Problem $5$:

Perform the following addition using IEEE$-754$ FP single precision format.

$13\mathrm{.}125+4\mathrm{.}75$

Show all your work in binary.

Give the result in binary and Hexadecimal.

Converting $13\mathrm{.}125$ into IEEE $754$ FP Representation:

let X $=13\mathrm{.}125$

$13=1101$

$0\mathrm{.}125=001$

$13\mathrm{.}125=1101\mathrm{.}001$

$=>$ X $=1\mathrm{.}101001$ x $2^3$

sign $=0$

$1.$ Single precision:

biased exponent $127+3=130$

$130=10000010$

Normalized mantissa $=101001$

we will add $0\text{'}$s to complete the $23$ bits.

The IEEE $754$ Single precision is:

$=01000001010100100000000000000000$

This can be written in hexadecimal form $41520000.$

Converting $4\mathrm{.}75$ into IEEE $754$ FP Representation:

Let Y $=4\mathrm{.}75$

$4=100$

$0\mathrm{.}75=11$

$4\mathrm{.}75=100\mathrm{.}11$

$=>$ Y $=1\mathrm{.}0011$ x $2^2$

sign $=0$

$1.$ Single precision:

biased exponent $127+2=129$

$130=10000001$

Normalized mantissa $=0011$

we will add $0\text{'}$s to complete the $23$ bits.

The IEEE $754$ Single precision is:

$=01000000100110000000000000000000$

This can be written in hexadecimal form $40980000.$

Procedure of Addition:

X $=1\mathrm{.}101001$ x $2^3$

Y $=1\mathrm{.}0011$ x $2^2=>$ Y $=0\mathrm{.}10011$ x $2^3$

$1.101001$ x $2^3(+)0.100110$ x $2^3=10.001111$ x $2^3$

$=>$ X$+$Y $=1\mathrm{.}0001111$ x $2^4$

Converting $17\mathrm{.}875$ i$.$e; X$+$Y into IEEE $754$ FP Representation:

sign $=0$

$1.$ Single precision:

biased exponent $127+4=133$

$133=10000101$

Normalized mantissa $=0001111$

we will add $0\text{'}$s to complete the $23$ bits.

The IEEE $754$ Single precision is:

$=01000010100011110000000000000000$

This can be written in hexadecimal form $428$F$0000.$