In my article “Binary Division” I showed how binary long division converts a fraction to a repeating bicimal. In this article, I’ll show you a well-known procedure — what I call the subtraction method — to do the reverse: convert a repeating bicimal to a fraction.
Bicimals
A bicimal is the base-two analog of a decimal; it has a bicimal point and bicimal places, and can be terminating or repeating.
A terminating bicimal has a finite number of bicimal places; a repeating bicimal (also known as a recurring bicimal or periodic bicimal) has an infinite number of bicimal places, due to a sequence of digits that repeat forever. There are two types of repeating bicimals: pure repeating (also known as immediate repeating or strictly repeating) and mixed repeating (also known as delayed repeating). In a pure repeating bicimal, the repeating part (also known as the repetend) starts immediately after the bicimal point. In a mixed repeating bicimal, a non-repeating part starts immediately after the bicimal point; it is then followed by a repeating part.
Here are some examples:
- 0.1101 (a terminating bicimal).
- 101.01 (a pure repeating bicimal).
- 11.1110 (a mixed repeating bicimal).
Like decimals, bicimals are created from fractions through long division. Also like decimals, bicimals can be converted back to fractions. You convert a bicimal to a fraction the same way you convert a decimal to a fraction — you just work in binary instead of decimal, and use powers of two instead of powers of ten.
A terminating bicimal is easy to convert to a fraction: the numerator of the resulting fraction is the bicimal itself, treated as an integer; the denominator is 2n, where n is the number of bicimal places. For example, 0.1101 = 1101/24 = 1101/10000 (which in decimal equals 13/16 = 0.8125). (Notice how I mixed decimal and binary numerals. Although potentially confusing, it is common practice, and actually makes things easier. I will be doing this throughout the article.)
Repeating bicimals take more work to convert. There are several methods; I will describe the subtraction method.
Subtraction Method
In the subtraction method, you take a bicimal b and create an expression that subtracts out its repeating part, and then rewrite that expression so that b is expressed as a fraction. Specifically, you create two nonnegative power of two multiples of b, mb and nb, such that mb – nb = i, where i is an integer. Using simple algebra, you rewrite this expression as a fraction: b(m-n) = i, or b = i/(m-n).
Here is a straightforward way to pick m and n. Let p be the length of the non-repeating part, and let r be the length of one cycle of the repeating part. Choose m = 2p+r, and choose n = 2p. These values of m and n shift b left by differing numbers of places, creating new bicimals with identical fractional parts but different integer parts. (For pure repeating bicimals, p will be 0 and thus n will be 1, so nb = b; that is, it’s not shifted.)
Examples
Example 1: 101.01 = 10000/11
Consider the pure repeating bicimal b = 101.01. Picking m = 22 = 4 shifts b left by two places, giving 10101.01. Now if we subtract b from that, we will remove the fractional part (leaving b as is means we’ve picked n = 20 = 1). Let’s show the subtraction algebraically, in mixed decimal and binary numerals:
mb – nb
4b – b = 10101.01 – 101.01
3b = 10000
11b = 10000
b = 10000/11
which in decimal equals 16/3 = 5.3.
Example 2: 11.1110 = 101111/1100
Consider the mixed repeating bicimal b = 11.1110. Picking m = 24 = 16 will shift b left by four places — pulling out the non-repeating part and one cycle of the repeating part — giving 111110.10. Picking n = 22 = 4 will shift a copy of b left by two places — pulling out just the non-repeating part — giving 1111.10. Now let’s do the algebra:
mb – nb
16b – 4b = 111110.10 – 1111.10
12b = 101111
1100b = 101111
b = 101111/1100
which in decimal equals 47/12 = 3.916.
Examples Redone Using Only Binary Numerals
You can perform the subtraction method using only binary numerals (thanks James for the comment); I will redo the two examples in this way.
Example 1: 101.01 = 10000/11
mb – nb
100b – b = 10101.01 – 101.01
11b = 10000
b = 10000/11
Example 2: 11.1110 = 101111/1100
mb – nb
10000b – 100b = 111110.10 – 1111.10
1100b = 101111
b = 101111/1100
A Way to Avoid Binary Arithmetic on the Left Hand Side
The binary subtraction on the left hand side of the equation can be avoided; let’s manipulate the left hand side algebraically (remember that m = 2p+r and n = 2p):
mb – nb
2p+rb – 2pb
2p(2rb – b)
2pb(2r – 1)
(2r – 1)2pb
You don’t need to do any arithmetic to compute (2r – 1)2p: it is a string of r 1s followed by p 0s. For example 1, r = 2 and p = 0 gives 11; for example 2, r = 2 and p = 2 gives 1100.
On Going Directly to a Fraction Written in Binary
If you look around the Web (for example, Wikipedia), you’ll see this procedure is typically used to convert a bicimal straight to a fraction written in decimal numerals. For example 2, the steps would be:
mb – nb
16b – 4b = 111110.10 – 1111.10
12b = 47
b = 47/12
which you’d then say is 101111/1100 in binary.
Conceptually, I see the process as bicimal -> fraction written in binary -> fraction written in decimal -> decimal, which is why I prefer my presentation.
Other Methods
Read my articles about the direct method and the series method — two other ways to convert a bicimal to a fraction.
Good method! Am I right in saying that this should work in all bases?
Decimal:
3.4’83 (apostrophe signifies start of repetend)
md – nd
1000d – 10d = 3483.’83 – 34.’83
990d = 3449
d = 3449/990
Correct!
Dozenal/uncial (0123456789XƐ):
3.4’83
mu – nu
1000u – 10u = 3483.’83 – 34.’83
ƐƐ0u = 344Ɛ
u = 344Ɛ/ƐƐ0
Correct!
Hexadecimal (0123456789ABCDEF):
3.4’83
mx – nx
1000x – 10x = 3483.’83 – 34.’83
FF0x = 344F
x = 344F/FF0
Correct!
(Checked on W|A).
@James,
Yes, it’s good for any base (of course you must have the means — like Wolfram Alpha — to do arithmetic and base conversion in that base).
You do the method slightly differently than I do. You do the left hand side arithmetic in the non-decimal base. For example, here is your base 12 example (copied here in my formatting):
mu – nu
1000u – 10u = 3483.83 – 34.83
ƐƐ0u = 344Ɛ
u = 344Ɛ/ƐƐ0
I would have done it this way (I’m using 123 = 1728 and 121 = 12):
mu – nu
1728u – 12u = 3483.83 – 34.83
1716u = 344Ɛ
ƐƐ0u = 344Ɛ
u = 344Ɛ/ƐƐ0
I think I like your way better: all numerals are in one base, and the required left shifting is more apparent.
Your examples also highlight the form the denominators take: decimal, 9s followed by 0s; dozenal, Ɛs followed by 0s; hexadecimal, Fs followed by 0s. I will talk about this more in my next article, which will discuss what I call the
denominator methoddirect method.As usual, thanks for the comment.