How come the population standard deviation formula doesn't give the correct standard deviation compared to the sample standard deviation? For example, 
If I had the numbers 4, 6, and 8 and I use to sample standard deviation formula I get 2 as the standard deviation.
But If I use the population standard deviation I end up with approximately 1.63 as the standard of deviation. 
 A: If those 3 values are the population, what makes the population standard "deviation incorrect"?  Even in a sample, what makes the sample variance formula you're using "correct" (other than convention)? It has some convenient properties, but it's not the only possible choice for sample variance.
Note that if you have the population (rather than a sample), the sample variance formula you used doesn't correctly compute the second moment (about the population mean) - by definition, the population variance.
The population variance formula is a given - that's the quantity that's defined, it follows directly from the definition of the variance of a random variable applied a finite set of values all equally likely to be chosen. 
It's actually the sample variance that we have to justify.
[So why in sample variance do people divide by n-1 instead of n? At heart because we're trying to estimate the population variance from a sample. If we knew the population mean, the sample variance about that mean would be correct on average -- but we don't know the population mean, and the sample mean is 'too close' to the data -- indeed it's the value for the mean that makes the variance about it as small as possible. We can make it right on average by multiplying that "too small on average" estimate by $\frac{n}{n-1}$, which is the same as replacing the n-denominator by n-1. That issue is explained in more detail in other questions here on math.SE and on stats.SE] 
Consequently, each formula is "correct" for the purpose it is designed for.
