> that only supports 15. it supports somewhere between 15 and 16, actually. There are 52 bits of precision in a double's mantissa[1]. This gives you log_10(2^5

> that only supports 15. it supports somewhere between 15 and 16, actually. There are 52 bits of precision in a double's mantissa[1]. This gives you log_10(2^52) = 52*log(2)/log(10) ~ 15.654 digits of precision, that is, the 16th decimal place is accurate about 65% of the time. You can also exactly represent in decimal the binary value stored, because as long as you have 52 decimal places, you have enough 2's in the denominator for all of the bits that you can represent (each 10 in the denominator pairs up with each 2 from each bit). -- [1] Plus an implicit bit, but not in the fractional part of the mantissa.