Converting recurring decimal to fraction when '0' is part of recurring pattern

Simonsky

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Got stuck on this question:

Write 0.0123456789 as a fraction where that whole sequence from 0-9 recurs.

The problem I've got is that in order to get the fraction you need to align the recurring bit so you can subtract it.

The only way I can think of doing it is by ( after designating the number as x) move everything before the point but that would make: 10000000000x =123456789 then subtract x so we have: 999999999x =123456789 so we then have:

123456789/9999999999 but that seems nuts-there must be an easier way I can't think of!
 
Got stuck on this question:

Write 0.0123456789 as a fraction where that whole sequence from 0-9 recurs.

The problem I've got is that in order to get the fraction you need to align the recurring bit so you can subtract it.

The only way I can think of doing it is by ( after designating the number as x) move everything before the point but that would make: 10000000000x =123456789 then subtract x so we have: 999999999x =123456789 so we then have: But 123456789.123456789123456789.... - 0.0123456789 is NOT 123456789 as the decimal number do NOT cancel out. Just look and you'll see that!

123456789/9999999999 but that seems nuts-there must be an easier way I can't think of! I do not know about it being nuts, but it is wrong
1st please see the red above.

Let x = 0.0123456789123456789123456789...

You want two numbers: One number where one part of the repetition is to the left of the decimal and another number where the repetition starts at the decimal.

So we have 10000000000x = 123456789.123456789123456789...
10x = .123456789123456789...

Now subtract and solve for x....
 
Last edited:
1st please see the red above.

Let x = 0.0123456789123456789123456789...

You want two numbers: One number where one part of the repetition is to the left of the decimal and another number where the repetition starts at the decimal.

So we have 10000000000x = 123456789.123456789123456789...
10x = .123456789123456789...

Now subtract and solve for x....


Thanks for response Jomo. I can see the mistake I made but haven't you made a mistake yourself bu saying: 10000000000x = 123456789.123456789123456789... surely it would be: 10000000000x = 123456789.01234567890123456789...as the nought recurs as well?
 
Thanks for response Jomo. I can see the mistake I made but haven't you made a mistake yourself bu saying: 10000000000x = 123456789.123456789123456789... surely it would be: 10000000000x = 123456789.01234567890123456789...as the nought recurs as well?

Actually, I think my original post works:

x= 0.0123456789

10000000000x = 123456789.0123456789......... then: 10000000000x - x = 9999999999x = 123456789 so x = 123456789/9999999999 = 1/81
 
Stuck on another one of similar ilk

Which is greater: 0.1704 (with 704 recurring) or 7/40

So I started by converting the recurring decimal to a fraction as follows:

0.1704 (704 bit recurring) = 1/10 + 0.704 /10 recurring which = 704/999 x 1/10 = 704/9990

so we have 1/10 + 704/9990 = 999/9990 + 704/9990 = 1703/9990 which has no common factors however I can still compare it to 7/40 by finding LCM which is 39960

so we have : 6812/39960 compared to 6993/39960 so the latter is bigger making 7/40 greater than 0.1704 (704 recurring)

I think that's correct?
 
Which is greater: 0.1704 (with 704 recurring) or 7/40

So I started by converting the recurring decimal to a fraction as follows:

0.1704 (704 bit recurring) = 1/10 + 0.704 /10 recurring which = 704/999 x 1/10 = 704/9990

so we have 1/10 + 704/9990 = 999/9990 + 704/9990 = 1703/9990 which has no common factors however I can still compare it to 7/40 by finding LCM which is 39960

so we have : 6812/39960 compared to 6993/39960 so the latter is bigger making 7/40 greater than 0.1704 (704 recurring)

I think that's correct?
I would do above problem this way:

o.1704704704.... < 0.170500000

7/40 = 0.175

0.175 > 0.1705 > 0.1704704704.....
 
Thanks for response Jomo. I can see the mistake I made but haven't you made a mistake yourself bu saying: 10000000000x = 123456789.123456789123456789... surely it would be: 10000000000x = 123456789.01234567890123456789...as the nought recurs as well?
Ooops, i thought you wrote that 0 is not part of the recurring pattern. The reason I thought that is because your problem had 2 zeros!

Then the result will come from subtracting the following.

Let x =.012345678901234567890123456789...
Then 10,000,000,000x = 123456789.01234567890123456789...

You say this is hard, but it is real easy. If you only have a repeating number and it starts at the decimal (not something like .23789789789... or 12.3123123...) then the answer is the repeating part with out a decimal divided by a sting of 9's. The number of 9s is the same as the number of digits that are repeating

Some examples: .123123123123....= 123/999. .0101010101010...= 01/99 or 1/99. .123451234512345...=12345/99999
 
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