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Exercise 1.1 · Q3

Q.Simplify the following and write in decimal notation: (1000101111100)2+(1100101000100)2(1000101111100)_2 + (1100101000100)_2, (11101100100)2(11101100100)_2.

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Each binary numeral is converted to decimal by summing dk⋅2kd_k\cdot 2^k over its digits; the first two are then added.

[!FORMULA]

(dndn−1…d1d0)2=∑k=0ndk⋅2k(d_n d_{n-1}\ldots d_1 d_0)_2=\sum_{k=0}^{n} d_k\cdot 2^{k}, where each digit dk∈{0,1}d_k\in\{0,1\} is counted from the rightmost bit (k=0k=0).

  1. Convert A=(1000101111100)2A=(1000101111100)_2 (13 bits, powers 2122^{12} down to 202^{0}): the 1-bits sit at positions 12,8,6,5,4,3,212,8,6,5,4,3,2. A=212+28+26+25+24+23+22=4096+256+64+32+16+8+4=4476A = 2^{12}+2^{8}+2^{6}+2^{5}+2^{4}+2^{3}+2^{2} = 4096+256+64+32+16+8+4 = 4476.
  2. Convert B=(1100101000100)2B=(1100101000100)_2: the 1-bits sit at positions 12,11,8,6,212,11,8,6,2. B=212+211+28+26+22=4096+2048+256+64+4=6468B = 2^{12}+2^{11}+2^{8}+2^{6}+2^{2} = 4096+2048+256+64+4 = 6468.
  3. Add the decimal values: A+B=4476+6468=10944A+B = 4476+6468 = 10944.
  4. Convert C=(11101100100)2C=(11101100100)_2 (11 bits): the 1-bits sit at positions 10,9,8,6,5,210,9,8,6,5,2. …

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