1. ## divisibility

Prove that 1 and -1 are only divisible by 1 and -1.

I think that I'm making this too complicated. Any help would be greatly appreciated.

2. I do it with $\displaystyle 1$, do it for $\displaystyle -1$.
A number $\displaystyle a \in \mathbb{N}$ is divisible by a number $\displaystyle b$ if and only if $\displaystyle a=bq+r$ with $\displaystyle 0\leq r <b$.
It's obvious that $\displaystyle a\geq b$.
So in your case you have that $\displaystyle a=1$. That means that if $\displaystyle a$ was divisible by a number $\displaystyle b$, it would necessarily be $\displaystyle 1$. We can try to check if it works : $\displaystyle 1=1q+r=1\cdot1 +0$, which works.
I think what I did is not wrong...
Good luck. (All you have to do is to generalize what I did for the numbers $\displaystyle \in \mathbb{N}$ for any number $\displaystyle \in \mathbb{Z}$. I suggest you to check this : Division algorithm - Wikipedia, the free encyclopedia out.)

3. Originally Posted by arbolis
I do it with $\displaystyle 1$, do it for $\displaystyle -1$.
A number $\displaystyle a \in \mathbb{N}$ is divisible by a number $\displaystyle b$ if and only if $\displaystyle a=bq+r$ with $\displaystyle 0\leq r <b$.
The division algorithm just says that given $\displaystyle a$ and $\displaystyle b$, there exists a unique $\displaystyle q$ and $\displaystyle r$, $\displaystyle 0 \leq r < b$ such that $\displaystyle a = qb + r$. Doesn't mention about a dividing b or vice versa but we can see that b divides a when r = 0.

For this question, just go by definition:
$\displaystyle a \mid 1 \iff ac_{1} = 1 \iff a = \frac{1}{c_{1}} \ \ c_{1} \in \mathbb{Z}$

If $\displaystyle \mid c_{1} \mid > 1$, we have that the numerator is less than the denominator which cannot be an integer. Thus $\displaystyle \mid c_{1} \mid = 1 \ \Rightarrow \ c_{1} = 1, -1$. So $\displaystyle a$ must be ...