floating-pointieee-754floating-point-precision# Is it 52 or 53 bits of floating point precision?

I keep on seeing this nonsense about 53 bits of precision in 64-bit IEEE floating point representation. Would someone please explain to me how in the world a bit that is stuck with a 1 in it contributes ANYTHING to the numeric precision? If you had a floating point unit with bit0 stuck-on with 1, you would of course know that it produces 1 less bit of precision than normally. Where are those sensibilities on this?

Further, just the exponent, the scaling factor without the mantissa, completely specifies exactly where the leading bit of the number is, so no leading bit is ever used. The 53th bit is about as real as the 19th hole. It is merely a (useful) crutch to aid the human mind and the logic for accessing such values in binary. To claim otherwise is double counting.

Either all the books and articles claiming this 53rd bit nonsense are wrong, or I am an idiot. But a stuck bit is a stuck bit. Let's hear the arguments to the contrary.

Solution

The mathematical significand^{1} of an IEEE-754 64-bit binary floating-point object has 53 bits. It is encoded with the combination of a 52-bit field exclusively for the trailing bits of the significand and some information from the exponent field that indicates whether the leading 53^{rd} bit is 0 or 1.

Since the trailing significand field is 52 bits, some people refer to the significand as 52 bits, but this is sloppy terminology. The significand field does not contain all the information about the significand, and the complete significand is 53 bits.

It is not true that the leading bit of the significand is never used (as anything other than 1). When the encoding of the exponent is zero, the leading bit of the significand is 0 instead of the more frequent 1.

^{1} “Significand” is the preferred term, not “mantissa.” A significand is linear, a mantissa is logarithmic.

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