Discount in Infinity I fully understand that $0.9999...$ mathematically will equal 1.0 exactly as the repeating decimal continues infinitely. This would be contrary to conventional logic where such a fraction is considered approximately 1.0 and not exactly 1.0. Here, of course, the conventional logic fails to understand what it mathematically means when the decimal is repeated in infinity. 

However can this principle be translated to an infinite discount as well? 

Say a number is reduced by 10% constantly (n-10%-10%-10%...). Will this too mathematically equal zero (as 0.0), or will it only ever get near zero?

Again, conventional logic will say that this will never actually be zero, it will only ever get closer to zero. Is conventional logic mathematically correct in such a case?

Thank you.
The phrase "is reduced by e.g. 10% in infinity" is perhaps ambiguous, but let me try to make it more precise:

If you reduce a quantity $A$ by $10\%$ (i.e., a proportion of $\frac{1}{10}$, then the remaining value is
$$\left(1 - \frac{1}{10}\right) A = \frac{9}{10} A,$$
and, inductively, if you do so $n$ times, then the remaining values is
$$\frac{9}{10}\left( \cdots\frac{9}{10}\left(\frac{9}{10} A\right) \cdots \right) = \left(\frac{9}{10}\right)^n A.$$
But for any small number $\epsilon &gt; 0$, we can make this quantity small than $\epsilon$. (In fact, a little algebra shows that taking $n \geq \left\lceil \frac{\log \epsilon - \log A}{\log r}\right\rceil$ will do.) On the other hand, for any $n$, we have $\left(\frac{9}{10}\right)^n A \geq 0$, and so the as $n \to \infty$, the remaining quantity $\left(\frac{9}{10}\right)^n A$ goes to zero.
In a sense, it's equal to zero in the same way your example is equal to 1. Both converge on to their target, but only after an infinite number of digits/reductions.