How to Prove the Statement [x-1][x+1]+1 = ([x])^2 for the Floor Function

[dgamma3]

Homework Statement

Ok, So I've tried to prove this statement unsuccessfully.
how would you guys do it? (assume [] means floor)

[x-1][x+1]+1 = ([x])^2

 

[HallsofIvy]

Let x= a+ r for some integer a and r between 0 and 1. [x-1]= a-1, [x+ 1]= a+1, and [x]= a.

 

[5hassay]

I'll assume were working with $x \in \mathbb{R}$.

Just to be clear of the definition of the floor function, it is the function $\left\lfloor \right\rfloor : \mathbb{R} \longrightarrow \mathbb{Z}$, with the mapping $x \in \mathbb{R} \longmapsto \mathrm{max} \left\{ y \in \mathbb{Z} : y \leq x \right\}$, i.e. the floor of a real number is the maximum integer in that set.

Notice that you have a few cases, from different numbers giving different results for the floor function:

1. $x$ being an integer
2. $x$ not being an integer, but greater than zero
3. $x$ not being an integer, but less than zero

The first case requires only a correct application of the definition of the floor function and some algebraic manipulations; just start with the left-hand side of the equation and work your way to the right-hand side (same for the other two cases), or the other way.

The second and third cases are more difficult. As already suggested, write $x$ in the way $x = \alpha + \beta$, where $\alpha \in \mathbb{Z}$, $\beta \in \mathbb{Q} - \mathbb{Z}$, i.e. $\beta$ to be a rational number that is not an integer. For the second case, what additional conditions would and must you apply to $\alpha$ and $\beta$? Similarly for the third case? Hint: $2.7 = 2 + 0.7$.