Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Anyone here who can give me some help about creating formula with these rules?

  • $X$ is given to find $Y$
  • $Y ≤ X$
  • $Y$ cannot be equal $0$ -
share|improve this question

closed as not a real question by mixedmath, Qiaochu Yuan Aug 29 '11 at 18:46

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

You state that if X = 1, then Y = 1, but you also state that Y must be less than X. Isn't this impossible, since 1 is not less than 1? –  Tanner Swett Aug 22 '11 at 2:30
What is the domain (i.e., what values is $X$ allowed to take)? –  Arturo Magidin Aug 22 '11 at 2:32
You have not stated, but probably are thinking, that the function should be continuous. Then $y \ne 0$ means that $y$ is always the same side of $0$. If $y \gt 0$ you must have $x \gt 0$. I suspect Gerry Myerson's solution is not what you are thinking, but explaining why may help the definition. –  Ross Millikan Aug 22 '11 at 4:47
@Ross, I also suspect my solution is not what Ran has in mind, but I'm hoping it will stimulate Ran to clarify. –  Gerry Myerson Aug 22 '11 at 7:32
it can be y <= x.. –  Ran Gualberto Aug 22 '11 at 10:14

1 Answer 1

$f(x)=1$ if $x=1$, else $f(x)=x-1$. Satisfies all the properties except the contradiction between $y\lt x$ and $x=1$ implies $y=1$.

EDIT: Now that the question has been edited, this answer satisfies ALL the conditions. So, Ran, is this the kind of answer you want? or are there some other conditions you've forgotten to mention?

share|improve this answer

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