Today we are going to learn about new methods to do computations with recurring decimals.
I. Adding Two Recurring Decimals
Let us say we have find the sum of two recurring decimals $3.\ \overline{68}$ and $1.\ \overline{783}$.
One of the approaches would be to convert both of them to fractions using the method explained , add them as fractions and convert them back to decimal format.. But that would required a large amount of calculation and therefore, will increase the possibility of errors.
Let us look at another method, which is much faster and less error prone.
For both the given numbers let us ignore the integer part for now and we can easily add them back later.
So let us find the sum of $0.\overline{68}$ and $0.\overline{783}$.
Let us observe that the recurrence length of the first number is $2$ and that of the second one is $3$.
Now if we take the $LCM$ of $2$ and $3$ we get $6$, hence, for both fractions we can for a repeating pattern of $6$ digits.
$0.\overline{68} = 0. 686868\ 686868\ 686868\ ...$ and
We can take any one group and add the two numbers like simple integer additions:
$\ \ \ 686868$
$\underline{+ 783783}$
$(1)470651$
You can see that we have written the leftmost $1$ within brackets, that is because it is the carry over $1$ and is making the length of the result $7$ digits.
For every group of recurring pattern there will be a carry over of $1$ which will get added to the rightmost digit of the next group.
So, the new $6$ digit group of the result will become $470652$ and the last carry over $1$ will get added to the non-recurring part, that is the part on the left of the recurring part, of the result.
In our case:
$\ \ \ 0. 686868\ 686868\ 686868\ ...$
$\underline{+ 0.783783\ 783783\ 783783\ ...}$
$(1). 470652\ 470652\ 470652\ ...$
Now we can easily add the integer parts $3$ and $1$, which we had left out earlier, and the carried over $1$, to get our final answer as:
Let us try to add the numbers $3.1\ \overline{796}$ and $4.\ \overline{85}$
We should observe that for the first number, the recurrence starts after one place of decimal, while for the second number the recurrence starts immediately after the decimal point.
To take care of that we can rewrite the second number as:
$4.8\ \overline{58}$
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Now, taking the $LCM$ of the two recurrence lengths $2$ and $3$ as $6$ we can write our problem as:
$\ \ \ 3.1\ 796796\ 796796\ ...$
$\underline{+ 4.8\ 585858\ 585858\ ...}$
Adding the two recurring groups we get $796796 + 585858 = (1)382654$
We can add the carry over $1$ to the rightmost digit of each group and we get our recurring group as $382655$
We can add the non-recurring part of our numbers to get:
$3.1 + 4.8 = 7.9$
And we need to add the carry over from the recurring group to the rightmost digit of this number to get $8.0$
Therefore, our final answer is $8.0\ 382655\ 382655\ ... = 8.0\ \overline{382655}$
II. Subtracting Recurring Decimals
Now, let us see how to subtract one recurring decimal number from another one:
Let us subtract $2.4\ \overline{859}$ from $7.\ \overline{46}$
As before, first we will need to rewrite the second number, so that the recurring pattern start after one place of decimal.
$7.\ 46\ 46\ 46\ ... = 7.4\ 64\ 64\ 64\ ...$
Now let us write the both numbers using a recurring length of $6$, and we get our sum as:
$\ \ \ 7.4\ 646464\ 646464\ ...$
$\underline{- 2.4\ 859859\ 859859\ ...}$
We need to subtract the group $859859$ from $646464$, but we cannot do it without borrowing. So we will borrow $1$ from the preceding group.
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Thus, we can subtract $859859$ from $1646463$ and we get $1646463 - 859859 = 780604$ (remember we borrow $1$ from each preceding groups last digit $4$ and placed it before the succeeding group, thus $646464$ became $1646463$
Now, we can subtract the non-recurring part as $7.3 - 2.4 = 4.9$ (remember we borrowed $1$ from the $4$ of $7.4$ for the next group, that is why $7.4$ became $7.3$).
And we get the result as: $4.9\ 780604\ 780604\ ... = 4.9\ 780604$
The methods explained above are much easier to use and compared to the conversion method. As an exercise we recommend that you try using both the conversion method and the methods explained above to solve the problems given in the examples and develop a good feeling of the method.