How is $0.11 \cdot 2^{1}$ converted to $1.5$ in base 10? I am unable to understand the theory behind this, since I believe $0.11 * 2^{1}$ corresponds to $1.1$ but I dont see the conversion process?
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$0.11\cdot 2^1$ does indeed correspond to $1.1$. In binary. Which is to say, it means $$ 1\cdot 2^0 + 1\cdot 2^{-1} $$ as per standard positional number system conventions. Writing this out in base ten does get you $1.5$.
Usually, when we write numbers in different bases close to one another, there is postential for such a misunderstanding. In those cases, it's common to see a subscript denoting which base we're using (this subscript uses base ten). So with that, the above paragraph would look like
$0.11_2\cdot 2^1$ does indeed correspond to $1.1_2$. In binary. Which is to say, it means $$ 1\cdot 2^0 + 1\cdot 2^{-1} $$ as per standard positional number system conventions. Writing this out in base ten does get you $1.5_{10}$.
Arthur
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Thanks. One question, I see that indeed it is $1.1$ in binary, does the $2^{-1}$ come from the $.1$ after $1.1$ (1 after the decimal)? – ysmflsd May 25 '20 at 12:50
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When we write $1.1_2$, then that means one whole, and one half. This is how western positional number systems always work. The leftmost $1$ denotes the whole, and the rightmost $1$ (the one after the "decimal" point) denotes the half. This is entirely analoguous to how $1.5_{10}$ means one whole and five tenths. – Arthur May 25 '20 at 12:51
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Ack, so $0.00111_2 \cdot 2^2$ would be equal to $2^{-1} + 2^{-2} + 2^{-3} = 0.875_{10}$? – ysmflsd May 25 '20 at 12:52
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@ysmflsd Yes, that's exactly what it would equal. $0.00111_2\cdot 2^2$ is equal to $0.111_2$, and you expanded that correctly, and then added them correctly together in base ten. – Arthur May 25 '20 at 12:53
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