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?
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Muhammad
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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 $\frac{b_n}{a_n}$.
PierreCarre
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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.
Greg Martin
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