due to a poor choice of scaling factor. Choosing a better scale factor Most data sets will not have a perfect scale factor; most likely, there will be some error introduced by the scaling process. However, it may be possible to choose a better scale factor. The ideal scale factor may not be the smallest, but rather one that preserves as much precision as possible. Dividing a number by a power of two is the same as shifting all the bits to the right once for each power of two. (This is the binary equivalent to shifting all decimal digits to the left or right when, respectively, multiplying or dividing by powers of ten.) The pattern of bits does not change, it just moves the number of places equal to the binary exponent (for instance, 3 places to the right when dividing by 8 = 23). On the other hand, when dividing by a number that is not an integer power of two in binary, the bit pattern changes. This is likely to produce a bit pattern with more bits to the right of the binary point, artificially introducing required precision. This is especially true when the fractional part has a denominator that is not a power of two, as all fractions not reciprocals of powers of two recur in binary. Therefore, it is almost always preferable to use a scale factor that is a power of two. It may still be possible to lose bits that get shifted right off the end of the field as a result of truncation, but this avoids introducing new bits that will be imprecise (due to rounding error) or truncated. As an illustration the use of powers of two in the scale factor, a scale factor of can be applied to the above data set. The binary values for the original data set are given below: 154 = 1001 1010 101 = 0110 0101 54 = 0011 0110 3 = 0000 0011 0 = 0000 0000 160 = 1010 0000 Being integers between 0 and 255, these all can be represented precisely with 8 bits. Scaling these by is the same as dividing by 16, which is the same as shifting the bits 4 places to the right. In this case, scaling is done by inserting a binary point between the first 4 bits and last 4 bits of each number. That happens