Proving Negative Divisors of an Integer

[tonit]

Homework Statement

Show that negative divisors of an integer, are just the negatives of the positive divisors.

The Attempt at a Solution

having an integer $n$ and a negative divisor $k$, we get the positive divisor by multiplying $(-1)k$, thus each negative divisor of an integer, is the negative of the positive divisor of $n$.

I have the idea but don't know if this kind of proof is correct.

 

[Mark44]

Homework Statement

Show that negative divisors of an integer, are just the negatives of the positive divisors.

The Attempt at a Solution

having an integer $n$ and a negative divisor $k$, we get the positive divisor by multiplying $(-1)k$, thus each negative divisor of an integer, is the negative of the positive divisor of $n$.

I have the idea but don't know if this kind of proof is correct.

It looks to me like what you showed is that if k is a negative number, then -k is a positive number. There's nothing in your proof that is related to division. What does it mean to say that a number is a divisor of some other number.

If k is a negative number that divides n, how do you know that -k also divides n? That's the part that's missing.

 

[tonit]

By a theorem, if $m|p$, and $n|p$, while $gcd(m,n) = 1$, then $mn|p$. Since $(-1)|n$ and $k|n$, and $gcd(-1,k) = 1$, then $(-1)k|n$.

Is this correct?

 

[Mark44]

Looks OK, but is more complicated than it needs to be. What I was thinking was more along these lines: if k|n, then for some integer a, n = ak. Using this basic concept, it's easy to show that if k|n, then -k|n as well, without the need to invoke any other theorem.

 

[tonit]

Hmmm what about this one:
if $k|n$, then for some integer $a$, $n = ak = ak1 = ak(-1)(-1) = a(-k)(-1)$ which shows that $(-k)|n$

So if $k$ is a negative divisor, then $(-k)$ which is positive, also divides $n$. So for each negative divisor $k$, we have the positive $(-k)$ integer. Now if $k$ is positive, then $(-k)$ is negative, which shows that for each positive divisor $k$, $(-k)$ which is a negative integer also divides $n$. So the converse is also true, and the statement holds.

 

[HallsofIvy]

Hmmm what about this one:
if $k|n$, then for some integer $a$, $n = ak = ak1 = ak(-1)(-1) = a(-k)(-1)$

It seems to me easier just to note that n= ak= (-a)(-k). Since so a is an integer, so is b= -a and you have shown that n= b(-k).

which shows that $(-k)|n$

So if $k$ is a negative divisor, then $(-k)$ which is positive, also divides $n$. So for each negative divisor $k$, we have the positive $(-k)$ integer. Now if $k$ is positive, then $(-k)$ is negative, which shows that for each positive divisor $k$, $(-k)$ which is a negative integer also divides $n$. So the converse is also true, and the statement holds.

 

[tonit]

yeah, I just wrote it the way it came on my mind. Thanks for noting it