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I'm trying to understand properties of the greatest integer function and I am struggling to find the value of $\lfloor x+y \rfloor$ where $x \in \mathbb{R}$, $y \in \mathbb{Z}$, and prove that it is correct value.

I don't really know how to prove this, but I have been dividing it into different cases. I think that it equals $\lfloor x \rfloor + y$ when $x,y$ are both positive but not sure how to prove it. Depending on if one or both $x$ and $y$ are negative, and their ultimate sum, I get different values. I am having trouble determining when exactly this happens though and then proving the results. Any help would be great, thanks!

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  • $\begingroup$ The general result is $\lfloor x+y \rfloor = \lfloor x \rfloor + y$ for any $x \in \mathbb{R}$ and $y \in \mathbb{Z}$ $\endgroup$ – user62089 Apr 2 '13 at 3:01
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It is always $\lfloor x \rfloor + y$. Write $x = \lfloor x \rfloor + \{ x \}$ where $0 \le \{ x \} \lt 1$. Then you just throw away the $\{ x \}$ because the rest is the integer. Remember that $\lfloor -2.3 \rfloor = -3$, not $-2$

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Note, lets use $x$ as the real and $n$ as the integer.

Since $x - 1 \le \lfloor x \rfloor \le x$, it follows that $-x \le - \lfloor x \rfloor \lt -x+1.$

Combining this inequality with $x+n - 1 \lt \lfloor x+n \rfloor \le x + n$, we obtain $n-1 \lt \lfloor x+n \rfloor - \lfloor x \rfloor \lt n+1.$

Hence $\lfloor x+n \rfloor - \lfloor x \rfloor = n.$

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  • $\begingroup$ $\color{blue}{\bf \ddot\smile}\color{red}{\bf +1}$ $\endgroup$ – Namaste Apr 7 '13 at 0:39
  • $\begingroup$ @amWhy: This is one of those examples where a bunch of excellent answers were given, but not really ack'ed, so never sure if the OP got it. Oh well! $\endgroup$ – Amzoti Apr 7 '13 at 0:40
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You can work directly from the definition. Let $n=\lfloor x\rfloor$; then $n\le x<n+1$. Now add the integer $y$ to the inequality to get (after very minor rearrangement of the righthand side) $$n+y\le x+y<(n+y)+1\;.$$ But since $n+y$ is an integer, that means precisely that $\lfloor x+y\rfloor=n+y$, by the definition of the floor function.

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Hint $\ $ Using the universal property of floor makes such proofs mechanical, e.g.

$$\rm n \le \color{#0A0}{x + k}\iff n\!-\!k \le x \iff n\!-\!k\le \lfloor x\rfloor\iff n\le \color{#C00}{\lfloor x\rfloor \!+\!k} $$

Therefore we deduce $\rm\ \lfloor \color{#0A0}{x + k}\rfloor = \lfloor \color{#C00}{\lfloor x\rfloor \!+\!k}\rfloor = \color{}{\lfloor x\rfloor \!+\!k}$

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