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