Consider the floating-point representation31-24 : Exponent23-0  : MantissaThe exponent is in 2's complement representation and mantissa is in the sign magnitude representation. the range of the magnitude of the normalized numbers in this represntaion is$(a)\ 0\ to \ 1$$(b)\ 0.5\ to\ 1$$(c)\ 2^{-23}\ to \ 0.5$$(d)\ 0.5\ to \ (1-2^{-23})$My approach:as the normalized number in floating point representation has as implicit 1.Hence smallest mantissa would be $1.0000\cdots [24\ 0's]=1$Largest mantissa would be $1.1111\cdots [24\ 1's]=1+(1-2^{-24})\approx2$So my ans coming as 1 to 2, Which is not in the option. What mistake I'm doing..
Problem on Floating Point Representation
floating point
First, this is not the way standard IEEE P754 single precision floating point numbers are represented :Exponent is not two's complement, there is instead an offset, for IEEE normalised : F=2^(Exponent - 127) * 1.Mantissa * SignNormalised numbers have an exponent different from 00 (being denormals) and FF (for infinites, NaN)If this peculiar format has no implicit 1, mantissa should be between 0.10[..]0000 and 0.11[..]1111 :  Answer (D) or between 1.0000 and 1.11111, which is your proposal.(IIRC, Intel double extended x87 80bits format does not have the implicit 1 and works as you describe. It is not striclty IEEE compliant, the 8087 being older than the standard.)