For a two’s complement representation, the truncation error is always negative and falls in the range

The ones complement of (1100)2 is (0011)2. Thus the two complement of this number is obtained as (0011)2+(0001)2=(0100)2.

The binary floating point representation commonly used in practice, consists of a mantissa M, which is the fractional part of the number and falls in the range 1/2<M<1, multiplied by the exponential factor 2E, where the exponent E is either a negative or positive integer. Hence a number X is represented as X= M.2E(1/2<M<1).

The number (-3/8) is stored in the computer as the 2’s complement of (3/8)

According to the IEEE 754 standard, for a 32-bit machine, single precision floating point number is represented as X=(-1)s.2E-127(M).

We can represent the numbers in binary float point as

Let the mantissa be represented by 23 bits plus a sign bit and let the exponent be represented by 7 bits plus a sign bit.

The round-off error is independent of the type of fixed point representation. The maximum error that can be introduced through rounding is 0.5(2-b+2-bm) and this can be either positive or negative, depending on the value of x. Therefore, the round-off error is symmetric about zero and falls in the range

According to the IEEE 754 standard, for a 32-bit machine, single precision floating point number is represented as X=(-1)s.2E-127(M).

According to the IEEE 754 standard, for a 32-bit machine, single precision floating point number is represented as X=(-1)s.2E-127(M).