# Digit binary number can be expanded as b n b n 1 b 2

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digit binary number can be expanded as b n b n - 1 ...b 2 b 1 b 0 | {z } base 2 representation = b n · 2 n + b n - 1 · 2 n - 1 + · · · + b 2 · 2 2 + b 1 · 2 1 + b 0 · 2 0 | {z } base 10 representation where the i th binary digit b i is either 0 or 1 for each i 2 { 0 , 1 , ...n } . Then, with this is mind, we have 11111 = 1 · 2 4 + 1 · 2 3 + 1 · 2 2 + 1 · 2 1 + 1 · 2 0 = 16 + 8 + 4 + 2 + 1 = 31 1000000 = 1 · 2 6 + 0 · 2 5 + 0 · 2 4 + 0 · 2 3 + 0 · 2 2 + 0 · 2 1 + 0 · 2 0 = 64 1001101101 = 1 · 2 9 + 0 · 2 8 + 0 · 2 7 + 1 · 2 6 + 1 · 2 5 + 0 · 2 4 + 1 · 2 3 + 1 · 2 2 + 0 · 2 1 + 1 · 2 0 To finish this exercise, use the concept of a dot product from linear algebra to concisely illustrate the binary to decimal conversion process: 10101010 = 2 7 2 6 2 5 2 4 2 3 2 2 2 1 2 0 2 6 6 6 6 6 6 6 6 6 6 4 1 0 1 0 1 0 1 0 3 7 7 7 7 7 7 7 7 7 7 5 = 2 7 + 2 5 + 2 3 + 2 1 000011110000 = 2 11 2 10 2 9 2 8 2 7 2 6 2 5 2 4 2 3 2 2 2 1 2 0 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 0 0 0 0 1 1 1 1 0 0 0 0 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 = 2 7 + 2 6 + 2 5 + 2 4
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Solution 6: Parts (f) - (j) Write out a table of the powers of two up through 2 11 :
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