Difference of two greatest integer functions

Question

If $[x]$ is the greatest integer less than or equal to x, then from the graph of the greatest function, and splitting into a case where x is an integer, and another where it is not, it is "obvious" that $$[x]-[x-1]=1$$

However, is there a more formal/rigourous proof possible?

Answer 1

Sure. Let $[x] = n$. That means $n \le x < n+1$. Subtract $1$ from each of the terms and that tells us $n-1 \le x-1 < n$. And that means that $[x-1]=n$. So $[x]-[x-1]=n - (n-1) = 1$.

If you don't want to introduce the variable $n$ just do

$[x] \le x < [x] + 1$ so

$[x]-1 \le x - 1 < [x]$ so

So $[x-1] = [x]-1$ so

$[x]-[x-1] = 1$.

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Actually, in hindsight if you want to be really slick:

$[x]-[x-1] = 1 \iff $

$[x-1] = [x] - 1 \underbrace{\iff}_{\text{by definition of greatest integer function}} $

$[x]-1 \le x-1 \le ([x]-1) + 1= [x]\iff$

$[x] \le x \le [x]+1$

which is true as it is the very definition of the greatest integer function.

Answer 2

A more formal proof could be derived from the intuition of the graph.

By definition, $[x - 1]$ is the greatest integer less than or equal to $x-1$, so we have: $[x - 1] \leq x - 1$.

Then adding $1$ to both sides: $[x - 1] + 1 \leq x$, so $[x] \geq [x - 1] + 1$, because $[x]$ is the greatest integer that is less than or equal to $x$, and $[x - 1] + 1$ is such an integer that is less than or equal to $x$.

On the other hand, $[x] \leq x$, so $[x] - 1 \leq x - 1$, which means $[x] \leq [x - 1] + 1$, because of the similar reasoning above: $[x - 1]$ is the greatest integer that is less than or equal to $x$, and $[x] - 1$ is such an integer.

Combine the two inequalities we got so far, we get what we want: $[x] - 1 = [x - 1]$.

Answer 3

This can be seen to follow from the more general property that $[x+n]=[x]+n$ for all real $x$ and integer $n$. Namely one sees that $[x]-[x-1]=[x]-([x]-1)=1$.

This more general property can be derived by noting that for $A$ the set of all integers no greater than $x$ and $B$ the set of all integers no greater $x+n$ that \begin{equation}a\in A\Leftrightarrow a\in\mathbb{Z}\text{ and }a\leq x\Leftrightarrow a+n\in\mathbb{Z}\text{ and }a+n\leq x+n\Leftrightarrow a+n\in B. (1) \end{equation} Now since $[x]\in A$ it follows by $(1)$ that $[x]+n\in B$.
If $b\in B$, then by $(1)$ one has that $b-n\in A$ and therefore, as $[x]$ is by definition the greatest element of $A$, that $b-n\leq [x]$ and that $b\leq [x]+n$.
Consequently $[x]+n$ is the greatest element of $B$ and hence is $[x+n]$.

In more techincal language the map $A\to B$ given by $a\mapsto a+n$ is an isomorphism of ordered sets and hence carries the greatest element $[x]$ of $A$ onto the greatest element $[x+n]$ of $B$.