I have to prove the following property of Floor function:

For any real number $x$, $x$ not being an integer, $\lfloor x \rfloor + \lfloor -x \rfloor = -1$.

Now, we know from the definition of floor that $\lfloor x \rfloor$ is the unique integer $n$ such that $n \leq x < n+1$. The trouble is writing $\lfloor -x \rfloor$. If I imagine a negative real number on the number line, it is obvious that $-n-1 \leq -x < -n$. Then, simply adding the two yields -1.

My problem is that I can't seem to arrive at $\lfloor -x \rfloor$ from the definition. For instance, if $x$ is a postive real number, then the floor is given by

$$n \leq x < n+1$$

Multiplying by -1 throughout,

$$-n \geq -x > -n-1$$

$$\Rightarrow -n-1 < x \leq -n$$

Feels like I'm almost there, but this does not match the definition (the $\leq$ has appeared on the upper bound). What am I doing wrong?

  • 2
    $\begingroup$ And where do you use the fact that $x$ is not an integer? $\endgroup$ – Mark Bennet Jul 19 '13 at 11:32
  • $\begingroup$ Well, I have to admit I don't feel very confident. However, if $x$ were an integer, the answer would be 0 because the floor in both the directions would equal the number in magnitude. I guess the whole point of working with the inequality is to preserve generality? $\endgroup$ – ankush981 Jul 19 '13 at 11:35
  • $\begingroup$ And what happens in the case of equality you don't like? $\endgroup$ – Mark Bennet Jul 19 '13 at 11:38
  • $\begingroup$ I'm not sure I follow you. Could you please provide a more direct hint? $\endgroup$ – ankush981 Jul 19 '13 at 11:41
  • $\begingroup$ Ah! After reading the answer below, I think I see what you mean. Because $x$ is not an integer, we can't have the equality sign, right? But that leaves me in even more trouble, I suspect. Let me think some more and get back. $\endgroup$ – ankush981 Jul 19 '13 at 11:48

Put $$\lfloor x\rfloor=:n,\qquad \lfloor-x\rfloor=: n'\ .$$ Then the definition of $\lfloor\cdot\rfloor$ implies together with $x\notin{\mathbb Z}$ that we have $$n<x<n+1\qquad\wedge\qquad n'<-x<n'+1\ .$$ Adding the left parts of these two squeezes gives $n+n'<0$, and adding the right parts leads to $n+n'>-2$. All in all only $n+n'=-1$ remains possible.

  • $\begingroup$ Beautiful answer! Thank you very much! :) $\endgroup$ – ankush981 Jul 19 '13 at 14:29

Let us show that $-1-[-x]$ satisfies the definition of $[x]$, for $x$ not an integer. The definition of $[x]$ would be $$[x]\leq x<[x]+1,$$

i.e. we need to check that $$-1-[-x]\leq x<-[-x].$$

The inequality on the right, $[-x]<-x$, is the first inequality of the definition of $[-x]$, i.e. $[-x]\leq-x<[-x]+1$, and therefore it is true. Notice the equality doesn't hold on the left, because it is not an integer.

The inequality on the left is $-x\leq [-x]+1$, which is equivalent to $-x<[-x]+1$, again because it is not an integer. And this is the other part of the definition of $[-x]$, and therefore it is true.

This proves that $$-1-[-x]=[x].$$

  • $\begingroup$ Thanks for the effort, but I think I'm getting confused following your answer. Could you please label the inequalities as (1), (2), etc., and rewrite the answer? Here are a few examples of how I'm getting confused: "The equality on the right, $[−x] < −x$" Did you mean inequality? "The inequality on the right is $−x \leq [−x]+1$" Where does this come from? Also, is $x$ is not an integer, how can we aim to establish that $-1-[-x] \leq x$, as the left part is an integer? $\endgroup$ – ankush981 Jul 19 '13 at 13:14

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