Mathematicians

There are two methods here. The essential elements appear to be the same. Repeating decimals can be converted into fractions by the use of simultaneous equations. The trick is to identify how many numbers or decimal places are involved in the "repeating part" of the repeating decimal. Let's do one, and let's make it a familiar one. We'll take 0.3333333... which is 1/3 as you already know. First, there is only one digit that repeats. It's the 3. So we'll need to set up two equations, and the first one is that X = 0.3333333... In that equation, "X" is the decimal equivalent of the number. Now we'll create a second equation to do the simultaneous equation thing. Since there is only one digit that repeats, we're going to move the decimal one place by multiplying that whole equation by 101 which is 10. [X = 0.3333333...] x 10 = 10X = 3.333333... Notice that both sides of the equation were multiplied by 101 here. This is key. We are going to have to multiply our repeating decimal by a factor of 10 that shifts all of the repeating sequence to the left of the decimal. If we'd had 0.47474747.... for our fraction, we'd have had to multiply the equation by 102 because the "repeating part" there, which is '47' is two digits long. If it had been 0.5678567856785678.... we'd be stuck multiplying by 104 because the "repeating part" there, which is '5678' is four digits long. Back to our problem at hand. The 10 times "X" equals 10X and the 0.3333333 times 101 = 3.333333... Now we have two equations, and we're going to subject one from the other. So let's do that. 10X = 3.333333... X = 0.3333333... 10X - X = 9X, and 3.333333... - 0.3333333... = 3 Notice how the "repeating part" of the decimal "dropped out" or "disappeared" there? That's why we built two equations and subtracted one from another. We need to get the "repeating part" to get lost. Look at what's left. It this equation: 9X = 3 Can you handle that? Sure you can. Divide both sides by 9 and you'll have X = 3/9 which reduces to 1/3 and presto! Problem solved. If we'd done 0.474747474747... it would look like this: X = 0.474747474747... 100X = 47.4747474747... Note that we multiplied by a factor of 10 enough to shift a "whole block" of the repeating part to the left of the decimal. Now subtract the top equation from the bottom one. 100X = 47.4747474747... X = 0.474747474747... 100X - X = 99X and 47.47474747... - 47.4747474747... = 47 99X = 47 and X = 47/99 As 47 is a prime number, we can't reduce this fraction. There are some simple rules that apply when converting any repeating decimal to a fraction. First we'll perform the "construction" of an initial equation where we set the fraction ("X") equal to the repeating decimal. Then we'll "manufacture" a second equation from the first by multiplying the first equation by 10n where n = the number of digits in the "repeating part" of the repeating decimal so they all kick over to the left of the decimal. Then we solve the simultaneous equations, and lastly reduce our fraction. Try a few and you'll be able to slam dunk this bad boy every time you see it. This is made quite easy with the following observations:

5/9 = 0.5555

7/9 = 0.7777

12/99 = 0.121212

23/99 = 0.232323

456/999 = 0.456456456

78/999 = 078/999 = 0.078078078

So we can see that fractions with a denominator of 9, 99, 999, 9999, are pretty useful for making repeating decimals. All we have to do is to reduce the fraction to its lowest terms. For example, let's take the repeating decimal 0.027027027...

Clearly this is 27/999 = 1/37 (having divided top and bottom of the fraction by 27)

Now let's try something more tricky. Take a look at 0.4588888888...

This isn't simply something divided by 99.. since the 45 bit doesn't repeat. What we need to do is move the decimal point over to the start of the repeating bit. In this case we multiply by 100 to get 45.888888...

Now we know the fraction part is 8/9. In total we have 45 + 8/9 = 413/9 (changing into a top-heavy fraction will make things easier for us).

So 45.88888... = 413/9

Now just divide both sides by 100 to get:

0.458888... = 413/900

Pascal's triangle. It is named after the French mathematician, Blaise Pascal in much of the Western world, although other mathematicians studied it centuries before him in India, Greece, Iran, China, Germany, and Italy. The rows of Pascal's triangle are conventionally enumerated starting with row n = 0 at the top. The entries in each row are numbered from the left beginning with k = 0 and are usually staggered relative to the numbers in the adjacent rows. (Wikipedia)

The top of the triangle -

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

And you can continue on, using the same rules.