Convert the Number 2 532.129 83 to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number. Detailed Explanations

Number 2 532.129 83(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: 2 532.
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;
  • 2 532 ÷ 2 = 1 266 + 0;
  • 1 266 ÷ 2 = 633 + 0;
  • 633 ÷ 2 = 316 + 1;
  • 316 ÷ 2 = 158 + 0;
  • 158 ÷ 2 = 79 + 0;
  • 79 ÷ 2 = 39 + 1;
  • 39 ÷ 2 = 19 + 1;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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.


2 532(10) =


1001 1110 0100(2)


3. Convert to binary (base 2) the fractional part: 0.129 83.

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) 0.129 83 × 2 = 0 + 0.259 66;
  • 2) 0.259 66 × 2 = 0 + 0.519 32;
  • 3) 0.519 32 × 2 = 1 + 0.038 64;
  • 4) 0.038 64 × 2 = 0 + 0.077 28;
  • 5) 0.077 28 × 2 = 0 + 0.154 56;
  • 6) 0.154 56 × 2 = 0 + 0.309 12;
  • 7) 0.309 12 × 2 = 0 + 0.618 24;
  • 8) 0.618 24 × 2 = 1 + 0.236 48;
  • 9) 0.236 48 × 2 = 0 + 0.472 96;
  • 10) 0.472 96 × 2 = 0 + 0.945 92;
  • 11) 0.945 92 × 2 = 1 + 0.891 84;
  • 12) 0.891 84 × 2 = 1 + 0.783 68;
  • 13) 0.783 68 × 2 = 1 + 0.567 36;
  • 14) 0.567 36 × 2 = 1 + 0.134 72;
  • 15) 0.134 72 × 2 = 0 + 0.269 44;
  • 16) 0.269 44 × 2 = 0 + 0.538 88;
  • 17) 0.538 88 × 2 = 1 + 0.077 76;
  • 18) 0.077 76 × 2 = 0 + 0.155 52;
  • 19) 0.155 52 × 2 = 0 + 0.311 04;
  • 20) 0.311 04 × 2 = 0 + 0.622 08;
  • 21) 0.622 08 × 2 = 1 + 0.244 16;
  • 22) 0.244 16 × 2 = 0 + 0.488 32;
  • 23) 0.488 32 × 2 = 0 + 0.976 64;
  • 24) 0.976 64 × 2 = 1 + 0.953 28;

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:


0.129 83(10) =


0.0010 0001 0011 1100 1000 1001(2)


5. Positive number before normalization:

2 532.129 83(10) =


1001 1110 0100.0010 0001 0011 1100 1000 1001(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 11 positions to the left, so that only one non zero digit remains to the left of it:


2 532.129 83(10) =


1001 1110 0100.0010 0001 0011 1100 1000 1001(2) =


1001 1110 0100.0010 0001 0011 1100 1000 1001(2) × 20 =


1.0011 1100 1000 0100 0010 0111 1001 0001 001(2) × 211


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): 11


Mantissa (not normalized):
1.0011 1100 1000 0100 0010 0111 1001 0001 001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


11 + 2(8-1) - 1 =


(11 + 127)(10) =


138(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;
  • 138 ÷ 2 = 69 + 0;
  • 69 ÷ 2 = 34 + 1;
  • 34 ÷ 2 = 17 + 0;
  • 17 ÷ 2 = 8 + 1;
  • 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) =


138(10) =


1000 1010(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. 001 1110 0100 0010 0001 0011 1100 1000 1001 =


001 1110 0100 0010 0001 0011


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 1010


Mantissa (23 bits) =
001 1110 0100 0010 0001 0011


The base ten decimal number 2 532.129 83 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 1010 - 001 1110 0100 0010 0001 0011

(32 bits IEEE 754)

Number 2 532.129 82 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

Number 2 532.129 84 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

Convert to 32 bit single precision IEEE 754 binary floating point representation standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

Available Base Conversions Between Decimal and Binary Systems

Conversions Between Decimal System Numbers (Written in Base Ten) and Binary System Numbers (Base Two and Computer Representation):


1. Integer -> Binary

2. Decimal -> Binary

3. Binary -> Integer

4. Binary -> Decimal