavy air quotes), and in the second case, you left-multiply by $a^{-1}$ ("left-divide by a")
$a\circ a=a\iff c\circ a\circ a=c\circ a$. Now we choose $c=a^{-1}$ which is fine by the axioms. Therefore $a^{-1}\circ a\circ a=a^{-1}\circ a\iff e\circ a=a=e$?
@saturatedexpo It gets easier. Just remember that groups are just nice enough that you can use cancellation and other stuff from normal algebra--you just can't rearrange the order of things unless your group is abelian.
It's just in that case, since you're already using the normal $+$ to define your new operation, calling your new operation $+$ is something you have to be very careful with.
I have to go now, but you should think about proving that in a group $G$ the identity element is unique, that is if $ae=ea=a$ for all $a\in G$ and $af=fa=a$ for all $a\in G$ then $e=f$
