Almost AIME 2

What typo?

Nothing has been "assumed" here.

The question says "Find the integer part of $P$ ", i.e., the answer is the integral part of $P$ . Now, $P=1+\sqrt{2016}$ is clearly a non-negative real. Hence, by definition of floor function, the answer is $\lfloor P\rfloor$ .

Here's an brief explanation: Consider a non-negative real $x=k+t$ where $k$ is the integral part of $x$ and $t$ is the fractional part $\in [0,1)$ . Then, from basic properties of floor function, we have $\lfloor x\rfloor = \lfloor k+t\rfloor=k+\lfloor t\rfloor$ . Since $t\in [0,1)$ by construction, $\lfloor t\rfloor = 0$ and hence $k=\lfloor x\rfloor$ . Recall that $k$ was defined to be the integral part of non-negative $x$ and hence, the integral part of a non-negative real is $\lfloor x\rfloor$ .

Note: Notice that for negative $x$ , the integral part is given by $\lceil x\rceil$ (try to prove this one on your own, use a method similar to the above).