In two’s complement method, why is there one more negative number than there are

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let's take the casee of 8 bits,from 8 bits we can form numbers from 10000000 to 01111111. here 10000000 is a negative numbers which represents -128 . 01111111 is a positive number indicating 127. in signed representation 0 indicates positive value and 1 indicates negative value. so by simply inspecting the MSB of a number we can tell that a number is positive or negative. to illustrate why there is one more negative number than positive numbers lets take the maximum number possible with 8 bits i.e. 01111111 which is +127. now adding 1 to +127 will give us 10000000 which is a negative number since its MSB is 1.how's that possible? since we have the maximum storage of 8 bits in which the MSB is the sign bit,rest of the bits denotes the magnitude of the number.so when adding 1 to 127 we get 128 but the result is wrapped around to give us the result -128. so we can not have any positive number of the form 2^(n-1) with n bits in sign -complement representation. now consider the case when -128 is decreased by 1 unit. i.e. (-128)-1 => (-128)+(-1) -128 : 10000000 in 2's complement form -1 : 11111111 in 2's complement form the sum produces the result 101111111 which is of 9 bits so the end carry is discarded.and the result is 01111111 which is +127. in this case also the result is wrapped around when we wanted to decrease -128 by one unit. hence this functionality of 2's complement causes number of postive numbers are one less than number of negative numbers.

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