MHB Proof concerning the greatest integer function

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The discussion focuses on proving the inequalities involving the greatest integer function, specifically that $$\lfloor x \rfloor +\lfloor y \rfloor \leq \lfloor x+y \rfloor \leq \lfloor x \rfloor + \lfloor y\rfloor +1$$. A key insight shared is that for an integer $$n$$ and a real number $$t$$, the relationship $$\lfloor n+t\rfloor = n+\lfloor t\rfloor$$ can simplify the proof. By expressing $$x$$ and $$y$$ in terms of their integer and fractional parts, the first inequality is established. The second inequality can be proven using a similar approach, leveraging the properties of the floor function. This method provides a more elegant solution than splitting into multiple cases.
Ragnarok7
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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 $$\lfloor x \rfloor$$ means the floor of $$x$$ or the greatest integer less than or equal to $$x$$.)

Prove that $$\lfloor x \rfloor +\lfloor y \rfloor \leq \lfloor x+y \rfloor \leq \lfloor x \rfloor + \lfloor y\rfloor +1 $$.

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!
 
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Ragnarok said:
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 $$\lfloor x \rfloor$$ means the floor of $$x$$ or the greatest integer less than or equal to $$x$$.)

Prove that $$\lfloor x \rfloor +\lfloor y \rfloor \leq \lfloor x+y \rfloor \leq \lfloor x \rfloor + \lfloor y\rfloor +1 $$.

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?
 
Thank you! I knew that fact but I didn't think of doing it like that. I can do the other one now.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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