Convert the Number 633.599 975 585 98 to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number. Detailed Explanations

Number 633.599 975 585 98(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: 633.
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;
  • 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.


633(10) =


10 0111 1001(2)


3. Convert to binary (base 2) the fractional part: 0.599 975 585 98.

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) 0.599 975 585 98 × 2 = 1 + 0.199 951 171 96;
  • 2) 0.199 951 171 96 × 2 = 0 + 0.399 902 343 92;
  • 3) 0.399 902 343 92 × 2 = 0 + 0.799 804 687 84;
  • 4) 0.799 804 687 84 × 2 = 1 + 0.599 609 375 68;
  • 5) 0.599 609 375 68 × 2 = 1 + 0.199 218 751 36;
  • 6) 0.199 218 751 36 × 2 = 0 + 0.398 437 502 72;
  • 7) 0.398 437 502 72 × 2 = 0 + 0.796 875 005 44;
  • 8) 0.796 875 005 44 × 2 = 1 + 0.593 750 010 88;
  • 9) 0.593 750 010 88 × 2 = 1 + 0.187 500 021 76;
  • 10) 0.187 500 021 76 × 2 = 0 + 0.375 000 043 52;
  • 11) 0.375 000 043 52 × 2 = 0 + 0.750 000 087 04;
  • 12) 0.750 000 087 04 × 2 = 1 + 0.500 000 174 08;
  • 13) 0.500 000 174 08 × 2 = 1 + 0.000 000 348 16;
  • 14) 0.000 000 348 16 × 2 = 0 + 0.000 000 696 32;
  • 15) 0.000 000 696 32 × 2 = 0 + 0.000 001 392 64;
  • 16) 0.000 001 392 64 × 2 = 0 + 0.000 002 785 28;
  • 17) 0.000 002 785 28 × 2 = 0 + 0.000 005 570 56;
  • 18) 0.000 005 570 56 × 2 = 0 + 0.000 011 141 12;
  • 19) 0.000 011 141 12 × 2 = 0 + 0.000 022 282 24;
  • 20) 0.000 022 282 24 × 2 = 0 + 0.000 044 564 48;
  • 21) 0.000 044 564 48 × 2 = 0 + 0.000 089 128 96;
  • 22) 0.000 089 128 96 × 2 = 0 + 0.000 178 257 92;
  • 23) 0.000 178 257 92 × 2 = 0 + 0.000 356 515 84;
  • 24) 0.000 356 515 84 × 2 = 0 + 0.000 713 031 68;

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:


0.599 975 585 98(10) =


0.1001 1001 1001 1000 0000 0000(2)


5. Positive number before normalization:

633.599 975 585 98(10) =


10 0111 1001.1001 1001 1001 1000 0000 0000(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 9 positions to the left, so that only one non zero digit remains to the left of it:


633.599 975 585 98(10) =


10 0111 1001.1001 1001 1001 1000 0000 0000(2) =


10 0111 1001.1001 1001 1001 1000 0000 0000(2) × 20 =


1.0011 1100 1100 1100 1100 1100 0000 0000 0(2) × 29


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): 9


Mantissa (not normalized):
1.0011 1100 1100 1100 1100 1100 0000 0000 0


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


9 + 2(8-1) - 1 =


(9 + 127)(10) =


136(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;
  • 136 ÷ 2 = 68 + 0;
  • 68 ÷ 2 = 34 + 0;
  • 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) =


136(10) =


1000 1000(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 0110 0110 0110 0110 00 0000 0000 =


001 1110 0110 0110 0110 0110


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 1000


Mantissa (23 bits) =
001 1110 0110 0110 0110 0110


The base ten decimal number 633.599 975 585 98 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1000 1000 - 001 1110 0110 0110 0110 0110

(32 bits IEEE 754)

Number 633.599 975 585 97 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

Number 633.599 975 585 99 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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Available Base Conversions Between Decimal and Binary Systems

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1. Integer -> Binary

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3. Binary -> Integer

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