Elementary rounding 
Mr. Brown rounded $14.486$ to the nearest whole number by rounding $14.486$ to $14.49$ by the "over $5$" rule. Then he rounded $14.49$ to $14.5$ by the same rule. Then he rounded $14.5$ to $15$ by the rule. Unfortunately this is wrong. Why is his answer wrong? How can using the "over $5$" rule be misleading in some cases? Using a number line, show why his answer is wrong. Explain your thinking.

 A: The answer is wrong since $14.486$ is closer to $14$ than it is to $15$.  As you can see from the number line below, $14.486$ is less than $1/2$ unit from $14$ and more than $1/2$ unit from $15$.
Specifically, $14.486 - 14 = 0.486 < 0.514 = 15 - 14.486$.  Hence, the nearest integer to $14.486$ is $14$, so we round $14.486$ to $14$ when we round to the nearest integer.
By convention, if $n$ is an integer and $x$ is a real number satisfying the inequalities $n \leq x < n + 1$, we round $x$ to $n$ if $n \leq x < n + 0.5$ and round $x$ to $n + 1$ if $n + 0.5 \leq x < n + 1$.  
In your example, $x = 14.486$, so $n = 14$.  Since $14 \leq 14.486 < 14.5$, we round $14.486$ to $14$ when we round to the nearest integer. 
What went wrong when Mr. Brown successively applied the "over $5$'' rule?  He converted a number ($14.486$) that should be rounded down to $14$ into one ($14.5$) that should be rounded up to $15$.  Specifically, successively rounding $14.486$ to the nearest hundredth to obtain $14.49$, then rounding $14.49$ to the nearest tenth to obtain $14.5$ gave him a number that had to be rounded up to $15$ rather than down to $14$.  The error was caused by rounding a number that had already been rounded rather than rounding the original number. 
A: Maybe try the same process with $14.4999$ and $14.5001$. You should notice that one should round "all at once" and not step by step as Mr Brown did.
A: Let $r_n(x)$ denote the number $x$ rounded to the $n$-th decimal.  Then $r_n(x)=\lfloor 10^nx+0.5\rfloor/10^n$.  Is there any reason that in general $r_0\Bigl(r_1\bigl(r_2(x)\bigr)\Bigr)=r_0(x)$?
