How to prove that $\lfloor a\rfloor+\lfloor b\rfloor\leq \lfloor a+b\rfloor$

Question

We have the floor function $F(x)=[x]$ such that $F(2.5)=[2.5]=2, F(-3.5)=[-3.5]=-4, F(2)=[2]=2$.

How can I prove the following property of floor function: $$[a]+[b] \le [a+b]$$

Answer 1

Consider the algebraic representation of $\mathrm{floor}(x)=\lfloor x\rfloor$ (usually the floor of $x$ is represented by $\lfloor x\rfloor$, not $[x]$): $$ x-1<\color{blue}{\lfloor x\rfloor\leq x}. $$ Hence, for arbitrary $a$ and $b$, we have $$ \lfloor a\rfloor\leq a\tag{1} $$ and $$ \lfloor b\rfloor\leq b.\tag{2} $$ Now simply add $(1)$ and $(2)$ together to get $$ \lfloor a\rfloor+\lfloor b\rfloor\leq a+b.\tag{3} $$ Finally, take the floor of both sides of $(3)$: $$ \lfloor a+b\rfloor\geq\Bigl\lfloor\lfloor a\rfloor+\lfloor b\rfloor\Bigr\rfloor=\lfloor a\rfloor+\lfloor b\rfloor. $$ Hence, we have that $\lfloor a\rfloor+\lfloor b\rfloor\leq \lfloor a+b\rfloor$, as desired.

Answer 2

Define $\left\{ a\right\} ,\left\{ b\right\}$ the fractional part of $a$ and $b$ and assume $\left\{ a\right\} +\left\{ b\right\} <1$ . Then$$\left\{ a+b\right\} =\left\{ \left[a\right]+\left[b\right]+\left\{ a\right\} +\left\{ b\right\} \right\} =\left\{ \left\{ a\right\} +\left\{ b\right\} \right\} =\left\{ a\right\} +\left\{ b\right\}.$$ So$$\left[a+b\right]=a+b-\left\{ a+b\right\} =a+b-\left\{ a\right\} -\left\{ b\right\} =\left[a\right]+\left[b\right].$$ Assume now $\left\{ a\right\} +\left\{ b\right\} \geq1$ , then$$\left\{ a+b\right\} =\left\{ \left[a\right]+\left[b\right]+\left\{ a\right\} +\left\{ b\right\} \right\} =\left\{ \left\{ a\right\} +\left\{ b\right\} \right\} =\left\{ a\right\} +\left\{ b\right\} -1$$ so$$\left[a+b\right]=a+b-\left\{ a+b\right\} =a+b-\left\{ a\right\} -\left\{ b\right\} +1>\left[a\right]+\left[b\right].$$

Answer 3

First, note that we always have $[x] \le x < [x] + 1$. Thus, $0 \le x - [x] < 1$ for all real $x$. To show your inequality, let $u = a - [a]$ and $v = b - [b]$. My claim is that, if $u + v < 1$, then $[a] + [b] = [a + b]$, and if $u + v \ge 1$, then $[a] + [b] < [a + b]$. Can you show that?

Answer 4

Working along similar lines to an answer I gave to another question, note that $f:\mathbb R\rightarrow\mathbb Z$ given by $f(x)=\lfloor x\rfloor$ is an increasing function, that is the identity on the integers. Also note that $f(x)\leq x$. Thus $$ f(a)+f(b)\leq a+b $$ and applying the increasing function $f$ on both sides, noting that the LHS is an integer, we have $$ f(f(a)+f(b))=f(a)+f(b)\leq f(a+b) $$

Answer 5

Let $a,b \in \Bbb{R}$ such that $a = n+x$ and $b = m+y$ where $x,y \in [0,1)$ and $n,m \in \Bbb{Z}$. Then $$\lfloor a\rfloor+\lfloor b\rfloor = \lfloor n+x\rfloor+\lfloor m+y\rfloor \\ = n+m$$

Case 1: $x+y\geq 1$. Then $x+y = 1+z$ for $z \in [0,1)$ and $\lfloor a+b\rfloor = n+m+1$, demonstrating that $$\lfloor a\rfloor+\lfloor b\rfloor < \lfloor a+b\rfloor $$

Case 2: $x+y<1$. Then $\lfloor a+b\rfloor = n+m$ so $$\lfloor a\rfloor+\lfloor b\rfloor = \lfloor a+b\rfloor $$ The combination of cases has shown that $$\lfloor a\rfloor+\lfloor b\rfloor \leq \lfloor a+b\rfloor $$ for all $a,b \in \Bbb{R}$.

Answer 6

By the definition of $[x]$ as the greatest integer less than or equal to $x$, we have

$$[a]\le a\quad\text{and}\quad [b]\le b$$

and thus

$$[a]+[b]\le a+b$$

Now $[a]+[b]$ is certainly an integer (because it's the sum of two integers). Therefore, by the definition of $[x]$ as the greatest integer less than or equal to $x$, we have

$$[a]+[b]\le [a+b]$$