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Let $x \in \R$ be a real number.

Let $y \in \R$ such that:

$y = \floor {x + \dfrac 1 2}$

Then $y$ is defined as $x$ rounded to the nearest integer.

Treatment of Half

Consider the situation when $\dfrac x {10^n} + \dfrac 1 2$ is an integer.

That is, $\dfrac x {10^n}$ is exactly midway between the two integers $\dfrac x {10^n} - \dfrac 1 2$ and $\dfrac x {10^n} + \dfrac 1 2$.

Recall that the general philosophy of the process of rounding is to find the closest approximation to $x$ to a given power of $10$.

Thus there are two equally valid such approximations:

$\dfrac x {10^n} - \dfrac 1 2$ and $\dfrac x {10^n} + \dfrac 1 2$

between which $\dfrac x {10^n}$ is exactly midway.

There are a number of conventions which determine which is to be used.


$72 \cdotp 8$ to Nearest Integer

$72 \cdotp 8$ rounded to the nearest integer is $73$.

This is because $72 \cdotp 8$ is closer to $73$ than it is to $72$.