Processing Math: Done
Solution 1.1:7b
From Förberedande kurs i matematik 1
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- | + | A rational number always has a decimal expansion which, after a certain decimal place, repeats itself periodically. | |
{{NAVCONTENT_STEP}} | {{NAVCONTENT_STEP}} | ||
- | + | In our case, the sequence 1416 is repeated indefinitely | |
- | <center><math>3{ | + | <center><math>3\textrm{.}\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math></center> |
- | + | In other words, the number is rational. | |
{{NAVCONTENT_STEP}} | {{NAVCONTENT_STEP}} | ||
- | + | 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. | |
{{NAVCONTENT_STEP}} | {{NAVCONTENT_STEP}} | ||
- | + | If we write | |
- | ::<math>\insteadof[right]{10000x}{x}{} = 3\,\ | + | ::<math>\insteadof[right]{10000x}{x}{} = 3\,\textrm{.}\,\underline{1416}\ \underline{1416}\ \underline{1416}\,\ldots</math> |
- | + | then | |
- | ::<math>\insteadof[right]{10000x}{10x}{} = 31\,\ | + | ::<math>\insteadof[right]{10000x}{10x}{} = 31\,\textrm{.}\,4161\ 4161\ 4161\,\ldots</math> |
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- | ::<math>\insteadof[right]{10000x}{100x}{} = 314\,\ | + | ::<math>\insteadof[right]{10000x}{100x}{} = 314\,\textrm{.}\,1614\ 1614\ 161\,\ldots</math> |
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- | ::<math>\insteadof[right]{10000x}{1000x}{} = 3141\,\ | + | ::<math>\insteadof[right]{10000x}{1000x}{} = 3141\,\textrm{.}\,6141\ 6141\ 61\,\ldots</math> |
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- | ::<math>\insteadof[right]{10000x}{10000x}{} = 31416\,\ | + | ::<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 expansion of 10000''x'' is in phase with the decimal expansion of ''x'', i.e. they have the same | |
+ | decimal expansion. | ||
{{NAVCONTENT_STEP}} | {{NAVCONTENT_STEP}} | ||
- | + | Therefore, | |
- | ::<math>10000x-x = 31416\,{ | + | ::<math>10000x-x = 31416\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots - 3\,\textrm{.}\,\underline{1416}\ \underline{1416}\,\ldots</math> |
{{NAVCONTENT_STEP}} | {{NAVCONTENT_STEP}} | ||
- | ::<math>\phantom{10000x-x}{}= 31413\quad</math>( | + | ::<math>\phantom{10000x-x}{}= 31413\quad</math>(The decimal parts cancel out each other) |
{{NAVCONTENT_STEP}} | {{NAVCONTENT_STEP}} | ||
- | + | and as <math>10000x-x = 9999x</math> we get that | |
::<math>9999x = 31413\,\mbox{.}</math> | ::<math>9999x = 31413\,\mbox{.}</math> | ||
{{NAVCONTENT_STEP}} | {{NAVCONTENT_STEP}} | ||
- | + | 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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- | <!--<center> [[ | + | <!--<center> [[Image:1_1_7b-1(2).gif]] </center> |
- | <center> [[ | + | <center> [[Image:1_1_7b-2(2).gif]] </center>--> |
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
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
x=3.1416 1416 1416...
then
10x=31.4161 4161 4161...
100x=314.1614 1614 161...
1000x=3141.6141 6141 61...
10000x=31416.1416 1416 1...
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,
10000x−x=31416.1416 1416...−3.1416 1416...
=31413 (The decimal parts cancel out each other)
and as
9999x=31413.
Solving for x in this relationship we find x as a quotient between two integers
x=999931413 =333310471
.