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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?

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You should find it straightforward. Note the $\lambda$ can be different for different ordered pairs... – David Mitra Mar 29 '12 at 17:03
The key here (1) there is a $\lambda$ (2) is that $\lambda \neq 0.$ This allows you to use $\frac{1}{\lambda}$ for reflexivity and $\frac{\lambda_1}{\lambda_2}$ for transitivity. – user2468 Mar 29 '12 at 18:28
@J.D. thank you, this was my main point of confusion. Problem solved. – Dominick Gerard Mar 30 '12 at 14:17
up vote 2 down vote accepted

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.

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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 $\mu$?

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$\alpha = \lambda \mu$ – grayQuant Jan 26 '15 at 22:56
@grayQuant: Yes, exactly. – Brian M. Scott Jan 26 '15 at 23:04


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 $(a,b)\sim(e,f)$

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