I'm very new to this and I'm not sure if this is a proper proof. It's within the integer set so I can't use the inverse multiplication proof.
Want to show that if $a,b \in \mathbb{Z}$ and $ab=0,$ then either $a=0$ or $b=0$.
Let $a,b \in \mathbb{Z}$
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Case 1: If $a \neq 0$ then:
We can rewrite $ab = 0$ as $ab = a\cdot0$.
Then $ab + a\cdot0 = 0 \Rightarrow ab + 0 = 0 \Rightarrow a(b+0)=0$
Since $a \neq 0$, then $a(b+0)=0 \iff b=0$.
Thus, when $a \neq 0$ and $ab = 0$, then $b=0$.
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Case 2: If $a = 0$ then:
$0 \cdot b = (0+0) \cdot b$
$(0 \cdot b) + 0 = 0 \cdot b + 0 \cdot b$
$0 \cdot b = 0$.
Which shows that $a = 0$ implies $ab = 0$.
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Therefore, if $a,b \in \mathbb{Z}$ and $ab=0,$ then either $a=0$ or $b=0$. \qedsymbol
Show $ab\ne 0$ for $a,b\in\Bbb N$. Do the rest by case distinction according to signs (using $(-a)b=a(-b)=-ab$)