An exponent field of 0 is used to encode numerical values that lie below the normal range.
If the exponent and significand fields of an encoding are
both 0, then the encoded value itself is 0 and the encoding is said to be a *zero*. If the exponent
field is 0 and the significand field is not, then
the encoding is either *denormal* or *pseudo-denormal*:

*(a) If
, then is a zero encoding for F.*

*(b) If
and either is implicit or
,
then is a denormal encoding for F.*

*(c) If is explicit and
, then is a pseudo-denormal encoding for F.*

Note that a zero can have either sign:

There are two differences between the decoding formulas for denormal and normal encodings:

- For a denormal encoding for an implicit format, the integer
bit is taken to be 0 rather than 1.
- The power of 2 represented by the zero exponent field of a denormal
or pseudo-denormal encoding is
rather than
.

We also define a general decoding function:

*(a)
.*

*(b)
.*

*(c)
.*

PROOF: (a) is trivial; (b) and (c) follow from Lemmas 4.1.13 and 4.1.14.

The class of numbers that are representable as denormal encodings is recognized by the following predicate.

*(a) ;*

*(b)
;*

*(c) is
-exact.*

If a number is so representable, then its encoding is constructed as follows.

Next, we examine the relationship between the decoding and encoding functions.

PROOF: Let , , , and . Since ,

This establishes that is representable as a denormal.

Now by Definition 5.3.3,

Therefore, by Definitions 5.3.1, 5.3.6,
and 5.1.4 and Lemmas 2.4.9 and 2.2.5,

PROOF: Let , , , and . By Lemma 2.4.1, is a -bit vector and by Lemma 2.4.7,

Since is -exact,

The smallest positive denormal is computed by the following function:

For any format ,

*(a)
;*

*(b)
is representable as a denormal in ;*

*(c) If is representable as a denormal in , then
.*

PROOF: Let , , and . It is clear that is positive. To show that is -exact, we need only observe that

Every number with a denormal representation is a multiple of the smallest positive denormal.

PROOF: Let and . For , let . Then and

Suppose that is representable as a denormal for some , . Then is -exact, and by Lemma 4.2.16, so is . But since , it follows from Lemma 4.2.5 that is also -exact. Since

Now suppose that is representable as a denormal. Let . Clearly, , and . It follows from Lemma 4.2.17 that , and consequently, is -exact. Thus, by Lemma 4.2.16, .

David Russinoff 2017-08-01