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At first I think of triangle inequalities but it is totally different. Then I consider that is this bracket symbolize anything in maths or these were just normal square brackets. I am really not getting anything please help me with the proof.

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marked as duplicate by Martin R, StubbornAtom, dantopa, Lee David Chung Lin, Alex Provost Mar 20 at 2:21

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  • $\begingroup$ If $ a $ is a real number, then there is a unique $k \in \mathbb Z$ such that $k \le a <k+1$. Then $[a]:=k.$ $\endgroup$ – Fred Mar 19 at 9:40
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    $\begingroup$ Confirm that your brackets denote the floor function. (If you are asked this, you must be aware.) $\endgroup$ – Yves Daoust Mar 19 at 9:41
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In usual notation $[x]$ is the largest integer less than or equal to $x$. It is uniquely determined by the inequalities $[x] \leq x <[x]+1$. If $[a]=n$ and $[b]=m$ then $n \leq a <n+1$ and $m \leq b <m+1$. Adding these we get $n+m \leq a+b <n+m+2$. Hence either $n+m \leq a+b <n+m+1$ or $n+m+1 \leq a+b <n+m+2$. This tells us that $[a+b]=n+m$ or $n+m+1$. In either case $[a+b] \geq [a]+[b]$.

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You can pull an integer out of the floor function. Then $$\left\lfloor a+b\right\rfloor=\left\lfloor\lfloor a\rfloor+\{a\}+\lfloor b\rfloor+\{b\}\right\rfloor=\left\lfloor\{a\}+\{b\}\right\rfloor+\lfloor a\rfloor+\lfloor b\rfloor.$$

It is immediate that

$$\left\lfloor\{a\}+\{b\}\right\rfloor\ge0.$$


We can add that equality holds when

$$\{a\}+\{b\}<1.$$

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Hint

$$a+b \geq [a]+[b]$$

Himt 2 $[a+b]$ is the largest integer smaller than $a+b$.

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  • $\begingroup$ I don't see how this works. $u\ge v$ and $w\le u$ does not imply $w\ge v$. $\endgroup$ – Yves Daoust Mar 19 at 9:52
  • $\begingroup$ @YvesDaoust it works because $\lfloor a\rfloor+\lfloor b\rfloor$ is an integer $\leq a+b$, so it can't be bigger than the largest integer $\leq a+b$. $\endgroup$ – Especially Lime Mar 19 at 10:08
  • $\begingroup$ @EspeciallyLime: agreed. $\endgroup$ – Yves Daoust Mar 19 at 10:14
  • $\begingroup$ @YvesDaoust The largest integer with some property is greater or equal than any fixed integer with that property ;) $\endgroup$ – N. S. Mar 19 at 14:17
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The floor function is non-decreasing, so

$$\left\lfloor a+b\right\rfloor\ge\left\lfloor\lfloor a\rfloor+\lfloor b\rfloor\right\rfloor$$

and $$\left\lfloor\lfloor a\rfloor+\lfloor b\rfloor\right\rfloor=\lfloor a\rfloor+\lfloor b\rfloor.$$

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