Rounding to nearest

Question

I have the number 0.101 in binary. I want to round it to 2 places after the radix point using the algorithm rounding to the nearest. Here is how I do it.

If I want to have two places, then I will have one of the two possible resulting numbers: 0.10 or 0.11. Let's see how by how much each is different from the original:

1) 0.101 - 0.10 = 0.11

2) 0.101 - 0.11 = 0.10

The second one gives us less difference then the first, so using the algorithm rounding to the nearest, 0.101 should be rounded to 0.11. Is this how the algorithm works?

Answer 1

First of all your differences are wrong: 0.101 - 0.10 = 0.001 etc. Second rounding to nearest must be completed with a tie breaking rule if there are two representable numbers with the same difference to the number to be rounded; in your case both candidate numbers are $\pm 1/8$ from the original number:

0.101 - 0.10 =  0.001 = 1/8
0.101 - 0.11 = -0.001 = -1/8

Normally we would use round to nearest even (i.e. take the number with the least significant bit $0$) and in your case the result would be 0.10 because 0.11 is odd (the last bit is $1$).