Nombers, coding, numerical analysis,

+Numbers and coding in numerical analysis


++Codage et nombres dans le électronique numérique

1-A- Exemple 1:

123: integer nombre  (base 10) = 1111011?

Demonstration :

Démonstration :

The integer nombre123 base 10 is (le nombre entier 123 à base 10 est) :1*(10)^2   + 2*(10)^1 + 1*(10)^0

=  1*(2)^6 + 1*(2)^5   + 1*(2)^4 + 1*(2)^3  + 0*(2)^2 + 1*(2)^1  + 1*(2)^0 = 0111,1011 (base 2, numéro binaire, binary nomber)


=   64      +    32      +  16       +  8          +  0       +  2         + 1 = 123


Fig-1

Fig-2

1-A- Exemple 2:

10: integer nomber (base 10) = 1010

 1*(10)^1 + 0*(10)^0

=  1*(2)^3  + 0*(2)^2 + 1*(2)^1  + 0*(2)^0 = 1010 (Base 2)

=   8          +  0       +  2         + 0  =  10 (base 10)


Fig-3

Fig-4


1-A- Exemple 3:

15: integer (base 10) = 1111

 1*(10)^1 + 5*(10)^0

=  1*(2)^3  + 1*(2)^2 + 1*(2)^1  + 1*(2)^0 = 1111 (base 2)

=   8          +   4       +   2         +  1 = 15 (Base 10)

1-A- Exemple 4:

127: integer (base 10) = 0111,1111

1*(10)^2   + 2*(10)^1 + 7*(10)^0


=            1*(2)^6     + 1*(2)^5  + 1*(2)^4 + 1*(2)^3   + 1*(2)^2 + 1*(2)^1  + 1*(2)^0 =0111,1111

=            +  64       + 32           +  16     +  8          +    4       +   2       +    1 = 127

1-A- Exemple 5:

128: integer (base 10) = 1000,0000

1*(10)^2   + 2*(10)^1 + 8*(10)^0


=          1*(2)^7     +    0*(2)^6     + 0*(2)^5  + 0*(2)^4 + 0*(2)^3   + 0*(2)^2 + 0*(2)^1  + 0*(2)^0 =1000,0000

=           128       + 0           +  0          +  0         +    0       +   0       +    0 = 128


128  = 1000;0000= 80 (base hexa)

128 = 010,000,000 = 200 (base 8)

1-A- Exemple 6:

255: integer  (base 10) = 1111,1111

2*(10)^2   + 5*(10)^1 + 5*(10)^0


=          1*(2)^7     +    1*(2)^6   + 1*(2)^5  + 1*(2)^4 + 1*(2)^3   + 1*(2)^2 + 1*(2)^1  + 1*(2)^0 =1111,1111 ( base 2)

=           128       +     64           +   32         +   16       +    8       +   4       +    2 +  1 = 255 ( Base 10)

1-A- Exemple 7:

255: integer  (base 10) = 1111,1111

2*(10)^2   + 5*(10)^1 + 5*(10)^0


=          1*(2)^7     +    1*(2)^6     + 1*(2)^5  + 1*(2)^4 + 1*(2)^3   + 1*(2)^2 + 1*(2)^1  + 1*(2)^0 =1111,1111

=           128       +     64           +   32         +   16       +    8       +   4       +    2 +  1 = 255

1111,1111 = FF (base hexa)
011,111,111 = 377 (base octa)



Merci thank you dunke my children and happy day.

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