Of the two values of the exponent field that lie outside of the range of normal encodings, the upper extreme is reserved for the encoding of infinities and other non-numerical entities, which will not be discussed here, while 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 with respect to a format with implicit MSB are both 0, with exponent field 0, then the encoded value itself is 0. If the exponent field is 0 and the significand field is not, then if the significand field is also 0, then the encoded value itself the encoding is said to be denormal.
Theere are two differences between the decoding formulas for denormal and normal representations:
PROOF: (a) is trivial; (b) and (c) follow from Lemmas 4.1.11 and 4.1.12.
The class of reals that are representable as denormal encodings is recognized by the following predicate.
|(b) , and|
|(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 . Since ,
and by Lemma 4.1.2,
which is equivalent to Definition 5.3.3(b). In order to prove (c), we must show, according to Definition 4.2.1, that
Now by Definition 5.3.2,
Therefore, by Definitions 5.3.1, 5.3.4, and 5.2.1 and Lemmas 2.4.9 and 2.2.5,
PROOF: Let . By Lemma 2.4.1, is a -bit vector and by Lemma 2.4.7,
Since is -exact,
and since ,
by Lemma 4.1.7, which implies
Finally, according to Definition 5.3.2,
The smallest positive denormal is computed by the following function:
|(c) If , , and , then .|
PROOF: It is clear that is positive. To show that is -exact, we need only observe that
holds and moreover, is the smallest positive that satisfies .
Every number with a denormal representation is a multiple of the smallest positive denormal.
PROOF: For , let . Then and
We shall show, by induction on , that for . First note that for all such ,
, i.e., , and hence, .
Now suppose that and . Let . Clearly, , and . It follows from Lemma 4.2.17 that , and consequently, is -exact. Thus, by Lemma 4.2.16, .