Proof problem involving set theory and relations Problem: If $R_1$ is defined on $\mathbb{R}$ by the relation $R_1=\{(a,b):1+ab>0\ , a, b \in \mathbb{R}\}$, then prove that $(a,b) \in R_1$ and $(b,c) \in R_1 \implies (a,c) \in R_1$ is not true for all $a,b,c \in \mathbb{R}$.
My attempt:
$$(a,b) \in R_1 \implies 1+ab>0 \tag{1}$$
$$(b,c) \in R_1 \implies 1+bc>0 \tag{2}$$
$$(a,c) \in R_1 \implies 1+ac>0 \tag{3}$$
But (1) and (2) do not imply (3). Therefore $(a,b) \in R_1$ and $(b,c) \in R_1 \implies (a,c) \in R_1$ is not true for all $a,b,c \in \mathbb{R}$.
My problem: Is this procedure correct. Also are there any alternate approaches to prove the same result?

That (1) and (2) do not imply (3) is what you were trying to prove - you cannot assume it midway through your proof. Put another way - just because you personally don't see a way to deduce (3) from (1) and (2) doesn't mean there isn't one.
In general, to show that $A$ does not imply $B$, you must give an example of $A$ occurring without $B$. In this case, that means that you need to supply an example of $a,b,c \in \mathbb{R}$ so that $1 + ab > 0$ and $1 + bc > 0$ (that's $A$) but $1 + ac \leq 0$ (i.e., $B$ does not hold).
The simplest approach is to choose an example that's the easiest to work with - I like using $0$ a lot. $a = 0$ or $c = 0$ won't work, because then $1 + ac$ will automatically be $1$. But we could try $b = 0$. Then we automatically get $1 + ab > 0$ and $1 + bc > 0$, so we just need to choose an $a$ and $c$ so that $1 + ac \leq 0$. $a = 1$ and $c = -1$ works, so my counterexample is $a = 1$, $b = 0$, and $c = -1$.
