I know the range of the $n$ digit numbers in $r$'s complement system is given as

$r^{n-1}-1$ to 0 to $-r^{n-1}$.

So for 3 bit 2's complement, its:

$(2^2-1)$ to $0$ to $-(2^2)$
that is
$3$ to $0$ to $-4$
that is
$(011)_{\text{2's comp}}$ to $(0)_{\text{2's comp}}$ to $(100)_{\text{2's comp}}$

And for 3 digit 8's complement, its

$(8^2-1)$ to $0$ to $-(8^2)$
that is
$63$ to $0$ to $-64$
that is
$(077)_{\text{8's comp}}$ to $(0)_{\text{8's comp}}$ to $(700)_{\text{8's comp}}$

Now I came to know that

the $r$'s complement of the smallest / most $-$ve number in the range will be the same number.

For example, taking 2's complement of most $-$ve / smallest number possible with 3 bit 2's complement system, $(100)_{\text{2's comp}}$ is itself:

1 0 0  
0 1 1 (1's complement)
1 0 0 (add 1 to get 2's complement)

But taking 8's complement of most $-$ve / smallest number possible with 3 digit 8's complement system, $(700)_{\text{8's comp}}$ does not seem to be itself:

7 0 0
0 7 7 (7's complement)
1 0 0 (add 1 to get 8's complement)

So 8's complement of 700 does not seem to be 700 itself but is in fact 100. So what I am missing? Or I am doing it all wrong?


2 Answers 2


First of all, if you are considering the $n$th digit of the number as being a sign digit, you can think of it as being either 0 or 1, no matter the value of $r$. In other words, it is always a bit (a binary digit), and that's how you would represent it in hardware, should that be the case.

$[1]73_{\text{8's comp}} = -5$

$[1]73_{\text{8's comp}} + [0]06_{\text{8's comp}} = [0]01_{\text{8's comp}}$

The complement of a negative number represented in $r$'s complement will obviously produce the positive version of that same number. The most negative number doesn't have a positive of the same magnitude that fits into the given number of digits (a weakness of this form of representation).

The number is still there, though, only unsigned:

$[1]00_{\text{8's comp}} = -64_{10}$

$100_8 = 64_{10}$


The smallest 3-digit 8's complement number is not 700, but 400.

Indeed, the assertion at the start of the question is already wrong: as long as $r$ is even, the actual range is from $-\tfrac r2r^{n-1}$ to $\tfrac r2r^{n-1}-1$.

  • $\begingroup$ $-\lfloor\frac{r^n}{2}\rfloor$ to $\lfloor\frac{r^n}{2}\rfloor - (1-r\pmod 2)$ should work for all values of $r$. It would also be interesting if you could explain to the OP why do you think the premise is wrong $\endgroup$ Commented Sep 29, 2016 at 0:03

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