powerless said:
It can easily be shown that the recurring decimal x = 1.123123... is rational, as follows:
10^{3}x-x = 1123.123...-1.123123...=1122 => x = \frac{1122}{999} \in Q
Show that the recurring decimals 0.3712437127... and 0.9999999...are rational numbers.
I presume you are missing a "3" and that should be 0.3712371237.. so that the "3712" is the recurring block.
3. The Attempt at a Solution
I'm not quite sure what the question is asking as I had never seen a question like this before!
Does the question mean what devided by what equals 0.3712437127... and 0.9999999...?
I don't know the method for this & I appreciate some guidance if anyone here knows how to do it.
They
gave an exaple of the method just before asking the quetion! But here's another example, with more detail:
0.234323432343... where it is the "2343" that repeats.
The first thing I do is note that the recurring block has 4 digits- and I know that I can "move the decimal point 4 places" by multiplying by 10
4= 10000. If I write x= 0.23432343... and multiply by 10000 I get 10000x= 2342.23432343... Notice that the first "2343" block has been moved in front of the decimal point and its place has been taken by the second "2343" block. And, because the repetition never stops,
every "2343" block is replaced by another identical block: the decimal part of 10000x is still just .23432343... . If we subtract the equation x= 0.23432343... from 10000x= 2343.23432343..., on the right we get 10000x= 999x and on the right the whole number 2343: 999x= 2343 so
x= 2343/999. That can be reduced- both numerator and denominator are divisible by 3- but at this point we have shown that the number can be written as a fraction.
Here's another, slightly harder example: 0.153221622162216... where it is the "2216" that repeats. The reason this is slightly harder is that we must take care of the "153" that we have
before the repeating part. That has 3 digits and multiplying by 10
3= 1000 will move the decimal point 3 places:
If we write x= 0.15322162216...
and multiply the equation by 1000 we get
1000x= 153.221622162216...
Now we have only the repeating block after the decimal point- and that repeating block has, again, 4 digits. If we multiply again by 10
4= 10000 we get
10000000x= 1532216.22162216...
Again, that each block in that infinitely repeating decimal has been replaced by the next one. The decimal parts of 1532216.22162216... and 153.22162216... are exactly the same.
Subracting 1000x= 153.22162216... from 10000000x= 1532216.22162216..., the decimal parts cancel and we have
(10000000- 1000)x= 1532216- 153 or
9999000x= 1532063.
Dividing both sides by 9999000 we get x= 1632064/9999000 which perhaps can be reduced but we have succeeded in writing the number as a fraction, proving that it is a rational number.
Now apply the same ideas to x= 0.3712371237.. where "3712" is repeating and
x= 0.99999999... where "9" is repeating.
