Let $a$ and $b$ be integers such that $ab = 0$. Then either $a = 0$ or $b = 0$ (or both $a=b=0$).
MY ATTEMPT
Let us consider that $a = m - n$ and $b = p - q$, where $m,n,p,q$ are natural nubmers.
Then we have that \begin{align*} ab = (m-n)(p-q) & = (mp + nq) - (mq + np) = 0 \end{align*}
but then I get stuck. Can somebody help me out?
On
Suppose that $a$ is non-zero. Then, by induction, every $ab$, $b\neq 0$ is non zero, in fact, $a \leq ab$.
For $b=1$, $ab = a$ by definition.
Suppose this is valid for $k$, so $a(k+1) = ak + a$. But then $a(k+1) \geq a + a > a$. So, every $ab$ is not zero. Thus, using the contrapositive, $b$ must be zero.
Edit: Now, that settles the matter for $b$ a natural number. If $b<0$ then let $b'=-b$. If $ab =0$ then $-(ab)=-0$ and then $a(-b)=0$, that is, $ab'=0$, but because $b'\geq 0$, it is a natural number and our previous result applies. So, if $a\neq 0$, we have $b'=0$ and then $b=-b'=0$.
You can try and prove the contrapositive, namely, that $a \neq 0$ and $b \neq 0$ implies $ab \neq 0$. Use trichotomy of integers for $a$ and $b$ and Lemma 2.3.3.
Or you can use contradiction and assume that $a \neq 0$ and $b \neq 0$. Then, use trichotomy of integers and see that it contradicts Lemma 2.3.3.
Or if you prefer the direct approach to exhaust all possible cases:
Let $a$ and $b$ be integers. Suppose that $ab = 0$. By Lemma 4.1.5 (Trichotomy of integers) on p. 77 we have for $a$ three cases, namely, $a = 0$, $a = n$ for positive natural number $n$, $a = -n$ for positive natural number $n$. Similarly, for $b$ we have three cases $b = 0$, $b = m$ for positive natural number $m$, $b = -m$ for positive natural number $m$. Then, check cases.
(1) $a = 0$ and $b = 0$.
(2) $a = 0$ and $b = m$.
(3) $a = 0$ and $b = -m$.
(4) $a = n$ and $b = 0$.
(5) $a = n$ and $b = m$. Hint use Lemma 2.3.3 on p. 30.
(6) $a = n$ and $b = -m$.
(7) $a = -n$ and $b = 0$.
(8) $a = -n$ and $b = m$.
(9) $a = -n$ and $b = -m$.