Solution 1.1:7b

A rational number always has a decimal expansion which, after a certain decimal place, repeats itself periodically.

In our case, the sequence is repeated indefinitely.

In other words, the number is rational.

The next problem is to rewrite the number as a fraction, for which we use the fact that multiplication by 10 moves the decimal point one place to the right.

If we write

\displaystyle \insteadof[right]{10000x}{x}{} = 3\,\color{red}{‚}\,\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots

then

\displaystyle \insteadof[right]{10000x}{10x}{} = 31\,\color{red}{‚}\,4161\ 4161\ 4161\,\ldots

\displaystyle \insteadof[right]{10000x}{100x}{} = 314\,\color{red}{‚}\,1614\ 1614\ 161\,\ldots

\displaystyle \insteadof[right]{10000x}{1000x}{} = 3141\,\color{red}{‚}\,6141\ 6141\ 61\,\ldots

\displaystyle \insteadof[right]{10000x}{10000x}{} = 31416\,\color{red}{‚}\,\underline{1416}\ \underline{1416}\ 1\,\ldots

Note that, in 10000x we have moved the decimal point sufficiently many places so that the decimal expansion of 10000x has come in phase with the decimal expansion of x, i.e. they have the same decimal expansion.

Therefore,

\displaystyle 10000x-x = 31416\,{,}\,\underline{1416}\ \underline{1416}\,\ldots - 3\,{,}\,\underline{1416}\ \underline{1416}\,\ldots

\displaystyle \phantom{10000x-x}{}= 31413\quad(The decimal parts cancel out each other)

and as \displaystyle 10000x-x = 9999x we get that

\displaystyle 9999x = 31413\,\mbox{.}

Solving for x in this relationship we find x as a quotient between two integers

\displaystyle x = \frac{31413}{9999}\quad\biggl({}= \frac{10471}{3333}\biggr)\,\mbox{.}