Proof concerning the greatest integer function

[Ragnarok7]

I'm unsure if this is a calculus or precalculus topic, but it's from a calculus book, so I'm putting it here. (Note \(\displaystyle \lfloor x \rfloor\) means the floor of \(\displaystyle x\) or the greatest integer less than or equal to \(\displaystyle x\).)

Prove that \(\displaystyle \lfloor x \rfloor +\lfloor y \rfloor \leq \lfloor x+y \rfloor \leq \lfloor x \rfloor + \lfloor y\rfloor +1 \).

I'm not really sure how to go about proofs with the floor function, at least not without splitting it up into a lot of tedious cases. Is there an elegant way to do this? Thank you!

 

[caffeinemachine]

I'm unsure if this is a calculus or precalculus topic, but it's from a calculus book, so I'm putting it here. (Note \(\displaystyle \lfloor x \rfloor\) means the floor of \(\displaystyle x\) or the greatest integer less than or equal to \(\displaystyle x\).)

Prove that \(\displaystyle \lfloor x \rfloor +\lfloor y \rfloor \leq \lfloor x+y \rfloor \leq \lfloor x \rfloor + \lfloor y\rfloor +1 \).

I'm not really sure how to go about proofs with the floor function, at least not without splitting it up into a lot of tedious cases. Is there an elegant way to do this? Thank you!

One useful and easy fact you should know about the floor function is that $\lfloor n+t\rfloor = n+\lfloor t\rfloor$ whenever $n$ is an integer and $t$ is any real.

To prove the first inequality, just write $x=m+f$ for some integer $m$ and some real $0\leq f<1$, and $y=n+g$ for some integer $n$ and some real $0\leq g<1$.

Now we have $\lfloor x+y\rfloor=\lfloor (m+n) + (f+g)\rfloor =m+n+\lfloor f+g\rfloor\geq m+n= \lfloor x\rfloor+\lfloor y\rfloor$.

The other one is similar. Can you try?

 

[Ragnarok7]

Thank you! I knew that fact but I didn't think of doing it like that. I can do the other one now.