Being In Z \mathbb{Z} Or Not?

Algebra Level 4

{ 1 if x Z 0 if x Z \large{\begin{cases} 1&\text{ if } x \in \mathbb{Z}\\ 0&\text{ if } x \notin \mathbb{Z} \end{cases}}

Which of these is an equivalent form of the expression above?

x x 1 \left \lfloor x \right \rfloor \cdot \left \lfloor x-1 \right \rfloor x 1 + 1 x \left \lfloor x-1 \right \rfloor + \left \lfloor 1-x \right \rfloor x x \left \lfloor x \right \rfloor - \left \lceil x \right \rceil x + 1 x \left \lfloor x \right \rfloor + \left \lfloor 1-x \right \rfloor x 1 1 x \left \lceil x-1 \right \rceil - \left \lceil 1-x \right \rceil

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1 solution

If x Z x \in \mathbb Z , the equations should result in 1 if the floor/ceiling operations are removed. This fails for four of the given options: ( x ) ( x ) = 0 ; ( x 1 ) ( 1 x ) = 2 x 2 ; ( x ) ( x 1 ) = x 2 x ; ( x 1 ) + ( 1 x ) = 0. (x) - (x) = 0; \\ (x-1) - (1-x) = 2x - 2; \\ (x) \cdot (x-1) = x^2 - x; \\ (x-1) + (1-x) = 0. What remains is x + 1 x \boxed{\lfloor x\rfloor + \lfloor 1 - x \rfloor} .

Let's check that this really works.

If x Z x \in \mathbb Z then x + 1 x = x + ( 1 x ) = 1 ; \lfloor x\rfloor + \lfloor 1 - x \rfloor = x + (1 - x) = 1;

Otherwise, if x = n + d x = n + d with n Z n \in \mathbb Z and 0 < d < 1 0 < d < 1 , then 1 x = ( n ) + ( 1 d ) 1 - x = (-n) + (1-d) , and x + 1 x = n + ( n ) = 0. \lfloor x\rfloor + \lfloor 1 - x \rfloor = n + (-n) = 0.

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