Question

Convert the following numbers to 32b IEEE 754 Floating Point format. Show bits in diagrams below. a) -769.0234375 Mantissa Exponent b) 8.111 Mantissa Exponent

Answer 1

1.)

-769.0234375

We start with positive version of the number: 769.0234375

First we will convert the integer part to binary as follows:

To convert from decimal to binary: Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero

• 769 ÷ 2 = 384 + 1;
• 384 ÷ 2 = 192 + 0;
• 192 ÷ 2 = 96 + 0;
• 96 ÷ 2 = 48 + 0;
• 48 ÷ 2 = 24 + 0;
• 24 ÷ 2 = 12 + 0;
• 12 ÷ 2 = 6 + 0;
• 6 ÷ 2 = 3 + 0;
• 3 ÷ 2 = 1 + 1;
• 1 ÷ 2 = 0 + 1;

​769 in binary = 1100000001₍₂₎

Now we will convert the fractional part to binary as follows:

Multiply it repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:

• 1) 0.023 437 5 × 2 = 0 + 0.046 875;
• 2) 0.046 875 × 2 = 0 + 0.093 75;
• 3) 0.093 75 × 2 = 0 + 0.187 5;
• 4) 0.187 5 × 2 = 0 + 0.375;
• 5) 0.375 × 2 = 0 + 0.75;
• 6) 0.75 × 2 = 1 + 0.5;
• 7) 0.5 × 2 = 1 + 0;

0.023 437 5₍₁₀₎ = 00000011₍₂₎

Hence given number in binary will be:

769.023 437 5₍₁₀₎ = 11 0000 0001.0000 011₍₂₎

Now we will normalize the number, by shifting the decimal mark 9 positions to the left so that only one non zero digit remains to the left of it:

769.023 437 5₍₁₀₎ = 1.1000 0000 1000 0011₍₂₎ × 2⁹

Since the number is negative, sign bit will be 1 in the IEEE format.

Unadjusted exponent = 9

Unnormalized Mantissa = 1.1000 0000 1000 0011

To get the adjusted exponent, we will add the bias 127 to it.

Hence adjusted exponent = 136

136 in binary = 1000 1000₍₂₎

To normalize the mantissa: remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then adjust its length to 23 bits, by adding the necessary number of zeros to the right

Normalized mantissa: 100 0000 0100 0001 1000 0000

Hence the three fields in IEEE 754 format are as follows:

Sign (1 bit) = 1 (a negative number)

Exponent (8 bits) = 1000 1000

Mantissa (23 bits) = 100 0000 0100 0001 1000 0000

S	Exponent	Mantissa
1	1 0 0 0 1 0 0 0	1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0

2.)

8.111:

First convert the integer part to binary:

• 8 ÷ 2 = 4 + 0;
• 4 ÷ 2 = 2 + 0;
• 2 ÷ 2 = 1 + 0;
• 1 ÷ 2 = 0 + 1;

8₍₁₀₎ = 1000₍₂₎

Now convert the franctional part to binary: Multiply it repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:

• 1) 0.111 × 2 = 0 + 0.222;
• 2) 0.222 × 2 = 0 + 0.444;
• 3) 0.444 × 2 = 0 + 0.888;
• 4) 0.888 × 2 = 1 + 0.776;
• 5) 0.776 × 2 = 1 + 0.552;
• 6) 0.552 × 2 = 1 + 0.104;
• 7) 0.104 × 2 = 0 + 0.208;
• 8) 0.208 × 2 = 0 + 0.416;
• 9) 0.416 × 2 = 0 + 0.832;
• 10) 0.832 × 2 = 1 + 0.664;
• 11) 0.664 × 2 = 1 + 0.328;
• 12) 0.328 × 2 = 0 + 0.656;
• 13) 0.656 × 2 = 1 + 0.312;
• 14) 0.312 × 2 = 0 + 0.624;
• 15) 0.624 × 2 = 1 + 0.248;
• 16) 0.248 × 2 = 0 + 0.496;
• 17) 0.496 × 2 = 0 + 0.992;
• 18) 0.992 × 2 = 1 + 0.984;
• 19) 0.984 × 2 = 1 + 0.968;
• 20) 0.968 × 2 = 1 + 0.936;
• 21) 0.936 × 2 = 1 + 0.872;
• 22) 0.872 × 2 = 1 + 0.744;
• 23) 0.744 × 2 = 1 + 0.488;
• 24) 0.488 × 2 = 0 + 0.976;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero

0.111₍₁₀₎ = 0.0001 1100 0110 1010 0111 1110₍₂₎

Hence

8.111₍₁₀₎ = 1000.0001 1100 0110 1010 0111 1110₍₂₎

Normalizing the binary form, by shifting the decimal mark 3 positions to the left so that only one non zero digit remains to the left of it:

8.111₍₁₀₎ = 1.0000 0011 1000 1101 0100 1111 110₍₂₎ × 2³

Sign: 0 (a positive number)

Exponent (unadjusted): 3

Mantissa (not normalized): 1.0000 0011 1000 1101 0100 1111 110

To adjust the exponent, add the bias of 127 to it.

Adjusted exponent = 127 + 3 = 130

130 in binary = 1000 0010₍₂₎

To normalize the mantissa: remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then adjust its length to 23 bits, by removing the excess bits, from the right

Mantissa (normalized) = 000 0001 1100 0110 1010 0111

Hence the three fields in IEEE 754 format are as follows:

Sign (1 bit) = 0 (a positive number)

Exponent (8 bits) = 1000 0010

Mantissa (23 bits) = 000 0001 1100 0110 1010 0111

S	Exponent	Mantissa
0	1 0 0 0 0 0 1 0	0 0 0 0 0 0 1 1 1 0 0 0 1 1 0 1 0 1 0 0 1 1 1

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