Question 14

In IEEE 754 single-precision (32-bit) floating-point format:

• Sign bit ($S$): 1 bit
• Exponent field ($E$): 8 bits, biased with bias = 127
• Fraction/Mantissa ($M$): 23 bits

An exponent field of 00000000 ($E = 0$) is reserved for zero and subnormal (denormalized) numbers. An exponent field of 11111111 ($E = 255$) is reserved for infinities and NaN.

For a normalized number, the exponent field ranges from $E = 1$ (00000001) to $E = 254$ (11111110). The smallest exponent for a normalized number is therefore $E = 1$ (00000001), which corresponds to an unbiased exponent of $1 - 127 = -126$.

The value of a positive normalized number is $(1.M) \times 2^{E - 127}$. To minimize this positive value, the mantissa bits must all be 0 (00000000000000000000000), yielding $(1.0)_2 \times 2^{-126} = 2^{-126}$.