Is this proof of the fact that only 1 and -1 divide 1 correct?

Question

The only numbers that divide $1$ are $1$ and $-1$.

If $x$ divides $1$, then \begin{equation} xk=1 \end{equation} for some $k$. To find all numbers that divide 1, we only need to find all integer solutions to the equation above. We argue by cases.

Clearly, if either $x$ or $k$ equals $0$ then the equation has no solutions.

Further, if $|x|=1$ and $|k|=1$, then $x= \pm 1$ and $k= \pm 1$. This gives four possibilities for $x$ and $k$. After checking each, we get two solutions $x=1$, $k=1$ and $x=-1$, $k=-1$.

If $|x|>1$ and $|k|=1$ then $|x||k| = |xk| > 1$. And $xk$ cannot equal 1.

If, on the other hand, $|x|=1$ and $|k|>1$, we again obtain $|xk|>1$ and so $xk \neq 1$.

Finally, if $|x|>1$ and $|k|>1$, then $|xk|>1$ and so again $xk$ cannot equal $1$.

Is this correct?

Why did you not take the definition? Only numbers with absolute value less than or equal to 1 can divide 1... — Parcly Taxel, Nov 03 '16 at 09:59
@drhab Of... divisors? And anyway the proof given looks correct. — Parcly Taxel, Nov 03 '16 at 10:02
@ParclyTaxel The OP started with a definition: $x|1$ if $xk=1$ for some $k$. Based on that he does the job. — drhab, Nov 03 '16 at 10:04
@ParclyTaxel The definition of "divisor" says nothing about absolute values. It's true that if $d$ divides $n$ then $|d| \le |n|$ but that's not part of the definition. — Najib Idrissi, Nov 03 '16 at 10:18
@NajibIdrissi Well I was a little confused there, between the definition and the properties that arise from it... $\ddot\frown$ But it's OK. — Parcly Taxel, Nov 03 '16 at 10:19

score 2 · Answer 1 · answered Nov 03 '16 at 10:03

Your proof is correct, if a little on the long and winding side.

You could shorten it by:

Saying that the case $|x|=1, |k|>1$ is equal to the case $|x|>1, |k|=1$.
Reducing the number of cases by looking simply at $|x|>1, |k|\geq 1$. That way, you check the cases $|x|>1, |k|=1$ and $|x|>1, |k|>1$ at the same time.

So, a shorter version (but in its essence the same thing) of the second part of the proof would be

Now let's assume that it is not true that $|x|=|k|=1$. Then, one of the two numbers is greater than $1$. Without loss of generality, let $|x|>1$. Then, since $|k|\geq 1$, we have

$$1=|xk| = |x||k|\geq |x| > 1,$$ a contradiction. Therefore, $|x|=|k|=1$.

Much shorter, no separation of cases, and I think it's still clear what's going on.

Is this proof of the fact that only 1 and -1 divide 1 correct?

1 Answers1