Dot Products of K-Vectors That Are Not Equal to Zero (Proof Verification)

Question

I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.

The Prompt:

If $k \ge{2}$ and $x ∈ \mathbb{R}^k$, prove that there exists $y ∈ \mathbb{R}^k$ such that $y \neq{0}$, but $x • y= 0$. Is this also true if k = 1 ?

My Proof:

Suppose $k \ge{2}$ and $x ∈ \mathbb{R}^k$. Suppose $x = 0$. Then there exists $y$ such that $y_i = 1$ for all $i$ and then $x • y= 0$. Now suppose that $x \neq{0}$. Define $y$ such that $y_i = -\frac{\sum_{j=1}^{i-1}{x_j} + \sum_{j=i+1}^{k}{x_j}}{x_i}$ for the last $i$ such that $x_i \neq{0}$ (and there must be at least one such $x_i$ if $x\neq{0}$). Now define all other $y_i=1$. Then:

$x • y = \sqrt{\sum_{a=1}^k{x_a}{y_a}} = \sqrt{\sum_{a=1}^{i-1}{x_a} + x_i(-\frac{\sum_1^{i-1}x_a + \sum_{i+1}^{k}x_a}{x_i}) + \sum_{a=i+1}^{k}{x_a}} = \sqrt{\sum_{a=1}^{i-1}{x_a} - \sum_{1}^{i-1}{x_a} - \sum_{i+1}^{k}{x_a} + \sum_{a=i+1}^{k}{x_a}} = 0$

So if $k \ge{2}$ and $x ∈ \mathbb{R}^k$, there exists $y ∈ \mathbb{R}^k$ such that $y \neq{0}$, but $x • y= 0$.

Now suppose $k=1$. If $x=0$, then there exists $y=1$ such that $x • y= 0 * 1 = 0$, so $x • y= 0$ is true. However, suppose $x\neq{0}$ and $x • y= 0$. Then:

$x • y= x*y = 0$

$y = \frac{0}{x} = 0$ since x is non-zero.

Thus, $y$ must be zero, so there does not exist a $y\neq{0}$ such that $x • y= 0$ when $k=1$.

Answer 1

It looks correct, but it is too long.

$k>1$: Either at least one of the coordinates of $x$ is $0$ or none is. In the first case, and if, say, the second coordinate of $x$ is $0$, take $y=(0,1,0,0,\ldots,0)$. Otherwise, if $x=(x_1,x_2,x_3,\ldots,x_k)$, take $y=(x_2,-x_1,0,\ldots,0)$.

$k=1$: If $x=(x_1)$ with $x_1\ne0$ and $y=(y_1)$ with $y_1\ne0$, then $x.y=(x_1y_1)$, and $x_1y_1\neq0$.