The Zero product property is:

if xy=0 then x=0 or y=0

Homework problemEdit

A student asks,

Using only the following laws prove that if mn=0, then either m=0 or n=0.

i) if m and n are in D then m+n and mn are in D.
ii) if m=p and n=q in D then m+n=p+q and mn=pq.
iii) m+n = n+m
iv) mn=nm
v) m+(n+p)=(m+n)+p
vi) m(np)=(mn)p
vii) m(n+p) = mn + mp
viii) if m =/= 0, then mn = mp ==> n=p.

The number 0 has the following properties. For each number m in D,

ix) m+0 = m
x) m.0 = 0
xi) there exists a number -m such that m+(-m) = 0

The number 1 =/= 0 has the property:

xii) m.1 = m for all numbers m in D.
MathHelp replies,

These are the field properties minus the multiplicative inverse, and with an additional axiom viii that allows you to cancel a common nonzero factor. Axiom ii is superfluous, and just states what is meant by the use of symbols. Axiom x is also superfluous, as it can be proved from the other axioms.

m.0 = m.(0+0) = m.0+m.0, and by adding (-m.0) to both sides, we get 0=m.0

Proof of zero product propertyEdit

We're being asked to prove the statement, if mn=0, then either m=0 or n=0. Let's prove instead the contrapositive, namely $ m\ne 0\text{ and }n\ne 0\implies mn\ne 0. $

While we're in a contrapositive mood, let's restate viii that way:

viii-contrapositive)   $ \text{if }m\ne 0,\text{ then }n\ne p\implies mn\ne mp. $

If we let p=0, and using the fact that m.0=0, then the statement of axiom viii becomes

viii-contra-with-p=0)   $ \text{if }m\ne 0,\text{ then }n\ne 0\implies mn\ne 0. $

Composing the implications, we get the statement we need, which is

$ \text{if }m\ne 0\text{ and }n\ne 0,\text{ then }mn\ne 0. $