# Proofs: If a|b then -a|b, a|-b, -a|-ab

by wubie
Tags: a|ab, proofs
P: n/a
Hello,

First I will post the question that I am working on.

 Prove: If a|b, then -a|b, a|-b, and -a|-ab.
I am not good at proofs (even elementry proofs such as these ones). I was wondering if someone could take a look at my work and perhaps confirm whether my proofs are adequate and/or make some suggestions.

First I will start off with a basic definition of a divisor:

An integer, a, not equal to zero, is called a divisor of an integer b if there exists an integer c such the b = a c.

i) If a|b then -a|b.

Assume a|b. Then b = a c for some integer c by def.

Let c = -k where -1, k are integers. Then

b = a (-k) = - a k.

Since k is an integer, then by def., if a|b then b = -a k.

Similarly,

ii) If a|b then a|-b.

Assume a|b. Then b = a c for some integer c by def.

Let c = -k where -1, k are integers. Then

b = a c = a -k = -a k = -(a k)
-b = --(a k)
-b = a k

Once again since k is an integer, then by def., if a|b then -b = a k.

Also

iii) If a|b then -a|-b.

I am sort of stuck on this one. I am not yet sure how to show

If a|b then -a|-b.

I thought

b = a c,
-b = -a c

By definition, -a|-b if -b = -a c for some integer c. Since c is an integer, then by def. if a|b then -a|-b.

Part iii) seems pretty weak to me. In fact all look pretty weak now.

Any help/insights are appreciated.

Thankyou.
P: 1,572
 Originally posted by wubie Hello, First I will post the question that I am working on. I am not good at proofs (even elementry proofs such as these ones). I was wondering if someone could take a look at my work and perhaps confirm whether my proofs are adequate and/or make some suggestions. First I will start off with a basic definition of a divisor: An integer, a, not equal to zero, is called a divisor of an integer b if there exists an integer c such the b = a c. i) If a|b then -a|b. Assume a|b. Then b = a c for some integer c by def. Let c = -k where -1, k are integers. Then b = a (-k) = - a k. Since k is an integer, then by def., if a|b then b = -a k.
this is fine but i would adjust it a bit:
since a|b, b=ac for some c&isin;Z. then b=(-a)(-c), implying that -a|b as -c&isin;Z.

(ie there is no need for the k)

 [/b] iii) If a|b then -a|-b. I am sort of stuck on this one. I am not yet sure how to show If a|b then -a|-b. I thought b = a c, -b = -a c By definition, -a|-b if -b = -a c for some integer c. Since c is an integer, then by def. if a|b then -a|-b. Part iii) seems pretty weak to me. In fact all look pretty weak now. Any help/insights are appreciated. Thankyou. [/B]
that's fine i think. here's how i would phrase it. since a|b, b=ac for some c&isin;Z. then -b=-ac, which implies that -a|-b.

 [/b]if If a|b then -a|-ab.[/b]
you want to find an integer q such that -aq=-ab. what might q be?
P: n/a
 you want to find an integer q such that -aq=-ab. what might q be?
Well, by cancellation I would say that q - b. But I don't understand your point.

Let me try this again.

iii)

Prove if a|b then -a|-b.

Assume

a|b.

Then by definition

b = a c for some integer c.

Let

-b = -a c for some integer c.

Then by definition

-a|-b.

Therefore

If a|b then -a|-b.

How would that be?

It seems a little stronger than what I had. But it feels like I am missing something inbetween.

P: 1,572

## Proofs: If a|b then -a|b, a|-b, -a|-ab

that seems fine except for the word "let."

q would be b.

-ab=-ab implies that -a|-ab.
 P: n/a That was a typo on my part. I meant q = b. 8) I am not sure I am getting this though. If you have -a q = -ab for some q in this case, I could then say -a|-ab by definition. I can see it better in the other direction: Say -a|g. Let g = -ab. Then -a|-ab. So by definition -ab = -a q from some q which is an element of the integers. If -ab = -aq, then b =q by cancellation. I feel like I am trying to run through a brick wall while the way through the brick wall is a door just a couple of feet to one side. 8(

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