Given integers a and b and the relation a <= b, intuitively I feel I can transform this inequality into a strict inequality like this:

a < b + 1

Conversely, I should be able to transform the strict inequality a < b to a non-strict one like this:

a + 1 <= b

My question is, is there any way I can do the same transformations if a and b are real numbers, WITHOUT introducing some kind of slack/surplus variables? I.e. is there a way to do the transformation by only using constants? I do not feel that it should be possible due to the nature of the real numbers, but I would just like to find out.

  • $\begingroup$ Excluding "dummy variables", I don't see how one could do it. But with those, I suppose you know that you can do $\forall \epsilon>0: a<b+\epsilon$ for $a\leq b$. $\endgroup$ – Raskolnikov May 6 '13 at 11:44
  • $\begingroup$ Yea, that was the only solution I could see. It complicates an algorithm I am working on though, so I just wanted to know if there was some other solution. $\endgroup$ – csvan May 6 '13 at 11:58
  • $\begingroup$ Maybe you can show us what the algorithm is and what you really want to achieve? There might be another way that doesn't involve what you're asking here. $\endgroup$ – Raskolnikov May 6 '13 at 12:04

With continous transforms, you can't.

Assume you had a continous $f$ and $b$, $c$ such that $$ x < b \text{ exactly if } f(x) \leq c \text{.} $$ Pick a sequence $(x_n)$ with $x_n \to b$ from below, i.e. $x_n < b$ for all $n$. You'd then have $$ f(x_n) \leq c \text{ for all $n$} $$ and because $f$ is continuous also $$ f(b) \leq c \text{,} $$ but that (according to the initial assumption) means $$ b < b $$ which of course is impossible.


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