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At this point, we can see that the fraction is repeating. Since we obtained the same value of 0.8 in steps 9 and step 1, we’ll continue to cycle through these values 0.8, 1.6, 1.2, 0.4, 0.8, …, which give us a repeating sequence of bits 0110… , represented with an overbar as (0(1(1(0. ................
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