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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?

Math1000
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scand1sk
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2 Answers2

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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.
Klaus Draeger
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Oh, well, I got to this counter-example :

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

scand1sk
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    You claiming this as a counter example leads me to believe you never wrote out the question properly because $-2 \not\in \mathbb{Z}^2$. – anak Jul 10 '15 at 12:54
  • Yes, I was meaning that $(x,y) \in \mathbb Z^2$. I forgot parentheses, sorry. – scand1sk Jul 10 '15 at 12:56