Hello: Sorry to have troubled you with the basic sql question below. Of course, it has no place here. The algebra will, of course, yield results in the same precision as the decimal types used in the calculation. George Hello Again: I'm trying to do some clinical research on large data sets and I ran into a problem represented in the following simplified example: geo-> select * from test2; +-----------+-----------+ | a | b | +===========+===========+ | 3.05 | 3.07 | | 3.06 | 3.04 | | 3.13 | 3.16 | | 3.22 | 3.22 | | 3.05 | 3.05 | | 3.07 | 3.06 | | 3.08 | 3.05 | | 3.12 | 3.09 | | 3.20 | 3.17 | | 3.17 | 3.20 | | 3.09 | 3.11 | | 2.74 | 2.73 | | 2.79 | 2.78 | | 2.80 | 2.81 | +-----------+-----------+ 14 rows geo-> select * from test2 where a=b; +-----------+-----------+ | a | b | +===========+===========+ | 3.22 | 3.22 | | 3.05 | 3.05 | +-----------+-----------+ 2 rows geo-> select * from test2 where (a/b)-1=0; +-----------+-----------+ | a | b | +===========+===========+ | 3.06 | 3.04 | | 3.22 | 3.22 | | 3.05 | 3.05 | | 3.07 | 3.06 | | 3.08 | 3.05 | | 3.12 | 3.09 | | 3.20 | 3.17 | | 2.74 | 2.73 | | 2.79 | 2.78 | +-----------+-----------+ <>9 rows geo-> select * from test2 where (b/a)-1=0; +-----------+-----------+ | a | b | +===========+===========+ | 3.05 | 3.07 | | 3.13 | 3.16 | | 3.22 | 3.22 | | 3.05 | 3.05 | | 3.17 | 3.20 | | 3.09 | 3.11 | | 2.80 | 2.81 | +-----------+-----------+ <>7 rows I thought that if a=b, then (a/b)-1 and (b/a)-1 must both also be equal to 0. Columns a and b are type decimal(7,2) and the data was entered to two decimal places by hand using insert statements. Thanks and best.......George ge I