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

Question

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.

Answer 1

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

Answer 2

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

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

Answer 3

Hint

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

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

Answer 4

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