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2. 6
5 13 . 4

− 10
3 4
− 30
4

Now, instead of stopping with a remainder of 4, we adjoin an insignificant 0 to the dividend, bring it
down, and continue the process:

2. 6
5 13 . 4 0
− 10
34
− 30
40

5goes into 40exactly 8times:

2. 6 8
5 13 . 4 0
− 10
34
− 30
40
− 40
0

The 0 remainder signals the end of the process. Thus 13.4 ÷ 5= 2.68.

Since we can adjoin as many insignificant 0’s as we need, it might seem that we can always continue
until we get a 0 remainder. But this doesn’t always happen. Sometimes, we never get a 0 remainder,
but instead, we get a sequence of non-zero remainders that repeats, forever. In this case, we get what
is know as a repeating decimal.

Example 168. Perform the division 3.2 ÷ 11.

Solution. We set up the long division process as usual.

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