Prove E is an equivalence relation over $ℝ^2$

Question

Consider the relation E over $ℝ^2$ defined as follows:

$(x_1, y_1) \space E \space (x_2,y_2)$ if $ \exists k \in ℝ (k \neq 0 \land (kx_1, ky_1) = (x_2, y_2)$

I was wondering if anyone has a better method of proving this and also if you can critique my writing to see where i need more work.

My working:

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Lemma 1: The binary relation $(kx_1, ky_1) E (x_2, y_2)$ is reflexive.

Proof:

Consider an arbitrary $x_2, y_2 \in ℝ^2$. We will prove that $(kx_1, ky_1) = (x_2, y_2)$ is reflexive, where $k$ \in ℝ.

Lets assume $x_1 = x_2, y_1 = y_2$ and $k_1=1$. Notice that with these values $(kx_1, ky_1) = (x_2, y_2)$. Therefore $(kx_1, ky_1) E (x_2, y_2)$ is reflexive.

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Lemma 2: The binary relation $(kx_1, ky_1) E (x_2, y_2)$ is symmetric.

Proof: Consider an arbitrary $x_2, y_2 \in ℝ^2$. We will prove that, $(kx_1, ky_1) E (x_2, y_2)$ is symmetric. Lets assume $(kx_1, ky_1) = (x_2, y_2)$ where $k$ is an integer. We will now prove $(ax_2, ay_2) E (x_2, y_2)$, where a is a real number.

Lets assume the real number, $a=1/k$. We can see that $(ax_2, ay_2) = (x_1, y_1)$. Therefore, we can conclude that $(x_1, y_1) \space E \space (x_2, y_2)$ is symmetric.

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Lemma 3: The binary relation $(x_1, y_1) \space E \space (x_2, y_2)$ is transitive.

Proof: Consider $x_2, y_2$ to be arbitrary integers, such that $x_2 \in ℝ^2 \cap y_2 \in ℝ^2 $. Lets assume there exists an integer, $k$ where $k \neq 0$ and $(kx_1, ky_1) = (x_2, y_2)$. We will prove that $(x_1, y_1) \space E \space (x_2, y_2)$ is transitive.

Lets assume $x_3, y_3$ are arbitary integers where $x_3 \in ℝ^2 \cap y_3 \in ℝ^2$. To prove that $(x_1, y_1) \space E \space (x_2, y_2)$ is transitive, consider $a$ to be an integer where $(ax_2, ay_2) = (x_3, y_3)$.

Notice that $(x_1, y_1) E (x_2, y_2) \cap (x_2, y_2) E (x_3, y_3)$. By multiplying $x$, and $y$, by the integers a and k we can see that $(x_1, y_1) E (x_3, y_3)$.

Therefore we can conclude that $(x_1, y_1) E (x_2, y_2)$ is transitive.

Answer 1

Just $1$ comment:

Lemma 1: The binary relation $(kx_1, ky_1) E (x_2, y_2)$ is reflexive.

You may want to write $E$ instead of the whole $(kx_1, ky_1) E (x_2, y_2)$ (in several other places as well).

The rest of your work seems legitimate to me.