Rounding half-integers to approximate product Given two half-integers $a,b\in\mathbb Z+\frac 1 2$, the integers nearest to each is given by $N_a=\{a-0.5,\ a+0.5\}$ and $N_b=\{b-0.5,\ b+0.5\}$.
Is there a general method to find, for the two given half-integers, the values $n\in N_a,m\in N_b$ where $\left|nm-ab\right|$ is minimal?

The motivation for this post came when I considered $1.5\times1.5=2.25$. Intuition told me that I could round either (or both) number either way and the result would be the same. Such a symmetry in fact does not exist. After doing a few other examples I simply could not find a pattern.
 A: What you're trying to do, is to find a number
$$(a\pm{1\over2})(b\pm{1\over2})$$
that's the closest to $ab$.
For start, let's assume $a$ and $b$ are both positive.
If they are both $+$ or both $-$ it's easy to see you are further than having one $+$ and one $-$. To see which one should be $+$ and which one $-$, calculate them, they are both the same distance from $ab+{1\over4}$. One is above and one is below (unless $a=b$, in which case they are both equal $ab+{1\over4}$). Clearly the one below is then closer to $ab$.
So one has a $+$ and one has a $-$, but which one has what depends on which one is larger.
If one (or both) is negative, work with $|a|$ and $|b|$ first. It's easy to figure out the rest.
A: We compute $$nm=ab\pm \frac{a}{2} \pm \frac{b}{2} \pm \frac{1}{4}$$
and hence we wish to minimize $$|nm-ab|=|\pm \frac{a}{2} \pm \frac{b}{2} \pm \frac{1}{4}|$$
In this expression, the three $\pm$ are not all freely chosen; there must be an even number of $-$'s chosen.  Let's assume without loss that $a\ge b\ge 0$ (if $a<b$, swap the roles of $a,b$).  For convenience, set $a'=\frac{a}{2}, b'=\frac{b}{2}$, with $a'\ge b'$.  Now there are four cases to consider:
$$|a'+b'+.25|,~ |a'-b'-.25|,~ |-a'+b'-.25|,~ |-a'-b'+.25|$$
Of these, it's clear that the first and last will never be smaller than the others.  Now, if $a'=b'$ then the middle two agree; otherwise $|a'-b'-.25|$ will be the smallest.  Hence, we have proved that the second one will always be smallest or tied for smallest.  This corresponds to $$n=a-0.5, ~~m=b+0.5$$
(keeping in mind that $a\ge b$).  
I leave the cases of one (or two) of $a,b$ negative open.
