MGMT 1030 Lecture 33: MGMT 1030 Lecture 33 Notes

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MGMT 1030 Lecture 33 Notes Single Zero
Introduction
Complements are found by subtracting the value from the modulus, in this case, 1000.
This ethod assues a sigle zeo, sie  −  od  is zeo.
Again, as with the previously discussed complementary methods, notice that the
complement of the complement results in the original value.
See the examples on the facing page.
Thee is a alteatie ethod fo opleetig a ’s opleet ue.
First, observe that 1000 = 999 + 1
You eall that the 9’s opleet as foud y sutatig eah digit fo 9: 9s op
= 999 − alue
Fo the peious euatio, the ’s opleet a e eitte as s op = 
− alue = 999 +  − alue = 999 − alue + 1 or, finally, 10s comp = 9scomp + 1
This gies a siple alteatie ethod fo oputig the ’s opleet alue
Fid the 9’s opleet, hih is easy, ad add  to the esult.
Either method gives the same result. You can use whichever method you find more
convenient.
This alternative method is usually easier computationally, especially when working with
binary numbers, as you will see.
Additio i ’s opleet is patiulaly siple.
Sie thee is oly a sigle zeo i ’s opleet, sums that cross the modulus are
unaffected.
Thus, the carry that results
EXAMPLE
Fid the ’s opleet of 7.
As a eide, ote that the uestio asks fo the ’s opleet of 7, ot the ’s
complement representation.
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