Apostol's Calulus: Prove that $[x+y] = [x]+[y]$ or $[x]+[y]+1$, where $[·]$ is the floor function.

Question

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.

Answer 1

$m+n\le x+y<m+n+2\implies\lfloor x+y\rfloor=m+n$ or $m+n+1$, because the only integers in $[m+n,m+n+2)$ are $m+n, m+n+1$

Answer 2

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 $\leq$.

Answer 3

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

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.

Answer 4

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

1. if $0\le\{x\}+\{y\}<1$, then $m+n\le x+y<m+n+1$ and so $\lfloor x+y\rfloor=m+n$;
2. if $1\le\{x\}+\{y\}<2$, then $m+n+1\le x+y<m+n+2$ and so $\lfloor x+y\rfloor=m+n+1$.

Answer 5

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 $[x] + [y] + 2\ge [x+y]+1$.

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.