+0 Let x be the binary number (0.001001001 . . .)2 and let y be the octal number (0.666666 . . .)8. What is x + y in decimal?
Please explain your answer.
+394 We can convert these two seperately.
0.001001001... when expressed in Base 2 is \(0*{2}^{-1}+0*{2}^{-2}+1*{2}^{-3}+0*{2}^{-4}+0*{2}^{-5}+1*{2}^{-6}...\)
The terms with the 0's are all 0, so this simplifies into \({2}^{-3}+{2}^{-6}+{2}^{-9}+{2}^{-12}...\)
Using the Infinite geometric series formula, (first term)/(1 - common ratio), we get \(\frac{{2}^{-3}}{1-{2}^{-3}}\)which simplifies to \(\frac{1}{7}\).
We do the second one similarly, simplifying 0.6666666666666 base 8 into \(6*{8}^{-1}+6*{8}^{-2}+6*{8}^{-3}+6*{8}^{-4}\), using the infinite geometric series formula, we get, \(6*\frac{{8}^{-1}}{1-{8}^{-1}}\)which simplifies to \(\frac{6}{7}\). Adding, 1/7 and 6/7 we get 1.
+394 We can convert these two seperately.
0.001001001... when expressed in Base 2 is \(0*{2}^{-1}+0*{2}^{-2}+1*{2}^{-3}+0*{2}^{-4}+0*{2}^{-5}+1*{2}^{-6}...\)
The terms with the 0's are all 0, so this simplifies into \({2}^{-3}+{2}^{-6}+{2}^{-9}+{2}^{-12}...\)
Using the Infinite geometric series formula, (first term)/(1 - common ratio), we get \(\frac{{2}^{-3}}{1-{2}^{-3}}\)which simplifies to \(\frac{1}{7}\).
We do the second one similarly, simplifying 0.6666666666666 base 8 into \(6*{8}^{-1}+6*{8}^{-2}+6*{8}^{-3}+6*{8}^{-4}\), using the infinite geometric series formula, we get, \(6*\frac{{8}^{-1}}{1-{8}^{-1}}\)which simplifies to \(\frac{6}{7}\). Adding, 1/7 and 6/7 we get 1.
