I get stuck in here.
Should it be 10 11 100 101 110?
Should "21/4" lie between 101 and 110 instead of 111?
Thanks for your help!
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Yes, you're right. The integer at the right of 101 should be 110.
The reason is that $110_2 = 5<\dfrac{21}{4}<6 = 110_2<7=111_2$.
There's nothing wrong in the equation that you underlined.
$$c_0 = d_0 + p d_1 + p^2 d_2 + \cdots + p^k d_k,$$
where $d_0,d_1,\dots,d_k \in \{0,1,\dots,p-1\}$ because the formula represents $c_0$ in base $p$.
In the picture, there're some digits $0$'s and $1$'s because the figure shows the representation of $\dfrac{21}4$ in the binary system.
GNUSupporter 8964民主女神 地下教會
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It is nothing wrong with that I underlined.Should "21/4" lie between 101 and110 instead of 111? – potter john Dec 01 '15 at 15:55
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Sorry, I overlooked the title of your question. – GNUSupporter 8964民主女神 地下教會 Dec 01 '15 at 15:57
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It doesn't matter – potter john Dec 01 '15 at 16:00
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um...I want to accept your answer...but this book is "kind of" famous one,I'm not sure about that........ – potter john Dec 01 '15 at 16:10
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Even famous book can have tiny printing errors. You may search "errata" for any published book on Google. – GNUSupporter 8964民主女神 地下教會 Dec 01 '15 at 16:13
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......google gave me nothing ......whatever,thank your very much! – potter john Dec 01 '15 at 16:20