So you may have noticed that when you divide an integer by 11, if it is not a multiple of 11, you get an alternating repeating decimal that looks like “.ABABABABABABAB…”. You may have also noticed that A and B always add up to 9. Let’s discuss why that is.
So let’s divide by 11, where is an integer less than 11. You only need to consider integers less than 11, because any improper fraction can be expressed as an integer plus a fraction where the numerator is less than the denominator.
Now if you convert that fraction into a decimal you get an alternating repeating decimal, , where 3 and 6 add to 9.
This decimal can also be expressed in another form by turning the decimal places into a sum:
We can then generalize this for any integer less than 11.
Where A and B are integers.
So now we want to prove that .
First let’s put the above equation into a summation form, since the sum is infinitely long, but can be expressed simply because A is multiplied by odd powers of 10, and B is multiplied by even powers of 10. This is done to simplify the expression so there is one instance of A and one instance of B.
Now we want to simplify this a bit more to get rid of the infinite sum.
First let’s pull out some factors of to make the sums in each term equivalent.
Now the sum can be reduced to a number:
So now we have:
which can be reduced to a nice expression:
From this we want to prove that , where are integers.
To prove this, let’s plug in a more general equation and solve for . The result is:
Now since and are integers, must also be an integer, meaning must be a multiple of 9. Since and can only hold one digit, the only options are 9 and 18. But only occurs when , which corresponds to .
So for the case , it must be that .