becomes, after conversion to 2’s complement by Eq. (2.14),
P
2
=A
2C
×(B
2C
)
=(2
n
− A
2
)×(2
n
−B)Mod 2
n
=2
n
×2
n
−2
n
A
2
−2
n
B
2
+ A
2
×B
2
Mod 2
=2
2n
−2
n
−2
n
+ A
2
×B
2
Mod 2
n
orP
2
=A
2
×B
2
,(2.23)
where2
2n
−2
n
−2
n
(Mod2
n
)=0.Thus,theproductoftwonegativebinarynumbers
in2’scomplementnotationisthepositiveproductof ...
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