Let $a$ and $b$ be integers such that $ab = 0$. Then either $a = 0$ or $b = 0$ or both.
So, I am doing exercise in a Terence Tao-Analysis 1 book and I came up with the above problem. At this point of the book the integers are represented as $x\in Z, x = a-b; a,b \in N$. That is every integer has the form of subtraction of two numbers from $N$.
The idea that I tried to implement while solving the problem was to break down into the cases for positive and positive, negative and negative, and positive and negative integers show that their products can not be $0$, hence the only remaining option is $a = 0$ or $b = 0$ or both. However, I don't succeded in doing so because at this point in the book I don't yet know that the cancelation law for the integers exists. But I do know that the cancelation law for $N$ exists. Therefore I decided to change the strategy.
take $a = x-y; x,y\in N$
take $b = c-d; c,d\in N$
$ab = (x-y)(c-d) = (xc+yd)-(yc+xd)$ by definition
if $ab = 0$ then $(xc+yd)-(yc+xd)=0=0-0$, so $xc+yd=yc+xd$-by definition
$xc-xd = yc-yd$
*then $x(c-d) = y(c-d)$
because for any $x \in N$ and $a-b \in Z$ $x(a-b) = (x-0)(a-b)=(xa+0b)-(0a-xb)=xa - (x(-b)) = xa-(x(b-0))=xa-xb$
since any natural number $n = n-0$
At this point my strategy is to use $x(c-d) = y(c-d)$ and induct on $c \in N$.
Base case: $c = 0$ $x(0-d) = y(0-d)$; $x(-d) = y(-d)$; $-xd=-yd$ at this point I will be good if I can proof that $(-1)a = -a$ for any $a \in Z$
Inductive step: Assume $x(c-d) = y(c-d)$ then $x((c++)-d) = y((c++)-d)$ then $xc+x-xd = yc+y-yd$, then $x+x(c-d)=y+y(c-d)$ by inductive hypothesis $x=y => a=0$.
Here also it seems to me like I am cheating. I think I can not make such an assumption from the inductive hypothesis since I am using cancelation law for integers which I don't know yet.
Please, can you clarify if the logic that I am using is plausible or not? If yes, can you help me to carry on?