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this post was submitted on 12 Mar 2024
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Asklemmy
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A 64 bit IEEE float has 53 significant bits (the “mantissa” or “significand”), and log~10~(2^53^) is 15.9546.
Isn't it just 15 significant figures then?
I would round up to 16.
Yeah I wasn't sure if it would be correct to throw out the exponent entirely or if it might end up contributing some amount to the final accuracy of the number. I hadn't spent a lot of time thinking about the problem.
Yeah the exponent just allows you to represent lots of magnitudes, but it wouldn’t contribute to the accuracy because you basically have 1.xyz * 2^exponent^. So the xyz significand is the only part that counts for significant digits. Although I guess in some sense you are partially right, because the exponent exists it is assumed that the first bit is always one, since otherwise you would just adjust the exponent to the first one, so only 52 bits have to be stored.