I'm supposed to covert hexadecimal value, $8D(16)$ into $8$-bit binary if signed magnitude representation is used.
$8D(16)$ $\to$ $1000$ $1101(2)$
For signed magnitude, the left most bit is used to signify whether the value is a $+$ or $-$ value. In this case, how do I represent the value 8 into signed magnitude binary? Isn't $1000$ equals to $-0$. I don't see any possible way to write $8D$ in binary
Best Answer
An 8-bit signed integer has a range of -128 to 127. But hexadecimal 8D is 141. So what you're trying to accomplish is impossible.