# $[a+b]\geq[a]+[b]$ for all a,b belongs to Real number [duplicate]

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.

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

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• If $a$ is a real number, then there is a unique $k \in \mathbb Z$ such that $k \le a <k+1$. Then $[a]:=k.$ – Fred Mar 19 at 9:40
• Confirm that your brackets denote the floor function. (If you are asked this, you must be aware.) – Yves Daoust Mar 19 at 9:41

## 4 Answers

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 $$n \leq a and $$m \leq b . Adding these we get $$n+m \leq a+b . Hence either $$n+m \leq a+b or $$n+m+1 \leq a+b . This tells us that $$[a+b]=n+m$$ or $$n+m+1$$. In either case $$[a+b] \geq [a]+[b]$$.

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

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

Hint

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

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

• I don't see how this works. $u\ge v$ and $w\le u$ does not imply $w\ge v$. – Yves Daoust Mar 19 at 9:52
• @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 $\leq a+b$. – Especially Lime Mar 19 at 10:08
• @EspeciallyLime: agreed. – Yves Daoust Mar 19 at 10:14
• @YvesDaoust The largest integer with some property is greater or equal than any fixed integer with that property ;) – N. S. Mar 19 at 14:17

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