Question

For which real numbers xand yis it true that(x+y) =

[x] + [y]?

Short answer

Sum of the fractional parts of x and y is at least 1 or if both x and y are an integer.

Step-by-step solution

Step 1:

Definitions:

Celling function [x]: smallest integer that is greater than or equal to x

Floor function [x]: largest integer that is less than or equal to x.

Step 2:

Given,

$[x+y]=\left[x\right]+\left[y\right]$

X and y are real numbers.

1) Let us assume x and y are integers

X+y is also integer because x and y are integers.

The floor function of an integer is also an integer

$[x+y]=x+y=\left[x\right]+\left[y\right]$

2) Let assume x is an integer and y is not an integer

There exits an integer n and real number s with 0<s<1

$y=n+s$

From previous solution,

$x+y=x+n+s$

The celling function of an integer is an integer itself

$[x+y]=x+n+1\ne x+n=\left[x\right]+\left[y\right]$

3) Let assume x is not an integer and y is a function

There exits an integer m and real number r with 0<r<1

$x=m+r$

Here, we get

$x+y=y+m+r$

The celling function of an integer is an integer itself

$[x+y]=m+y+1=(m+1)+y=\left[x\right]+\left[y\right]$

Step 3:

Let assume x and y are not integers, there exists real numbers m and n and real numbers r and s with 0<r<1, 0<s<1

From celling and floor function

$\left[x\right]=m+1$

$\left[y\right]=n$

Step 4:

Let us expression for x and y

$x+y=(m+n)+(r+s)$

$m+n$is an integer, m and n are integers

$0<r+s<2$

If $0<r+s\le 1,$then $[x+y]=m+n+1$

If $1<r+s<2,$then $[x+y]=m+n+2$

Then, $[x+y]=m+n+1=\left[x\right]+\left[y\right]$