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 represents an element of a group operated upon itself n-times. If then the result is the element itself. If then the result is the identity element. If 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,
.
and your relation should be:
where is the group's operator.
but here there are two distinct operations involved ,But wait! is an abelian group yet, how can that be?
as here and not .
RonL
Let me say it my way.Originally Posted by CaptainBlack
In the case of the meaning of,
is actually,
. Thus,
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!?!?