Does $k\cdot x < y$ imply that $x \ne y$?

Question

Sorry if the question is trivial, but I have trouble getting my head around it.

To keep short, does $\forall k \in \mathbb Z^*, \forall (x, y) \in \mathbb Z^2, k\cdot x < y \implies x \neq y$?

($\mathbb Z^* = \mathbb Z - \{0\}$)

I work in $\mathbb Z$ but a proof in $\mathbb R$ is problably easier. I conjecture that statement is true, but I cannot find a formal proof or counter-example.

[edit] The initial question was false as there is that trivial counter-example with $k = 2, (x, y) = (-1, -1)$. So I have an extension.

Given $k_x \cdot x < k_y \cdot y$ with $k_x, k_y \in \mathbb Z^*, x, y \in \mathbb Z$, is there a relation between $k_x$ and $k_y$ which implies that $x\neq y$? It is true for $k_x = k_y$, but are there other cases?

Answer 1

For the updated question (is there a relation between $k_x$ and $k_y$ besides $k_x=k_y$ which ensures that $k_x\cdot x<k_y\cdot y\to x\neq y$?), the answer is no.

Suppose $k_x\neq k_y$. Then either

• $k_x<k_y$, in which case $x=y=1$ is a counterexample, or
• $k_x>k_y$, in which case $x=y=-1$ is a counterexample.

Answer 2

Oh, well, I got to this counter-example :

$k = 2, x = -2, y=-2$.