Precisely

Algebra Level 3

⌊ x 2 ⌋ + ⌊ x + 1 2 ⌋ = ⌊ x ⌋ \large\left\lfloor \frac { x }{ 2 } \right\rfloor +\left\lfloor \frac { x +1 }{ 2 } \right\rfloor = \large\left\lfloor x \right\rfloor

Find the most precise solution of x x satisfying the equation above.

Notations:

For all x ∈ Z x \in \mathbb{Z} For all x ∈ R x \in \mathbb{R} For all x ∈ N x \in \mathbb{N}

Ravneet Singh by Ravneet Singh , 30, India

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1 solution

Chew-Seong Cheong
Feb 23, 2017

⌊ x 2 ⌋ + ⌊ x + 1 2 ⌋ = ⌊ x ⌋ ⟹ ⌊ ⌊ x ⌋ + { x } 2 ⌋ + ⌊ ⌊ x ⌋ + { x } + 1 2 ⌋ = ⌊ x ⌋ \begin{aligned} \left \lfloor \frac x2 \right \rfloor + \left \lfloor \frac {x+1}2 \right \rfloor & = \left \lfloor x \right \rfloor \\ \implies \left \lfloor \frac {\lfloor x \rfloor + \{x\}}2 \right \rfloor + \left \lfloor \frac {\lfloor x \rfloor + \{x\}+1}2 \right \rfloor & = \left \lfloor x \right \rfloor \end{aligned}

If ⌊ x ⌋ \lfloor x \rfloor is odd, let ⌊ x ⌋ = 2 n + 1 \lfloor x \rfloor = 2n + 1 , where n ∈ Z n \in \mathbb Z . Then

⌊ ⌊ x ⌋ + { x } 2 ⌋ + ⌊ ⌊ x ⌋ + { x } + 1 2 ⌋ = ⌊ x ⌋ ⌊ 2 n + 1 + { x } 2 ⌋ + ⌊ 2 n + 1 + { x } + 1 2 ⌋ = 2 n + 1 ⌊ n + 1 + { x } 2 ⌋ + ⌊ n + 1 + { x } 2 ⌋ = 2 n + 1 2 n + 1 = 2 n + 1 \begin{aligned} \left \lfloor \frac {\lfloor x \rfloor + \{x\}}2 \right \rfloor + \left \lfloor \frac {\lfloor x \rfloor + \{x\}+1}2 \right \rfloor & = \left \lfloor x \right \rfloor \\ \left \lfloor \frac {2n + 1 + \{x\}}2 \right \rfloor + \left \lfloor \frac {2n + 1 + \{x\}+1}2 \right \rfloor & = 2n+1 \\ \left \lfloor n + \frac {1 + \{x\}}2 \right \rfloor + \left \lfloor n+ 1 + \frac {\{x\}}2 \right \rfloor & = 2n+1 \\ 2n + 1 & = 2n + 1 \end{aligned}

Therefore, the equation is true for real x x , whose ⌊ x ⌋ \lfloor x \rfloor is odd.

If ⌊ x ⌋ \lfloor x \rfloor is even, let ⌊ x ⌋ = 2 n \lfloor x \rfloor = 2n , where n ∈ Z n \in \mathbb Z . Then

⌊ ⌊ x ⌋ + { x } 2 ⌋ + ⌊ ⌊ x ⌋ + { x } + 1 2 ⌋ = ⌊ x ⌋ ⌊ 2 n + { x } 2 ⌋ + ⌊ 2 n + { x } + 1 2 ⌋ = 2 n ⌊ n + { x } 2 ⌋ + ⌊ n + { x } + 1 2 ⌋ = 2 n 2 n = 2 n \begin{aligned} \left \lfloor \frac {\lfloor x \rfloor + \{x\}}2 \right \rfloor + \left \lfloor \frac {\lfloor x \rfloor + \{x\}+1}2 \right \rfloor & = \left \lfloor x \right \rfloor \\ \left \lfloor \frac {2n + \{x\}}2 \right \rfloor + \left \lfloor \frac {2n + \{x\}+1}2 \right \rfloor & = 2n \\ \left \lfloor n + \frac {\{x\}}2 \right \rfloor + \left \lfloor n + \frac {\{x\}+1}2 \right \rfloor & = 2n \\ 2n & = 2n \end{aligned}

Therefore, the equation is true for real x x , whose ⌊ x ⌋ \lfloor x \rfloor is even.

Therefore, the equation is true for all x ∈ R \boxed {\text{for all } x \in \mathbb R} .

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You was always there tremendous man

Ram Sita - 3 years, 9 months ago

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