The expression $\frac{r}{pq}$ with $p, q, r$ integer numbers, $p$ and $q$ different of $0$ is always positive if:
I) $\frac{r}{p} > 0$ and $q > 0$
II) $p*q < 0$ and $r$ not positive.
I answer that $I$ is true, because and integer multiplied by another integer, is always positive. $(\frac{r}{p}) * (\frac{1}{q}) = + * + = +$(Sign representation)
also, I answer that $II$ is true, because a negative integer multiplied by another negative integer is always positive. $(r) * (p * q) = - * - = + $( Sign representation)
But the correct answer should be only $I$ correct, and $II$ is incorrect, why ?
I) is correct because:
$q > 0 \implies \frac 1q > 0$ and $\frac rp > 0$ and $\frac 1q > 0 \implies \frac r{pq} > 0$.
II) is incorrect because
$r$ not positive means $r \le 0$. If $r= 0$ and $pq < 0$ then $\frac {r}{pq} = 0$.
The basic axioms/props you should have at your fingertips are:
A) If $x> 0$ and $y < z$ then $xy < xz$ and
B) If $y <z$ then $y + x < z + x$.
As well as i) $0*x = x$ and ii) $-(-x) = x$ and iii) $x*(-y) = (-x)*y = -xy$.
From those you know:
0) If $ x > 0$ then $x-x > 0 - x$ or $0 > -x$ or $-x >0$ and if $x< 0$ then $x -x < 0 -x = -x$ and so $-x > 0$.
1) If $x> 0$ and $y > 0$ then $xy > x*0 = 0$
2) If $x> 0$ and $y < 0$ then $xy< x*0 = 0$
3) If $x < 0$ and $y < 0$ then $-x>0$ and $-xy < -x*0 = 0$ and so $xy > 0$
4) If $x > 0$ then $x^2 = x^2 > 0$. And if $x < 0$ then $x^2 = x*x > 0$ and if $x = 0$ then $x^2 = 0*0=0$ so $x^2 > 0$ if $x \ne 0$ and $x^2=0$ if $x = 0$ so $x^2 \ge 0$ with equality holding only if $x = 0$.
5) So $1 = 1^2 > 0$.
6) If $q > 0$ and $q*\frac 1q = 1$ then since $\frac 1q = 0$ would mean $1=q*\frac 1q = 1*0 = 0$ and $\frac 1q < 0$ would mean $1 = q\frac 1q < 0$ and those are both nutty, we must know that $\frac 1q > 0$.