Divisors of $-1$ are only $1$ and $-1$?

Question

I'm working through a discrete math textbook and I've come across this question with answer:

Prove that the only divisors of $−1$ are $−1$ and $1$.

Answer:

We established that $1$ divides any number; hence, it divides $−1$, and any nonzero number divides itself. Thus, $1$ and $−1$ are divisors of $−1$. To show that these are the only ones, we take $d$, a positive divisor of $−1$. Thus, $dk = −1$ for some integer $k$, and $(−1)dk = d(−k) = (−1)(−1) = 1$; hence, $d\mid 1$, and the only divisors of $1$ are $1$ and $−1$. Hence, $d = 1$ or $d = −1$.

I understand everything stated in the answer except for the part: $(-1)dk = d(-k) = (-1)(-1) = 1$

Perhaps someone could help me understand where the $(-1)dk$ comes from? And how we go from $d(-k)$ to $(-1)(-1)$?

Answer 1

We are assuming that $d$ is a divisor of $-1$ that is

$$dk=-1$$

and multiplying each side by $-1$ we obtain

$$-1\cdot dk=-1\cdot (-1)=1 \iff d(-k)=1 \iff d=1,-1 $$

Answer 2

The $(-1)dk$ bit is basically use trying to establish that, whatever $d$ is, it divides $1$ - which, since the only divisor of $1$ is itself, means $d=1$. By establishing $d=1$, then we establish that no other divisors to $-1$ exist, other than $1$ and $-1$ - this is because we assumed that $d$ was some other, arbitrary divisor, but show that such an assumption means $d=1$ (an analogous argument can probably show that $d=-1$ under a slightly different construction).

So... from the fact that $d$ divides $-1$, we know there exists some integer $k$ such that $dk = -1$.

So, we begin with just considering $(-1)dk$, and we want to see where that takes us.

As multiplication is commutative, $(-1)dk = d(-1)k$.

$(-1)k = -k$, obviously, so $(-1)dk = d(-1)k = d(-k)$.

However, recall that, since $d|-1 \;\; \Rightarrow \;\; dk = -1$, we also have $(-1)dk = (-1)(-1)$.

Thus, $d(-k) = (-1)(-1)$.

We know $(-1)(-1) = 1$, obviously, so we thus have $d(-k)=1$.

$d$ and $k$ by assumption are both integers (and thus $-k$ is too). Thus, $d|1$ and $-k|1$.

Since $d$ divides $1$, $d$ is a factor of $1$ by definition. However, the only factors of $1$ are ... just $1$ itself. Thus, $d=1$.

Answer 3

Presumably the book has already proved or taken as axioms:

• $(ab)c=a(bc)$ and you can then write the result as $abc$, called associativity of multiplication
• $ab=ba$, call commutativity of multiplication
• $(-1)c=-c$
• $(-1)(-1)=1$
• the only divisors of $1$ are $1$ and $−1$, or at least that the only positive divisor of $1$ is $1$

Then using the first three points you have $(-1)dk = ((-1)d)k=(d(-1))k = d((-1)k)=d(-k) $

while, since $dk=-1$, you have $(-1)dk=(-1)(-1)=1$ from the fourth point

together implying $d(-k)=1$, and since $d$ is assumed to be positive it must be $1$ as the only positive divisor of $1$, leading to the conclusion that $-k=1$ and so $(-1)(-k)=(-1)1$, i.e. $k=-1$