# Find all real $x$ such that $1990[x] +1989[-x]=1$ (where $[x]$ is the floor function for $x$).

Find all real $x$ such that $1990[x] +1989[-x]=1$ (where $[x]$ is the floor function for $x$).

My effort

Rearranging our equation we have :

\begin{array}{c} 1990[x]+1989[-x]&=1 \\ 1989([x]+[-x])+[x] &=1 \\ \end{array}

Supposing that $x$ is an integer ,I have that $[x]+[-x]=0$ and the problem breaks down to $$[x]=1$$ which has the only solution $x=1$

Else ,$x$ is a real number with nonzero fractional part and $[x]+[-x]=-1$ which yields in our case

\begin{array}{c} -1989 + [x] &= 1 \\ [x] &=1990 \\ \end{array}

For this to happen we must therefore have that $x \in (1990,1991)$

Question

Is my effort complete and correct ?What would have been other ways to approach the problem ?

• Looks fine to me, and the approach differs only trivially from the one that I’d have used. – Brian M. Scott Feb 11 '16 at 19:05
• Where are you getting 1991 from in the ordered pair? It seems like from your own statements that the set of solutions is 1 and 1990. – The Great Duck Feb 13 '16 at 0:43
• @TheGreatDuck $x \in (a,b)$ is a fancy notation to indicate that we want $x$ to be in the interval between $1990$ and $1991$ but excluding those two endpoints of the interval(i.e $1990$ and $1991$ aren't counted but $1990,5$ is for example.) – Mr. Y Feb 13 '16 at 14:24
• I'm aware of range notation. I'm literally developing a notation for the general replacement of a range of a graph for dynamic shift translations. I just didn't realize the author meant it in that context. Unfortunately they were presenting it as a solution set. What they should have written is that x is in {1,(1990,1991)} then the range would make sense as something other than a random list of numbers. I apologize if I can't make sense of their math. As someone complained recently mathematics is not exact or precise at all. There's a gazillion different redundant ambiguous notations. Ugh. – The Great Duck Feb 14 '16 at 8:39