1. Originally Posted by Quick
The first compass originated in China. Although I'm not sure where it was first practically used.
If they were not sure where they were when the compass was first used it
could not have been working properly

RonL

2. The "compass" was actually a toy, a "magic spoon" if you will. It was a spoon with magnets on the head so that if you dropped it the head would always point in the same direction.
Originally Posted by CaptainBlack
If they were not sure where they were when the compass was first used it
could not have been working properly

RonL

3. Wow, this is stupid. Here is another one of my riddles.

If G is an abelian group then,
$\displaystyle (a+b)^n=a^n+b^n$.

But wait! $\displaystyle \mathbb{Z}$ is an abelian group yet, $\displaystyle (1+1)^2=1^2+1^2=2$ how can that be?

4. Originally Posted by ThePerfectHacker
Wow, this is stupid. Here is another one of my riddles.

If G is an abelian group then,
$\displaystyle (a+b)^n=a^n+b^n$.
What is the group operation here, and what does $\displaystyle a^n$ denote?

RonL

5. Originally Posted by CaptainBlack
What is the group operation here, and what does $\displaystyle a^n$ denote?

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.

6. Originally Posted by ThePerfectHacker
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.
Then does + represent the group operation, and do exponentiation and
addition still mean this when you write:

$\displaystyle (1+1)^2=1^2+1^2=2$

??

RonL

7. Originally Posted by CaptainBlack
Then does + represent the group operation,
Of course + is the group operation, surly multiplication does not form a group.

Originally Posted by CaptainBlank
and do exponentiation and
addition still mean this when you write:

$\displaystyle (1+1)^2=1^2+1^2=2$

??

RonL
I cannot answer that because it is as if I am giving an answer.

8. If G is an abelian group then...
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:
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.

9. Originally Posted by Rebesques
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:
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.
I was just about to say that

10. Originally Posted by Rebesques
I don't think this is enough. We need G to be a field...
Nope.
$\displaystyle (a+b)^n=(a+b)+...+(a+b)=(a+..+a)+(b+...+b)$ (because G is an abelian group). Thus, $\displaystyle a^n+b^n$.
But, nobody answer my problem, why is it that $\displaystyle (1+1)^2\not = 1^2+1^2$???

11. Originally Posted by ThePerfectHacker
Of course + is the group operation, surly multiplication does not form a group.

I cannot answer that because it is as if I am giving an answer.
These are directed questions to show you are confusing two different
operations in the case of the field $\displaystyle \mathbb{Z}$ with the one
you have in the case of a group.

RonL

12. Originally Posted by CaptainBlack
These are directed questions to show you are confusing two different
operations in the case of the field $\displaystyle \mathbb{Z}$ with the one
you have in the case of a group.

RonL
Say better. Exactly why it that wrong?

13. Originally Posted by ThePerfectHacker
Say better. Exactly why it that wrong?
If G is an abelian group then,
$\displaystyle (a+b)^n=a^n+b^n$.
Here there is but one operation:

$\displaystyle x^n=x \oplus x^{n-1};\ x^1=x$

and your relation should be:

$\displaystyle (a \oplus b)^n=a^n \oplus b^n$

where $\displaystyle \oplus$ is the group's operator.

But wait! $\displaystyle \mathbb{Z}$ is an abelian group yet, $\displaystyle (1+1)^2=1^2+1^2=2$ how can that be?
but here there are two distinct operations involved $\displaystyle +, \times$,
as here $\displaystyle x^2=x \times x$ and not $\displaystyle x+x$.

RonL

14. Originally Posted by CaptainBlack
Here there is but one operation:

$\displaystyle x^n=x \oplus x^{n-1};\ x^1=x$

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 $\displaystyle +, \times$,
as here $\displaystyle x^2=x \times x$ and not $\displaystyle x+x$.

RonL
Let me say it my way.
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.

15. 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!?!?

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