# Can I apply the floor function to the left- and right hand side this way?

I'm wondering if I can apply the floor function to both sides like below? This isn't all I want to do I just want to know if the operation is legal. Thanks! $$a + x < b + y \to \lfloor a + x \rfloor < \lfloor b + y \rfloor$$

EDIT: To clarify I'd only like to know if the operation is allowed. What I'd actually like to is prove $$a \leq b \to \lfloor a \rfloor \leq \lfloor b\rfloor$$ but I want to know if I can come up with this proof by applying the floor function operation to both sides kind of like adding 1 to both sides makes the inequality still hold.

• NO! For example 1.5 < 1.8 but [1.5]=1=[1.8] Aug 5 '19 at 15:23

If $$f$$ is a function, by definition, if $$a=b$$ then $$f(a)=f(b)$$. The floor function is an example of a function and so you have $$a=b$$ implies $$\lfloor a\rfloor = \lfloor b\rfloor$$.

Now... some functions will retain inequalities too. If a function is monotonic increasing that means by definition that if $$a\leq b$$ then $$f(a)\leq f(b)$$. The floor function is an example of a monotonic increasing function. That is to say, if $$a\leq b$$ then it is true that $$\lfloor a\rfloor \leq \lfloor b\rfloor$$.

Not every function is monotonic increasing however. Take for example $$f(x)=x^2$$. You have $$-3\leq 1$$ but $$(-3)^2\not\leq (1)^2$$.

Finally, some functions will retain strict inequalities as well. If a function is strictly monotonic increasing that means by definition that if $$a then $$f(a). The floor function is not strictly monotonic, for example how $$1.2<1.5$$ but $$\lfloor 1.2\rfloor \not\lt \lfloor 1.5\rfloor$$ since they both result in the same value of $$1$$.

As for a hint on how to prove that $$a\leq b\implies \lfloor a\rfloor \leq \lfloor b\rfloor$$, rewrite $$b$$ as $$a+(b-a)$$ and recognize that $$(b-a)\geq 0$$. Further, apply the definition of the floor function as being the largest integer less than or equal to its input.

Let $$x\geq 0$$. Then let $$c = \lfloor a\rfloor$$ be the largest integer less than or equal to $$a$$. It follows that $$c$$ is an integer which is less than or equal to $$a+x$$, whether it is the largest one or not... and so whatever the largest integer less than or equal to $$a+x$$ is will be at least as large as $$c$$. As such we have $$\lfloor a\rfloor \leq \lfloor a+x\rfloor$$, showing that the floor function is monotonically increasing.

• Thanks, this was helpful! Aug 5 '19 at 15:58

We can write, using the fractional part, that \eqalign{ & a \le b\quad \Rightarrow \quad \left\{ \matrix{ \left\lfloor a \right\rfloor + \left\{ a \right\} \le \left\lfloor b \right\rfloor + \left\{ b \right\} \hfill \cr 0 \le \left\{ a \right\},\left\{ b \right\} < 1 \hfill \cr} \right. \cr & \left\{ a \right\} - \left\{ b \right\} \le \left\lfloor b \right\rfloor - \left\lfloor a \right\rfloor \cr}

At the same time, you can write your inequality as \eqalign{ & a \le b\quad \Rightarrow \quad 0 \le b - a\quad \Rightarrow \cr & \Rightarrow \quad 0 \le \left\lfloor {b - a} \right\rfloor + \left\{ {b - a} \right\}\quad \Rightarrow \cr & \Rightarrow \quad 0 \le \left\lfloor {\left\lfloor b \right\rfloor - \left\lfloor a \right\rfloor + \left\{ b \right\} - \left\{ a \right\}} \right\rfloor + \left\{ {b - a} \right\}\quad \Rightarrow \cr & \Rightarrow \quad 0 \le \left\lfloor b \right\rfloor - \left\lfloor a \right\rfloor + \left\lfloor {\left\{ b \right\} - \left\{ a \right\}} \right\rfloor + \left\{ {b - a} \right\}\quad \Rightarrow \cr & \Rightarrow \quad 0 \le \left\lfloor b \right\rfloor - \left\lfloor a \right\rfloor - \left[ {\left\{ b \right\} < \left\{ a \right\}} \right] + \left\{ {b - a} \right\} \cr} where the square brackets denote the Iverson bracket

And, apart from obvious manipulations, I do not know other ways to put it for, general $$a, \, b$$.