2
$\begingroup$

For $x_1, x_2, x_3 \in \mathbb{Z}^+$, does there exist a function $f(\cdot)$ defined on $\mathbb{Z}^+$, not necessarily continuous or differentiable, such that:

$$f(x_1) > f(x_2) \\ f(x_2) > f(x_3) \\ f(x_3) > f(x_1) $$

My immediate thought is that no such function exists, since a function can only be a many-to-one or one-to-one relation, and the above would require a one-to-many relation. If so, is there a more formal way of showing that the above is impossible? Or is it trivial to observe?

If I am wrong and such a function does exist, what is an example of one?

$\endgroup$
2
$\begingroup$

A function has exactly one output.

If such a function exists, by transitivity, we would have $f(x_1) > f(x_1)$ which is a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.