# Dividing an integer by an infinite decimal that extends irregularly

The existance of the multiplicative inverse of a nonzero real number, proves the existance of fractions, such as, $$\frac{1}{\sqrt{2}}$$ and $$\frac{1}{\pi}$$.
So, how to compute such fractions, where an integer is divided by an infinite decimal extending without regular repetition?

An irrational number is the limit of a sequence of rational numbers. If you take two sequences of integers $$a_n, b_n$$ such that $$\frac{a_n}{b_n} \to \pi$$, the irrational number $$\frac{1}{\pi}$$ can be obtained as $$\frac{1}{\pi} = \lim \frac{b_n}{a_n}$$. so, the fractions can be computed, to arbitrary precision, considering higher and higher order terms of the sequence $$\frac{b_n}{a_n}$$.
What does "computing $$\frac1\pi$$" mean, exactly? Presumably it means having an algorithm that will produce any arbitrarily long initial segment of the decimal expansion of $$\frac1\pi$$ (since we don't expect any pattern, just as we've postulated no pattern to the decimal expansion of $$\pi$$ itself). But to compute the first 100 decimal places of $$\frac1\pi$$, it suffices to know the first 102 (or whatever) decimal places of $$\pi$$. So the apparent problem is not a problem at all.