Fraction to decimal
Terminating decimals
The way to convert a fraction to a decimal is of course through the familiar long division algorithm, which was already shown in a previous blog post of mine. The way I show it here is how I learned it at school, which is a fast method but which leaves nearly nothing explained in its working. Let's convert 43/5 to decimal form:
5 )4 3
We start by seeing how many times the denominator 5 goes into the first digit of the numerator, 4. It goes 0 times into it and leaves a remainder of 4. We write the integer quotient as the first digit at the top. The remainder we write in front of the next digit in the numerator.
0 5 )4 43
We now see how many times 5 goes into 43. It goes 8 times into it and leaves 3 as a remainder. We write the integer quotient as the second digit at the top. We now have no more digits in the numerators, so we add a decimal point and a zero after it to create more digits. We also add a decimal point to the quotient at the top.
0 8 . 5 )4 43 . 30
And repeat.
0 8 . 6 5 )4 43 . 30 00
Since the last remainder was 0, if we had to continue from here onwards we'd be adding nothing by zeros to the top quotient which would be pointless. So instead we declare the number as terminating decimal and stop there. The answer the top quotient, that is, 43/5 = 8.6
What's happening here is that we're first trying to find the tens digit of the quotient by seeing how many times 50 goes into 43 which is 0 (tens) remainder 43. Then we're trying to find the units digit of the quotient by seeing how many times 5 goes into 43 which is 8 (units) remainder 3. Then we're trying to find the tenths digit of the quotient by seeing how many times 0.5 goes into 3, or equivalently, how many times 5 goes into 30, which is 6 (tenths) remainder 0.
Recurring decimals
Let's use the previous method to convert 1/3 to decimal form:
3 )1
We start by seeing how many times the denominator 3 goes into the first digit of the numerator, 1. It goes 0 times into it and leaves a remainder of 1.
0 . 3 )1 . 10
We now see how many times 3 goes into 10. It goes 3 times into it and leaves 1 as a remainder.
0 . 3 3 )1 . 10 10
And repeat.
0 . 3 3 3 )1 . 10 10 10
As you can see, this process will continue indefinitely, meaning that the 3 at the top will keep on repeating itself. This makes the quotient a recurring decimal, the proof of which is that one of the remainders after the decimal point was reached appeared twice and hence the same number must come out again.
Decimal to fraction
Terminating decimals
To convert a terminating decimal to a fraction you simply multiply it by a power of 10 that is large enough to make it a whole number, then put that same power of 10 as the denominator under the whole number. For example, if you have 12.3456 to convert to fraction form:
12.3456 = 12.3456 x 10000 / 10000
= 123456/10000
= 7716/625
Recurring decimals
Of course the previous method cannot be used when the fractional part is infinitely long. But there is a trick we can use. The fraction 1/9 has the following decimal expansion:
0 . 1 1 ... 9 )1 . 10 10 ...
In other words, it gives a decimal number with an infinite sequence of 1s. We can use this to our advantage so that we can obtain infinite sequences of any digit by simply multiplying the digit by 1/9.
1/9 = 0.111...
2/9 = 0.222...
3/9 = 0.333...
etc
The interesting thing about this is that when you try to get an infinite sequence of 9s, you get the whole number 1. This is one of the proofs that 0.999... = 1.
But this is only good for single digit recurrences. For two digit recurrences we can use 1/99:
0 . 0 1 0 1 ... 99 )1 . 10 100 10 100 ...
This is useful. We can replace every 1 with any digit by multiplying 1/99 by that digit.
1/99 = 0.0101...
2/99 = 0.0202...
3/99 = 0.0303...
etc
We can also shift those digits one place to the left by multiplying the digit by 10:
10/99 = 0.1010...
20/99 = 0.2020...
30/99 = 0.3030...
etc
So we can control both digits by adding these two fractions together:
10/99 + 3/99 = 0.1010... + 0.0303... = 0.1313...
Which of course simplifies to
13/99 = 0.1313...
That is, you just multiply the 2 digit number you want to be repeated by 99.
Can this be extended to any number of digits? 1/999 gives the following expansion:
0 . 0 0 1 0 0 1 ... 999 )1 . 10 100 1000 10 100 1000 ...
So with 1/999 we can repeat any 3 digit recurrences. With each new 9 added to the denominator we are postponing the number of times the remainder must be moved before it is big enough to be divided by the denominator. That postponement results in another 0 added to the recurring decimal.
So in short, to convert a recurring decimal number to a fraction, just take the repeating part of the decimal and put it over a denominator consisting of as many 9s as there are digits in the repeated number:
0.01230123... = 0123/9999
But this will only work if the decimal number consists of nothing but repeating numbers. What if you have the following decimal:
210.67801230123...
In this case, the repeating part is the "0123" but it starts with other non-repeating digits. Not to worry, we just separate the two parts of the number into a sum:
210.67801230123... = 210.678 + 0.00001230123...
The recurring term can be multiplied by a power of 10 that will make it lose its leading zeros. Of course we can't just multiply it by a power of 10 without also dividing it by the same number in order to avoid changing its value:
210.67801230123... = 210.678 + (1000 x 0.00001230123... / 1000)
210.67801230123... = 210.678 + (0.01230123... / 1000)
We can now convert each term separately:
210.67801230123... = 210.678 + (0.01230123... / 1000)
= 210678/1000 + ((0123/9999) / 1000)
= 140437963/666600
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