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# Tag Info

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If the processor is only reading odd numbered bits and we assume that the right-most bit is bit $0$ then in effect the processor is doing a bitwise AND with bit string $101010$ before converting to decimal. $010111_2 \text{ AND } 101010_2 = 000010_2 = 2_{10}$

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I'm assuming positions are counted from $1$ starting from the least significant bit. In this case you're using the exponents as if the number was 0111. The question is asking to use the same exponents you would use for the odd-position digits of the number 010111. To break it down, from the least to the most significant bit, the contribution to the final ...

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The mantissa ranges from $$(1.\underbrace{0\cdots 0}_{m \text{ times}})_2 = 1$$ to $$(1.\underbrace{1\cdots 1}_{m \text{ times}})_2 = 2 - (0.\underbrace{0\cdots 0}_{m-1 \text{ times}}1)_2 = 2^{-m}.$$ For example, $(1.1)_2 = 1.5 = 2 - 1/2$ and $(1.11)_2 = 1.75 = 2 - 1/4$.

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