Prove that every number is divisible by $1$ and that the only numbers that divide $1$ are $1$ and $-1$
axiom: if $d \mid n, \, \exists k \in \mathbb{Z}$ such that $d \cdot k = n$
for $1 \cdot k = n, \forall n \,\exists k$ such that $k = n \longrightarrow \forall n, \, 1 \mid n$
For the second part for $d \cdot k = 1$ there are two cases:
case 1: $k = 1 \longrightarrow d = 1$
case 2: $k = -1 \longrightarrow d = -1$
Does anybody have any feedback on this and my use of notation? Thanks.
They all need work ...
Take any $n$. Since $1 \cdot n =n$, there exists some $k$ such that $1 \cdot k =n$. So, for any $n$: $1|n$
2 and 3. You need to show that there are no solutions for any other $k$.
Let $n$ divide $1$. That means that there exists some $k$ such that $n\cdot k =1$. Hence, $n = \frac{1}{k}$.
Now, if $k=0$, then $n\cdot k= n \cdot 0 =0 \not =1$. So, $k \not =0$.
If $k=1$, then $n=1$, so we have a solution for $k=1$ and $n=1$, so $1|1$.
If $k>1$, then from $n = \frac{1}{k}$ is not an integer, so there is no solution there.
If $k=-1$, then $n=\frac{1}{-1}=-1$, so solution there with $n=-1$.
If $k<-1$, then $n=\frac{1}{k}$ is not an integer.
This covers all possible values of $k$, so there are only two solutions: $n=1$ and $n=-1$. Hence, $1$ and $-1$ are the only numbers that divide $1$.
