Proving an Equivalence Relation Define a relation $\sim$ on $\mathbb{R} \times \mathbb{R} = \mathbb{R}^2$ by setting $(a,b) \sim (c,d)$ if there is a nonzero real number $\lambda$ such that $(a,b) = (\lambda c, \lambda d)$. Prove that $\sim$ is an equivalence relation.
I am confident that I can prove the reflexive property for this, but I can't seem to show symmetry or transitivity. What is the best way to do this?
 A: if $(a,b)\sim (c,d)$, then $(a,b)=(\lambda c, \lambda d)$. As $\lambda$ is nonzero, you have thus $(\frac{1}{\lambda}a,\frac{1}{\lambda}b)=(c,d)$, hence $(c,d)\sim(a,b)$ and $\sim$ is symmetric.
If $(a,b)\sim (c,d)$ and $(c,d) \sim (e,f)$, then there are $\lambda, \lambda'$ such that $(a,b)=(\lambda c,\lambda d)$ and $(c,d)=(\lambda' e, \lambda' f)$. It follows that $(a,b)=(\lambda\lambda' e,\lambda\lambda' f)$, hence $(a,b)\sim (e,f)$ and $\sim$ is transitive.
A: Suppose that $(a,b)\sim(c,d)$; you want to show that $(c,d)\sim(a,b)$.
Translate this into something more basic: by definition there is some non-zero $\lambda$ such that 
$$\begin{align*}a&=\lambda c\\
b&=\lambda d\;,
\end{align*}\tag{1}$$ and you want to show that there is a non-zero $\mu$ such that 
$$\begin{align*}c&=\mu a\\
d&=\mu b\;.
\end{align*}\tag{2}$$
Knowing that $\lambda\ne 0$, you should be able to use $(1)$ to solve $(2)$ for $\mu$ in terms of $\lambda$ without any real trouble.
You should approach transitivity the same way. Suppose that $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$. Then you know that there are non-zero $\lambda$ and $\mu$ such that 
$$\begin{align*}a&=\lambda c\\
b&=\lambda d\\
c&=\mu e\\
d&=\mu f\;,
\end{align*}\tag{3}$$
and you want to show that there is a non-zero $\alpha$ such that 
$$\begin{align*}
a&=\alpha e\\
b&=\alpha f\;.
\end{align*}$$
Can you see how to combine the information in $(3)$ to get this $\alpha$ in terms of $\lambda$ and $\mu$?
A: Transitivity:  
We need to show:
$(a,b)\sim(c,d),(c,d)\sim(e,f)\Rightarrow (a,b)\sim(e,f)$


*

*$(a,b)\sim(c,d)$ - exist $\lambda$ such that $(a,b)=(\lambda c,\lambda d)$

*$(c,d)\sim(e,f)$ - exist $\sigma$ such that $(c,d)=(\sigma e,\sigma f)$


From the two above we easily conclude that:
$(a,b)=(\lambda\sigma e,\lambda\sigma f)$, or $(a,b)\sim(e,f)$
