Convert the Number 60.259 to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number. Detailed Explanations

Number 60.259(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

The first steps we'll go through to make the conversion:

Convert to binary (to base 2) the integer part of the number.

Convert to binary the fractional part of the number.


1. First, convert to binary (in base 2) the integer part: 60.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 60 ÷ 2 = 30 + 0;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


60(10) =


11 1100(2)


3. Convert to binary (base 2) the fractional part: 0.259.

Multiply it repeatedly by 2.


Keep track of each integer part of the results.


Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.259 × 2 = 0 + 0.518;
  • 2) 0.518 × 2 = 1 + 0.036;
  • 3) 0.036 × 2 = 0 + 0.072;
  • 4) 0.072 × 2 = 0 + 0.144;
  • 5) 0.144 × 2 = 0 + 0.288;
  • 6) 0.288 × 2 = 0 + 0.576;
  • 7) 0.576 × 2 = 1 + 0.152;
  • 8) 0.152 × 2 = 0 + 0.304;
  • 9) 0.304 × 2 = 0 + 0.608;
  • 10) 0.608 × 2 = 1 + 0.216;
  • 11) 0.216 × 2 = 0 + 0.432;
  • 12) 0.432 × 2 = 0 + 0.864;
  • 13) 0.864 × 2 = 1 + 0.728;
  • 14) 0.728 × 2 = 1 + 0.456;
  • 15) 0.456 × 2 = 0 + 0.912;
  • 16) 0.912 × 2 = 1 + 0.824;
  • 17) 0.824 × 2 = 1 + 0.648;
  • 18) 0.648 × 2 = 1 + 0.296;
  • 19) 0.296 × 2 = 0 + 0.592;
  • 20) 0.592 × 2 = 1 + 0.184;
  • 21) 0.184 × 2 = 0 + 0.368;
  • 22) 0.368 × 2 = 0 + 0.736;
  • 23) 0.736 × 2 = 1 + 0.472;
  • 24) 0.472 × 2 = 0 + 0.944;

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 => FULL STOP (losing precision...)


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.259(10) =


0.0100 0010 0100 1101 1101 0010(2)


5. Positive number before normalization:

60.259(10) =


11 1100.0100 0010 0100 1101 1101 0010(2)


The last steps we'll go through to make the conversion:

Normalize the binary representation of the number.

Adjust the exponent.

Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Normalize the mantissa.


6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the left, so that only one non zero digit remains to the left of it:


60.259(10) =


11 1100.0100 0010 0100 1101 1101 0010(2) =


11 1100.0100 0010 0100 1101 1101 0010(2) × 20 =


1.1110 0010 0001 0010 0110 1110 1001 0(2) × 25


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)


Exponent (unadjusted): 5


Mantissa (not normalized):
1.1110 0010 0001 0010 0110 1110 1001 0


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


Exponent (unadjusted) + 2(8-1) - 1 =


5 + 2(8-1) - 1 =


(5 + 127)(10) =


132(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 132 ÷ 2 = 66 + 0;
  • 66 ÷ 2 = 33 + 0;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


132(10) =


1000 0100(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.


b) Adjust its length to 23 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 111 0001 0000 1001 0011 0111 01 0010 =


111 0001 0000 1001 0011 0111


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

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


Exponent (8 bits) =
1000 0100


Mantissa (23 bits) =
111 0001 0000 1001 0011 0111


The base ten decimal number 60.259 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 0100 - 111 0001 0000 1001 0011 0111

(32 bits IEEE 754)

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Available Base Conversions Between Decimal and Binary Systems

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1. Integer -> Binary

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