Question

Write down the binary representation of the decimal number 63.25 assuming it was stored using the single precision IBM format (base 16, instead of base 2, with 7 bits of exponent).

Short answer

Binary representation

Sign bit	7-bit Exponent	24-bit Fraction value
0	1000010	001111110100000000000000

Step-by-step solution

IBM format

In IBM Format representation of floating point number is 1 bit for the sign, 7 bit for the exponent, and 24 bit for the fraction

Convert (63.25)10 into binary

First, we convert integer part 63

${\left(63\right)}_{10}={\left(111111\right)}_2$

Then, convert fractional part 0.25 into binary

${(0.25)}_{10}={\left(0.01\right)}_2$

Now,

${(63.25)}_{10}={\left(111111.01\right)}_2$

Normalization of the binary representation of the number

The value can be normalized by moving the radix point by 4 bits once until LSB becomes zero i.e. 0.0011111101

The radix point is moved by 2 digits for normalizing value which creates an exponent of ${16}^2$. The correct bias is 64 and 2 is added to it which is 66 and binary for 66 is ${\left(1000010\right)}_2$. Thus, 7-bit exponent is $1000010$.

Binary representation of the number using the single precision IBM format

IEEE754 in single-precision IBM format:

$$\begin{gathered} {\left(-1\right)}^{sign}\times 0.significand\times {16}^{exponent-64} \\ ={\left(-1\right)}^0\times (0.001111101000000000000000)\times {16}^{\left(66-64\right)} \end{gathered}$$

The binary representation of63.25in IEEE 754 single precision in IBM format:

Sign bit	7-bit Exponent	24-bit Fraction value
0	1000010	001111110100000000000000