Is Dividing 1 by 3 an Invalid Problem?

[disregardthat]

No, it wasn't "chosen". 0.9999... means .9+ .09+ .009+ .0009+ .00009+ ...= .9(1+ .01+ .001+ .0001+ .00001+ ...)

That last is a "geometric series" which is taught in any good "precalculus" or "algebra II" class.
a(1+ r+ r^2+ r^3+ r^4+ ...) has sum a/(1- r) as long as |r|< 1.

Here, a= 0.9 and r= .1. a/(1- r)= .9/(1- .1)= .9/.9= 1.

The point is that it was chosen to mean exactly that, not that we chose to do our calculations correctly...