Generalizing floor sums 2

Algebra Level 1

Given that $x$ is a real number but not an integer, compute $\lfloor x \rfloor + \lfloor - x \rfloor$ .

1 -1 None of these choices 0

3 solutions

Shaun Leong
Jan 25, 2016

Since $x$ is not an integer, we have $a<x <a+1$ $-(a+1)<-x<-a$ for some integer a.

The floor function returns the greatest integer less than or equal to the argument, so the answer is $a-(a+1)=\boxed{-1}$

Typo: The second inequality should have $(-x)$ in the middle.

Prasun Biswas - 5 years, 4 months ago

Thanks, it has been edited

Shaun Leong - 5 years, 4 months ago

But the question said that x is an integer

Rindell Mabunga - 5 years, 4 months ago

The question now states that x is a real number but not an integer. The question previously (I believe) was $x \in \mathbb{R} - \mathbb{Z}$ which means the real numbers excluding the integers.

Shaun Leong - 5 years, 4 months ago

Gogul Raman Thirunathan
Feb 9, 2016

An elegant and easy solution

Gogul Raman Thirunathan - 5 years, 4 months ago

William Isoroku
Jan 25, 2016

Look at the graph of a floor function

But the thing is if the real number is an integer then output of the function is equal to input. I mean suppose x is equal to 2 then [x] will be equal to 2 and [-x] will be equal to -2 and in this case [x]+[-x] will be equal to 0. So, answer should be none of these choises as there can be two answer to this i.e. -1 if x is non-integer real number and 0 if x integer real number.

Rachit Chaudhary - 3 years, 7 months ago

But it is stated in the question that it is in fact not an integer.

Don Weingarten - 2 years, 4 months ago

Generalizing floor sums 2

3 solutions

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