Chapter 2, Problem 51E

a)

Explanation of Solution

IEEE-754 floating point single precision:

IEE-754 floating point single precision has 32 bits.

• One bit for sign, 8 bits for exponent, and 23 bits for significant bits.

Storing “12.5” using IEEE-754 single precision:

Step 1: Converting decimal to binary number:

Step (i): Divide the given number into two parts, integer and the fractional part. Here, the integer part is “12” and the fractional part is “.5”.

Step (ii): Divide “12” by 2 till the quotient becomes 1. Simultaneously, note the remainder for every division operation.

Step (iii): Note the remainder from the bottom to top to get the binary equivalent.

Step (iv): Consider the fraction part “.5”. Multiply the fractional part “.5” by 2 and it continues till the fraction part reaches “0”.

$0.5\times 2=1.0$

Step (v): Note the integer part to get the final result.

Thus, the binary equivalent for “12.5” is $1100.1$

Step 2: Normalize the binary fraction number:

Now the given binary fraction number should be normalized. To normalize the value, move the decimal point either right or left so that only single digit will be left before the decimal point.

$1100.1\to 1.1001\times 2^3$

Step 3: Convert the exponent to 8 bit excess-127:

To convert the exponent into 8-bit excess-127 notation, the exponent value should be added with 127. After addition, it is converted into binary equivalent.

$127+3=130$

Converting ${130}_{10}={10000010}_2$

Step 4: Convert the significant to hidden bit:

To convert the significant to hidden bit the leftmost “1” should be removed.

$1.1001\to 1001$

Step 5: Framing the number “12.5” in 32 bit IEEE-754 single precision

Sign bit(1 bit)	Exponent bit(8 bits)	Significant bit(23)
0	10000010	10010000000000000000000

Thus, the number “12.5” in 32 bit IEEE-754 single precision is represented as “$01000001010010000000000000000000$”.

b)

Explanation of Solution

Storing “-1.5” using IEEE-754 single precision:

Step 1: Converting decimal to binary number:

Step (i): Consider the fraction part “0.5”. Multiply the fractional part “.5” by 2 and it continues till the fraction part reaches “0”.

$0.5\times 2=1.0$

Step (ii): Note the integer part to get the final result.

Thus the binary equivalent for “1.5” is $1.1$

Step 2: Normalize the binary fraction number:

Now the given binary fraction number should be normalized. To normalize the value, move the decimal point either right or left so that only single digit will be left before the decimal point.

$1.1\to 1.1\times 2^0$

Step 3: Convert the exponent to 8 bit excess-127:

To convert the exponent into 8-bit excess-127 notation, the exponent value should be added with 127. After addition, it is converted into binary equivalent.

$127+0=127$

Converting ${127}_{10}={01111111}_2$

Step 4: Convert the significant to hidden bit:

To convert the significant to hidden bit the leftmost “1” should be removed.

$1.1\to 1$

Step 5: Framing the number “-1.5” in 32 bit IEEE-754 single precision

Sign bit(1 bit)	Exponent bit(8 bits)	Significant bit(23)
1	01111111	10000000000000000000000

Thus, the number “-1.5” in 32 bit IEEE-754 single precision is represented as “$10111111110000000000000000000000$”.

c)

Explanation of Solution

Storing “.75” using IEEE-754 single precision:

Step 1: Converting decimal to binary number:

Step (i): Consider the fraction part “.75”. Multiply the fractional part “.75” by 2 and it continues till the fraction part reaches “0”.

$\begin{array}{l}0.75\times 2=1.501\\ 0.50\times 2=1.001\end{array}$

Step (ii): Note the integer part to get the final result.

${0.75}_{{}_{10}}={0.11}_2$

Thus the binary equivalent for “.75” is ${0.11}_2$

Step 2: Normalize the binary fraction number:

Now the given binary fraction number should be normalized. To normalize the value, move the decimal point either right or left so that only single digit will be left before the decimal point.

$.11\to 1.1\times 2^{-1}$

Step 3: Convert the exponent to 8 bit excess-127:

To convert the exponent into 8-bit excess-127 notation, the exponent value should be added with 127. After addition, it is converted into binary equivalent.

$127+(-1)=126$

Converting ${126}_{10}={01111110}_2$

Step 4: Convert the significant to hidden bit:

To convert the significant to hidden bit the leftmost “1” should be removed.

$1.1\to 1$

Step 5: Framing the number “.75” in 32 bit IEEE-754 single precision

Sign bit(1 bit)	Exponent bit(8 bits)	Significant bit(23)
0	01111110	10000000000000000000000

Thus, the number “.75” in 32 bit IEEE-754 single precision is represented as “$10111111010000000000000000000000$”.

d)

Explanation of Solution

IEEE-754 floating point single precision:

IEE-754 floating point single precision has 32 bits.

• One bit for sign, 8 bits for exponent, and 23 bits for significant bits.

Storing “26.625” using IEEE-754 single precision:

Step 1: Converting decimal to binary number:

Step (i): Divide the given number into two parts, integer and the fractional part. Here, the integer part is “26” and the fractional part is “.625”.

Step (ii): Divide “26” by 2 till the quotient becomes 1. Simultaneously, note the remainder for every division operation.

Step (iii): Note the remainder from the bottom to top to get the binary equivalent.

Step (iv): Consider the fraction part “.5”. Multiply the fractional part “.625” by 2 and it continues till the fraction part reaches “0”.

$\begin{array}{l}0.625\times 2=1.251\\ 0.25\times 2=0.500\\ 0.50\times 2=1.001\end{array}$

Step (v): Note the integer part to get the final result.

${0.625}_{10}={0.101}_2$

Thus the binary equivalent for “26.625” is ${11010.101}_2$

Step 2: Normalize the binary fraction number:

Now the given binary fraction number should be normalized. To normalize the value, move the decimal point either right or left so that only single digit will be left before the decimal point.

$11010.101\to 1.1010101\times 2^4$

Step 3: Convert the exponent to 8 bit excess-127:

To convert the exponent into 8-bit excess-127 notation, the exponent value should be added with 127. After addition, it is converted into binary equivalent.

$127+4=131$

Converting ${131}_{10}={10000011}_2$

Step 4: Convert the significant to hidden bit:

To convert the significant to hidden bit the leftmost “1” should be removed.

$1.1010101\to 1010101$

Step 5: Framing the number “26.625” in 32 bit IEEE-754 single precision

Sign bit(1 bit)	Exponent bit(8 bits)	Significant bit(23)
0	10000011	10101010000000000000000

Thus, the number “12.5” in 32 bit IEEE-754 single precision is represented as “$01000001110101010000000000000000$”.