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The problem is to convert $32$ to $6$-bit two's complement or note that the range would overflow.

For $32$, I have: $100000$.

How does this overflow? Is it because the 1 means negative, so it would be $0100000$ but the space cannot accommodate this? But the notes say the max positive number is $2^5 - 1 = 31$.

TPace
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user855911
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  • The range you want to cover with 6 bits is from $-2^5$ to $2^5-1$, is it? – mike239x Feb 18 '18 at 22:33
  • Where does the power 5 come from? – user855911 Feb 18 '18 at 22:45
  • Well, 5 = 6-1. Just like 32-bit system stores numbers from $-2^{31}$ to $2^{31}-1$. – mike239x Feb 18 '18 at 22:49
  • My question was about following: with 6 bits you can either store numbers in range $[−2^5, 2^5−1]$, or numbers in range $[0,2^6-1]$, or technically speaking you can store an element from a fixed set with $2^6$ elements. So the question if the range would overflow is a bit weird without clarification, what your range is. – mike239x Feb 18 '18 at 22:52
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    @mike239x: the question says that the representation is 6-bit two's complement, so the range is $-2^5$ to $2^5-1$. The OP's explanation for the overflow is essentially correct: $32_{10} = 100000_{2}$ and $100000$ is interpreted as $-32_{10}$ in 6-bit two's complement. – Rob Arthan Feb 18 '18 at 23:02

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As you say, in twos complement the first bit represents the sign and the rest of the bits represent the number. Writing $32_{10}=100\ 000_2$ is not legal in twos complement as the leading bit being $1$ represents a negative number and $32_{10}$ is not negative. In $6$ bit twos complement the largest positive number you can represent is $31_{10}=011\ 111$. As you count down you get to $)_{10}=000\ 000$, then roll over to $-1_{10}=111\ 111$. As you keep counting down you finally get to $-32_{10}=100\ 000$, which is the most negative number you can represent. Your proposed representation of $32_{10}=100\ 000$ would be read as $-32_{10}$

Ross Millikan
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