Floating-point: "The leading 1 is 'implicit' in the significand." -- ...huh?

Question

I'm learning about the representation of floating-point IEEE 754 numbers, and my textbook says:

To pack even more bits into the significand, IEEE 754 makes the leading 1-bit of normalized binary numbers implicit. Hence, the number is actually 24 bits long in single precision (implied 1 and 23-bit fraction), and 53 bits long in double precision (1 + 52).

I don't get what "implicit" means here... what's the difference between an explicit bit and an implicit bit? Don't all numbers have the bit, regardless of their sign?

Answer 1

Yes, all normalised numbers (other than the zeroes) have that bit set to one ^(a), so they make it implicit to prevent wasting space storing it.

In other words, they save that bit totally, and reuse it so that it can be used to increase the precision of your numbers.

Keep in mind that this is the first bit of the fraction, not the first bit of the binary pattern. The first bit of the binary pattern is the sign, followed by a few bits of exponent, followed by the fraction itself.

For example, a single precision number is (sign, exponent, fraction):

<1>  <--8--->  <---------23---------->  <- bit widths
 s   eeeeeeee  fffffffffffffffffffffff

If you look at the way the number is calculated, it's:

(-1)sign x 1.fraction x 2exponent-bias

So the fractional part used for calculating that value is 1.fffff...fff (in binary).

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^(a) There is actually a class of numbers (the denormalised ones and the zeroes) for which that property does not hold true. These numbers all have a biased exponent of zero but the vast majority of numbers follow the rule.