Inspired by 'Floor Ceiling Ceiling Floor : JEE Style'

Algebra Level 4

$\large \displaystyle (\lceil x + 1 \rceil)^2 + (\lfloor x -1 \rfloor)^2 = (\lceil x -1 \rceil)^2 + (\lfloor x + 1 \rfloor)^2$

The solution set of the equation above is?

Notations :

$\mathbb{R}$ denote the set of real numbers.

$\mathbb{Z}$ denote the set of integers.

$\mathbb{N}$ denote the set of natural numbers.

$\mathbb{Q}$ denote the set of rational numbers.

Inspiration .

Try my set

x \in \mathbb{N}

x \in \mathbb{Z}

x \in \mathbb{R}

x \in \mathbb{Q}

1 solution

Chew-Seong Cheong
May 20, 2015

Let $f(x) = \lceil x+1 \rceil ^2 + \lfloor x-1 \rfloor^2 - \lceil x-1 \rceil ^2 - \lfloor x+1 \rfloor^2$

When $x \in \mathbb {Z} \Longrightarrow \lceil x+1 \rceil = \lfloor x+1 \rfloor$ and $\lceil x-1 \rceil = \lfloor x-1 \rfloor \Longrightarrow f(x) = 0$ .

When $x \in \mathbb{R} \notin \mathbb{Z}$ ,

$\begin{aligned} \Rightarrow f(x) & = \lceil x+1 \rceil ^2 - \lfloor x+1 \rfloor^2 - \lceil x-1 \rceil ^2 + \lfloor x-1 \rfloor^2 \\ & = (\lfloor x+1 \rfloor +1)^2 - \lfloor x+1 \rfloor^2 - (\lfloor x-1 \rfloor +1)^2 + \lfloor x-1 \rfloor ^2 \\ & = 2\lfloor x+1 \rfloor +1 - 2\lfloor x-1 \rfloor - 1 \\ & = 2\lfloor x \rfloor + 2 - 2\lfloor x \rfloor + 2 \\ & = 4 \end{aligned}$

$\Rightarrow f(x) = \begin{cases} 0 & \text{when } \boxed{x \in \mathbb{Z}} \\ 4 & \text{when } x \in \mathbb{R} \notin \mathbb{Z} \end{cases}$

Moderator note:

Wonderful work!

Brilliant Sir... Just loved the way you did it👍

Avinash Singh - 6 years ago

A slightly easier way (but still equivalent to what you did) is to set $\lfloor x \rfloor = n$ for non-integers, and then compare $(n+2)^2 + (n-1) ^2 - n^2 - (n+1) ^2$ .

Calvin Lin Staff - 6 years ago

same method....!!!

Deepansh Jindal - 4 years, 11 months ago

Same there

Aakash Khandelwal - 4 years, 11 months ago

Inspired by 'Floor Ceiling Ceiling Floor : JEE Style'

Inspiration .

Try my set

1 solution

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