Ytukyg

Prove that $[x+y] = [x]+[y]$ or $[x]+[y]+1$, where $[·]$ is the floor
function

I'm Having a little bit of trouble with the last part of this proof.

First, I will use the definition of floor function:

$[x] = m ≡ m ≤ x < m+1$

and

$[y] = n ≡ n ≤ y < n+1$

so, $[x]+[y] = m+n ≡ m+n ≤ x+y < m+n+2$

This is where I get stuck; I have that $[x+y] = t ≡ t ≤ x < t+1 $, so putting
$m+n ≤ x+y < m+n+2$ in that form, seems impossible, let alone the one that corresponds to $[x]+[y]+1$.

Could you help me with this last part?

Thanks in advance.

$m+nle x+y<m+n+2implieslfloor x+yrfloor=m+n$ or $m+n+1$, because the only integers in $[m+n,m+n+2)$ are $m+n, m+n+1$

You have $[x] = m$ and $[y] = n$ so $m+nle x + y < m+n+2$

But how does $x+y$ compare to $m + n + 1$? There are two possibilities.

1) $x + y < m+n + 1$ then

$m + n le x + y < m+ n + 1$. Then by definition you have $[x+y] = m+n = [x]+[y]$.

2) $x + y ge m+n + 1$ then

$m+n + 1 le x+y < m+n + 2=(m+n+1) + 1$ so by definition you have $[x+y] = m+n+1$.

That's it.

I didn't read the question exactly, but you have floor(x)=n iff. $n<x<n+1$, hence you get $...<m+n+2$ instead of $leq$.

Let $m=lfloor xrfloor$ and $n=lfloor yrfloor$; let ${x}=x-lfloor xrfloor=x-m$ and ${y}=y-lfloor xrfloor=y-n$. Then
$$
x+y=m+{x}+n+{y}
$$
Note that $0le{x}<1$ and $0le{y}<1$, so $0le{x}+{y}<2$. There are two cases:

1. if $0le{x}+{y}<1$, then $m+nle x+y<m+n+1$ and so $lfloor x+yrfloor=m+n$;


2. if $1le{x}+{y}<2$, then $m+n+1le x+y<m+n+2$ and so $lfloor x+yrfloor=m+n+1$.

Alternatively. By definition $[x+y]$ is the largest possible integer that this less than or equal to $x+y$. But $[x] le x$ and $[y] le y$ so $[x] + [y] le x+y$. So $[x]+[y] le [x+y]$.

Likewise $[x+y] + 1$ by definition is the smallest possible integer that is larger $x + y$. But $[x]+ 1 > x$ and $[y] + 1 > y$ so $[x]+[y] + 2 > x+y$. So $[x] + [y] + 2ge [x+y]+1$.

So $[x]+[y] + 1 ge [x+y]$.

So $[x]+[y] le [x+y] le [x] + [y] + 1$

As $[x]+[y]$ and $[x+y]$ and $[x] + [y]+1$ are all integers. And there is NO integer between $[x] +[y]$ and $[x]+[y]+1$ there are only two options $[x] + [y]= [x+y]$ or $[x+y] = [x] + [y]+1$.

.......

And a third way.

$[x] le x < [x]+1$ means $0 le x - [x] < 1$.

A) $0 le x-[x] < 1$ and $0 le y -[y] < 1$ so $0 le x+y -[x]-[y] < 2$.

B) $0 le x+y - [x+y] < 1$.

Reverse B) to get

B') $-1 < [x+y] - x - y le 0$.

Add B' and A to get:

$-1 < ([x+y] - x - y) + (x + y - [x]-[y]) < 2$ so

$-1 < [x+y] - [x] -[y] < 2$ or

$0 le [x+y] -[x] -[y] le 1$ or

$[x]+[y] le [x+y] le [x] + [y] + 1$.

.....

Basically: $x+y$ is between $[x+y]$ and $[x+y] + 1$; two integers that are only $1$ apart.

But $x + y$ is also between $[x] + [y]$ and $[x] + [y] + 2$; two integers that are only $2$ apart.

There are only so many choicese to find these integers $[x+y], [x]+[y], [x+y] + 1$ and $[x]+[y] + 2$ so that they all fit in such a tight range.

It doesn't matter how you prove it but you must have $[x+y]$ and $[x]+ [y]$ within one of each other and you must have $[x]+[y]le [x+y]$. There are only two ways that can happen.