Establishing an equivalence relation for a particular question

Question

Condition:For $x,y \in R^{n+1} \setminus \{0\} $ define: $x\sim y$ iff $y = \lambda x $ for some $\lambda \in \Bbb R$, $ \lambda \ne 0$

1. $x = \lambda x \implies \lambda = 1$ which is a scalar so $x\sim x$

2. $y = \lambda x \implies x = \frac{1}{\lambda} y \implies y\sim x$ because $\frac{1}{\lambda} $ is still a scalar

3. $x\sim y \implies y = \lambda x $ and if $y\sim z \implies z = \delta y $, therefore $z = \delta (\lambda x) = (\delta \lambda)x \implies x\sim z$

Did I establish the relations correctly?

Answer 1

The proofs are roughly correct, but are missing the proper sequence of logical steps. To show that a binary relation $\sim$ is an equivalence relation, we must show reflexivity, symmetry, and transitivity. Perhaps it is useful to see the proof written out in gory detail for one of the properties.

Reflexivity: The reflexivity property states that for any $x \in \mathbb{R}^{n+1} \setminus \{0\}$, we must have $x \sim x$. This is true if and only if for any such $x$ there exists $\lambda \in \mathbb{R} \setminus \{ 0 \}$ with $x = \lambda x$. By the properties of scalar multiplication, selecting $\lambda = 1$ implies $x = x$. Therefore, for any $x \in \mathbb{R}^{n+1} \setminus \{ 0 \}$ we have $x \sim x$. Note that this is exactly the converse implication from the one you wrote.

In general, you want to state the desired result, start from what is given (e.g. we are given arbitrary $x$ in the first property; $y, x,$ and $\lambda$ such that $y = \lambda x$ in the second property, etc.), and show through a sequence of implications that the desired result follows. The final result (e.g. $x \sim y, y \sim z \implies x \sim z$) should be the rightmost statement in a series of $\implies$, not the leftmost. Do you see how to fix the logical deductions for the other properties?