Is 1/113 a rational number?

Question

Before two days , one of my physics si told me if one wants to use value of pi more accurately 355/113 can be considered as value of pi to get more accurate result.

I want to know is 1/113 a rational number.

My view I have tried dividing till 22 digits after decimal but still I have not got the repeating numbers after decimal.

My observation Really result using value pi=355/113 is more accurate then 22/7 or 3.14.

If anyone find the reason behind these ,please answer. If someone gets repeating number please answer the number.

Answer 1

Any number of the form $\dfrac{p}q$, where $p \in \mathbb{Z}$ and $q \in \mathbb{Z} \backslash \{0\}$ is a rational. This is in fact the definition of rational number. Your number $1/113$ will repeat after $112$ digits.

Answer 2

$1/113$ is rational, yes. 355/113 is a more accurate approximation of $\pi$ than $3.14$ or $22/7$, but not exact:

$$\left|\pi-3.14\right|\approx1.59\cdot10^{-3}$$

$$\left|\pi-\frac{22}{7}\right|\approx1.26\cdot10^{-3}$$

$$\left|\pi-\frac{355}{113}\right|\approx2.66\cdot10^{-7}$$

Answer 3

$1/113=\scriptscriptstyle0.\overline{0088495575221238938053097345132743362831858407079646017699115044247787610619469026548672566371681415929203539823}$

Answer 4

The number $$ \frac{1}{10^n-1} $$ has a period of length $n$: $$ \frac{1}{10^n-1}= \frac{1}{\underbrace{999\dots999}_{\text{$n$ nines}}}= 0.\underbrace{000\dots000}_{\text{$n-1$ zeros}}1 $$ Thus finding numbers with a very long period should not be a surprise. But not finding a repetition in a long period of time tells you neither that the number is rational nor that it's irrational.

Actually, $355/113$ is rational by definition of rational number, so we are sure to find a repetition in its decimal expansion, because this can be proved: when performing the division algorithm, the set of possible remainders is finite, so at some point one of them must appear for the second time. However, the definition of rational number is not based on repetitions in the decimal expansion.