Finding set of non recurring non terminating decimals

Question

I need to find a set of two Integers P and Q such that P/Q comes out to be non recurring non terminating decimals . Such as 106951484895/47666297253.

How can we find such two numbers. Is there any formula for this?

I need the result to be a decimal that does not end, nor do the numberals repeat themselves in any recognizable sequence.

Example: 106951484895/47666297253 = 2.24375483430840090053512174785916761672572536877348541885492361677904883097801043287594504105040726394682939649060976328569701751152716078581955550664345621014373276853529464561869479096457226131837381633508104200383965998089941882484488017422971446512579402430136563475351346061895880989881427322957327423437070470786123178209350642720039431463073874604572487035088141630189988694987841412729755839421966691187327329614175517087337289424929857158670121550588652768665265722816433005640...

EDIT: From the comments I got that such numbers don't exist.

So how can I get Fractions which approximates Irrational numbers?

An example is Fraction such as 22/7 which is commonly used to approximate π

Answer 1

In the comments, OP clarifies, "I want a fraction with a particularly long period in its decimal expansion."

The longest the period of $m/n$ can be is $n-1$ (because when you calculate the decimal expansion, the period begins when you see the same remainder twice, and there are only $n-1$ nonzero remainders available when you are dividing by $n$).

The period $n$ can only be achieved when $n$ is prime (because remainders that have a common factor with $n$ can't be achieved, and if $n$ isn't prime there are numbers between 1 and $n-1$ that have a common factor with $n$).

But not every prime number $n$ works. E.g., $13$ is prime, but the period of $m/13$ is $6$, not $12$ (try it!).

The primes $n$ for which you get period $n-1$ are tabulated at https://oeis.org/A001913, and the first 10,000 of them are listed at https://oeis.org/A001913/b001913.txt. It is widely believed, but not yet proved, that there are infinitely many of them.

Answer 2

Continued fractions are a way to go. See the English Wikipedia article for Continued Fraction, perhaps especially the section "Continued fraction expansion of $\pi$".

Answer 3

Here's a method for taking any string of digits and finding the fraction which has that string as its repeating period.

First, the string can't have any leading zeros. If it needs to have some, just move them to the end of the string and divide the resulting fraction by a power of ten.

If $S$ is a string without leading zeroes, first we have to convert $S$ to an integer $N$, for instance if $S$ is $(1, 2, 4, 5)$, then $N$ is $1245$. Now if $l$ is the length of $S$, we have:

$$0.\overline S = \frac{N}{10^{\space l} - 1}$$

For example, $0.\overline{1234} = \frac{1234}{9999}$.