MHB Proof concerning the greatest integer function

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SUMMARY

The discussion focuses on proving the inequalities involving the greatest integer function, specifically that $$\lfloor x \rfloor +\lfloor y \rfloor \leq \lfloor x+y \rfloor \leq \lfloor x \rfloor + \lfloor y\rfloor +1$$. The proof utilizes the property that $$\lfloor n+t\rfloor = n+\lfloor t\rfloor$$ for an integer $$n$$ and real $$t$$. By expressing $$x$$ and $$y$$ in terms of their integer and fractional parts, the proof demonstrates the validity of both inequalities effectively.

PREREQUISITES
  • Understanding of the greatest integer function (floor function)
  • Familiarity with basic properties of real numbers
  • Knowledge of integer and fractional parts of numbers
  • Basic proof techniques in mathematics
NEXT STEPS
  • Study the properties of the floor function in depth
  • Learn about inequalities involving real numbers
  • Explore proof techniques in calculus and precalculus
  • Practice problems involving the floor function and related inequalities
USEFUL FOR

Students in calculus or precalculus, mathematics educators, and anyone interested in understanding the properties of the greatest integer function and its applications in proofs.

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.
 
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