If they were not sure where they were when the compass was first used itOriginally Posted by Quick
could not have been working properly
RonL
The accepted notion of $\displaystyle a^n$ represents an element of a group operated upon itself n-times. If $\displaystyle n=1$ then the result is the element itself. If $\displaystyle n=0$ then the result is the identity element. If $\displaystyle n<0$ then the result is the inverse of result when without the negative.Originally Posted by CaptainBlack
I don't think this is enough. We need G to be a field (so addition and multiplication are both defined) and we need it to be of characteristic n:If G is an abelian group then...
This means, ng=g+...+g=0 for all g in G.
Under these considerations, (a+b)^n=a^n+b^n+(remaining terms), and since the remaining terms are all multiples of n, we have that (a+b)^n=a^n+b^n.
Examples of such fields are the Z_p for p prime, as nx=0 for all x in Z_p.
Originally Posted by ThePerfectHackerHere there is but one operation:If G is an abelian group then,
$\displaystyle (a+b)^n=a^n+b^n$.
$\displaystyle x^n=x \oplus x^{n-1};\ x^1=x$
and your relation should be:
$\displaystyle (a \oplus b)^n=a^n \oplus b^n$
where $\displaystyle \oplus$ is the group's operator.
but here there are two distinct operations involved $\displaystyle +, \times$,But wait! $\displaystyle \mathbb{Z}$ is an abelian group yet, $\displaystyle (1+1)^2=1^2+1^2=2$ how can that be?
as here $\displaystyle x^2=x \times x$ and not $\displaystyle x+x$.
RonL
Let me say it my way.Originally Posted by CaptainBlack
In the case of $\displaystyle \mathbb{Z}$ the meaning of,
$\displaystyle (a+b)^n$ is actually,
$\displaystyle n(a+b)$. Thus,
$\displaystyle n(a+b)=na+nb$ are nothing is violated.
The purpose of my riddle was to confuse the exponent. The meaning of exponent in this case is operating an element by itself a certain number of times. However, I confused it with the meaning of regular exponents of multiplication.
Here I came up with another riddle, this one seems really easy.
Let X be a non-empty set and let ~ be a relation on X. Why is it that we say need to check reflexisve property if we can show it is it true from symettry and transitivity? Observe, if x~y then y~x (symettric) but then, x~x (transitive). Therefore it is reflexsive!?!?