Shumëzimi i numrave binarë

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Numrat Binare

y-bartja 2y- për shkak te numrave binare Psh. 3=1+2*1 dmth. bartja është 1, 3=z, kurse 1 është pozita (x).


a2a1a0

b2b1b0

x4x3x2x1x0


0=a0b0=z0=x0+2y1

1=y1+a1b0+a0b1=z1=x1+2y2

2=y2+a2b0+a1b1+a0b2=z2=x2+2y2

3=y3+a2b1+a1b2=z3=x3+2y4

4=y4+a2b2=z4=x4+2y5

5=y5



a2a1a0

b2b1b0

c2c1c0

___________________

x6x5x4x3x2x1x0



0=a0b0c0=z0=x0+2y1

1=y1+a1b0c0+a0b1c0+a0b0c1=z1=x1+2y2

2=y2+a2b0c0+a1b1c0+a1b0c1+a0b2c0+a0b1c1+a0b0c2=z2=x2+2y2

3=y3+a2b1c0+a2b0c1+a1b2c0+a1b1c1+a1b0c2+a0b2c1+a0b1c2=z3=x3+2y4

4=y4+a2b2c0+a2b1c1+a2b0c2+a1b2c1+a1b1c1+a0b2c2=z4=x4+2y5

5=y5+a2b2c1+a2b1c2+a1b2c2=z5=x5+2y6

6=y6+a2b2c2=z6=x6+2y7

7=y7

Sipas formules :

a2a1a0

b2b1b0

c2c1c0

_________________________

x6x5x4x3x2x1x0



0=a0b0c0=z0=x0+2y1

1=y1+a1b0c0+a0b1c0+a0b0c1=z1=x1+2y2

2=y2+a2b0c0+a1b1c0+a1b0c1+a0b2c0+a0b1c1+a0b0c2=z2=x2+2y2

3=y3+a2b1c0+a2b0c1+a1b2c0+a1b1c1+a1b0c2+a0b2c1+a0b1c2=z3=x3+2y4

4=y4+a2b2c0+a2b1c1+a2b0c2+a1b2c1+a1b1c1+a0b2c2=z4=x4+2y5

5=y5+a2b2c1+a2b1c2+a1b2c2=z5=x5+2y6

6=y6+a2b2c2=z6=x6+2y7

7=y7



120110

121100

121110

______________________

1716051403021100



0=101=0=0+20

1=0+000+111+101=1=1+20

2=0+101+011+001+111+111+101=2=0+21

3=1+111+101+011+011+001+111+111=4=0+22

4=2+111+111+101+011+011+111=5=1+22

5=2+111+111+011=4=0+22

6=2+111=3=1+21

7=1