Solution 1.1:7b

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Ett rationellt tal har alltid en decimalutveckling som från och med en viss decimal upprepar sig periodiskt.
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A rational number always has a decimal expansion which, after a certain decimal place, repeats itself periodically.
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I vårt fall så upprepas sekvensen 1416 i all oändlighet
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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>
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Med andra ord är talet rationellt.
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In other words, the number is rational.
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Nästa problem är att skriva om talet som ett bråktal och då utnyttjar vi att multiplikation med 10 flyttar decimalkommat ett steg åt höger.
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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.
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Om vi skriver
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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>
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så är därför
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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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Notera att i 10000''x'' har vi flyttat decimalkommat tillräckligt många steg så att decimalutvecklingen av 10000''x'' har kommit i fas med decimalutvecklingen av ''x'', dvs. de har samma decimalutveckling.
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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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decimal expansion.
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Därför är
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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>(decimalerna tar ut varandra)
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::<math>\phantom{10000x-x}{}= 31413\quad</math>(The decimal parts cancel out each other)
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och eftersom <math>10000x-x = 9999x</math> så har vi alltså sambandet
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and as <math>10000x-x = 9999x</math> we get that
::<math>9999x = 31413\,\mbox{.}</math>
::<math>9999x = 31413\,\mbox{.}</math>
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Löser vi ut ''x'' ur detta samband får vi ''x'' som en kvot mellan två heltal
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Solving for ''x'' in this relationship we find ''x'' as a quotient between two integers
::<math>x = \frac{31413}{9999}\quad\biggl({}= \frac{10471}{3333}\biggr)\,\mbox{.}</math>
::<math>x = \frac{31413}{9999}\quad\biggl({}= \frac{10471}{3333}\biggr)\,\mbox{.}</math>
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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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