Verify equivalence relation solution for $(a,b)R(c,d) \iff \frac{c}{a}=\frac{b}{d}, ac \neq 0$

Question

Question

Let $X$ be the set of all ordered pairs $(x,y)$ of real numbers such that $xy \neq 0$. Define a relation $R$ on $X$ as follows: $(a,b)R(c,d) \iff \frac{c}{a}=\frac{b}{d}, ac \neq 0$.

Prove that $R$ is an equivalence relation on $X$

Answer

For relexiveness; let $a,b \in \Bbb{R}$ s.t. $\frac{a}{a}=\frac{b}{b} \iff (a,b)R(a,b), ab \neq0$, thus $R$ is relexive.

For symmetry; let $a,b,c,d \in \Bbb{R}, ab \neq 0$ s.t. $(a,b)R(c,d) \iff \frac{c}{a}=\frac{b}{d}$. I need to show that $(c,d)R(a,b)$, therefore $$\frac{c}{a}=\frac{b}{d}\\ c=\frac{ba}{d} \\ dc=ba \\ d=\frac{ba}{c} \\ \frac{d}{b}=\frac{a}{c} \iff (c,d)R(a,b)$$ Hence $R$ is symmetric.

For Transitivity, I hit a block, which is where I need further help. Let $a,b,c,d,e,f \in \Bbb{R}$ s.t. $(a,b)R(c,d) \iff \frac{c}{a}=\frac{b}{d}$ and $(c,d)R(e,f) \iff \frac{e}{c}=\frac{d}{f}$. I need to show that $(a,b)R(e,f) \iff \frac{e}{a}=\frac{b}{f}$.

Since $\frac{c}{a}=\frac{b}{d} \Rightarrow c=\frac{ab}{d} \text{ and } d=\frac{ab}{c}$. And when substituted into $\frac{e}{c}=\frac{d}{f} \Rightarrow \frac{e}{\frac{ab}{d}}=\frac{\frac{ab}{c}}{f} \Rightarrow\frac{ed}{ab}=\frac{ab}{ef} \Rightarrow \frac{ed(ef)-ab(ab)}{abef}$. Somewhat stuck as to how to proceed from here. Any help appreciated. I assume it is transitive.

Answer 1

If $$X = \{(x,y) \in \mathbb R^2 : xy\neq 0\},$$ then, for $(x,y) \in X$ we have $x$ and $y$ both different than $0$, and you don't have to to say that $ac \neq 0$, in the definition of $R$, because $\mathbb R$ is a field and so $ac \neq 0$ if $a,c \in \mathbb R \setminus \{0\}$.

Concerning the proof that $R$ is an equivalence relation, what you need is

• To prove that $R$ is reflexive, show that $(x,y)R(x,y)$ for all $(x,y) \in X$.
  By definition, this is so iff $a/a = b/b$, which is the case.
• To prove that $R$ is symmetric, pick $(a,b),(c,d) \in X$ such that $(a,b)R(c,d)$; you need to prove that $(c,d)R(a,b)$.
  From $(a,b)R(c,d)$, it follows that $c/a=b/d$, whence $$a/c = \frac{1}{c/a} = \frac{1}{b/d}=d/b,$$ and thus, $(c,d)R(a,b)$.
• To prove transitivity, pick $(a,b),(c,d),(e,f) \in X$ such that $(a,b)R(c,d)$ and $(c,d)R(e,f)$, and prove that $(a,b)R(e,f)$. From the hypothesis, $c/a=b/d$ and $e/c=d/f$, and you must prove that $e/a=b/f$.
  Now, $$\frac{e}{a}=\frac{e}{c}\frac{c}{a}=\frac{d}{f}\frac{b}{d}=\frac{b}{f}.$$