# Thread: Zero product property, Help.

1. ## Zero product property, Help.

I'd like to know something...

How can I show that the Zero product property $ab=0$ implies $a=0$ or $b=0$ holds in the Integers?

I'd like to show that so I can show that the rationals are a field (without getting into the real numbers)...but all the proofs I've seen about the Zero product property are using field axioms. I don't know if I can show that (The only thing that seems to fail in the proof is the closure of the sum given the possibility that the denominator after the sum becomes zero if I don't prove that it doesn't happens in the integers).

So I don't know how to do it in a ring.

2. ## Re: Zero product property, Help.

Hey Leviathantheesper.

I don't know how formal this has to be, but one idea that comes to mind is to use the same sort of proof used in a vector space where 0*v = 0 vector.

Alternatively you can assume that a != 0 and b != 0 and use proof by contradiction.

3. ## Re: Zero product property, Help.

Hi,
I think you have to assume something. The natural numbers (non-negative integers) are described by the Peano axioms. In this development, one defines x > y to mean there is a non-zero z with x=y+z. In particular then for any non-zero natural number x, x > 0 since x = 0 + x (part of the definition of addition). It also follows from the axioms that x > 0 implies x is not 0. Also induction is one of the axioms. So a proof by induction that ab > 0 for non-zero natural numbers a and b can proceed by induction on b by:
$a\cdot1=a>0$
Assuming ab > 0, a(b+1)=ab+a>ab>0: a(b+1)=ab+a is part of the definition of multiplication, also transitivity of > can be proved.

I realize this may be a little mysterious if you've never studied the Peano axioms. Wikipedia has a very terse discussion of the axioms

Finally, you can extend the result to all integers by playing with signs.

4. ## Re: Zero product property, Help.

I hadn't thought about Induction. I had already thought to use the Natural numbers but I didn't know how to do it with the negatives. Thanks.

5. ## Re: Zero product property, Help.

Personally, I wouldn't use induction. Just a(b+ 0)= ab+ a0 (distributive law) and a(b+ 0)= ab (0 is the additive identity) is sufficient. Since both are equal to a(b+0), ab+a0= ab and subtracting ab from both sides (existence of additive inverse) a0= 0.

6. ## Re: Zero product property, Help.

But we are not proving the same thing...we must prove that (ab=0)→(a=0 or b=0)...
Not that for all a in Z, a0= 0...I think, the other option is that I didn't understand your proof.

Anyway using Induction I have already proven it.