Solution 1.1:7b

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A rational number always has a decimal expansion which, after a certain decimal place, repeats itself periodically.
A rational number always has a decimal expansion which, after a certain decimal place, repeats itself periodically.
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In our case, the sequence is repeated indefinitely.
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In our case, the sequence 1416 is repeated indefinitely
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<center><math>3{,}\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math></center>
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<center><math>3\textrm{.}\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math></center>
In other words, the number is rational.
In other words, the number is rational.
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If we write
If we write
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::<math>\insteadof[right]{10000x}{x}{} = 3\,\color{red}{&#130;}\,\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math>
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::<math>\insteadof[right]{10000x}{x}{} = 3\,\textrm{.}\,\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math>
then
then
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::<math>\insteadof[right]{10000x}{10x}{} = 31\,\color{red}{&#130;}\,4161\ 4161\ 4161\,\ldots</math>
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::<math>\insteadof[right]{10000x}{10x}{} = 31\,\textrm{.}\,4161\ 4161\ 4161\,\ldots</math>
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::<math>\insteadof[right]{10000x}{100x}{} = 314\,\color{red}{&#130;}\,1614\ 1614\ 161\,\ldots</math>
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::<math>\insteadof[right]{10000x}{100x}{} = 314\,\textrm{.}\,1614\ 1614\ 161\,\ldots</math>
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::<math>\insteadof[right]{10000x}{1000x}{} = 3141\,\color{red}{&#130;}\,6141\ 6141\ 61\,\ldots</math>
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::<math>\insteadof[right]{10000x}{1000x}{} = 3141\,\textrm{.}\,6141\ 6141\ 61\,\ldots</math>
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::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\color{red}{&#130;}\,\underline{1416}\ \underline{1416}\ 1\,\ldots</math>
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::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\ 1\,\ldots</math>
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Note that, in 10000''x'' we have moved the decimal point sufficiently many places so that the decimal
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Note that, in 10000''x'' we have moved the decimal point sufficiently many places so that the decimal expansion of 10000''x'' is in phase with the decimal expansion of ''x'', i.e. they have the same
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expansion of
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10000''x'' has come in phase with the decimal expansion of ''x'', i.e. they have the same
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decimal expansion.
decimal expansion.
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Therefore,
Therefore,
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::<math>10000x-x = 31416\,{,}\,\underline{1416}\ \underline{1416}\,\ldots - 3\,{,}\,\underline{1416}\ \underline{1416}\,\ldots</math>
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::<math>10000x-x = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots - 3\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots</math>
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::<math>\phantom{10000x-x}{}= 31413\quad</math>(The decimal parts cancel out each other)
::<math>\phantom{10000x-x}{}= 31413\quad</math>(The decimal parts cancel out each other)

Current revision

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

In our case, the sequence 1416 is repeated indefinitely

\displaystyle 3\textrm{.}\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots

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\,\textrm{.}\,\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots

then

\displaystyle \insteadof[right]{10000x}{10x}{} = 31\,\textrm{.}\,4161\ 4161\ 4161\,\ldots
\displaystyle \insteadof[right]{10000x}{100x}{} = 314\,\textrm{.}\,1614\ 1614\ 161\,\ldots
\displaystyle \insteadof[right]{10000x}{1000x}{} = 3141\,\textrm{.}\,6141\ 6141\ 61\,\ldots
\displaystyle \insteadof[right]{10000x}{10000x}{} = 31416\,\textrm{.}\,\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 is in phase with the decimal expansion of x, i.e. they have the same decimal expansion.

Therefore,

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