Decimal to 32 Bit IEEE 754 Binary: Convert Number 0.000 000 119 209 289 550 779 79 to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 0.000 000 119 209 289 550 779 79(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 119 209 289 550 779 79.

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) 0.000 000 119 209 289 550 779 79 × 2 = 0 + 0.000 000 238 418 579 101 559 58;
  • 2) 0.000 000 238 418 579 101 559 58 × 2 = 0 + 0.000 000 476 837 158 203 119 16;
  • 3) 0.000 000 476 837 158 203 119 16 × 2 = 0 + 0.000 000 953 674 316 406 238 32;
  • 4) 0.000 000 953 674 316 406 238 32 × 2 = 0 + 0.000 001 907 348 632 812 476 64;
  • 5) 0.000 001 907 348 632 812 476 64 × 2 = 0 + 0.000 003 814 697 265 624 953 28;
  • 6) 0.000 003 814 697 265 624 953 28 × 2 = 0 + 0.000 007 629 394 531 249 906 56;
  • 7) 0.000 007 629 394 531 249 906 56 × 2 = 0 + 0.000 015 258 789 062 499 813 12;
  • 8) 0.000 015 258 789 062 499 813 12 × 2 = 0 + 0.000 030 517 578 124 999 626 24;
  • 9) 0.000 030 517 578 124 999 626 24 × 2 = 0 + 0.000 061 035 156 249 999 252 48;
  • 10) 0.000 061 035 156 249 999 252 48 × 2 = 0 + 0.000 122 070 312 499 998 504 96;
  • 11) 0.000 122 070 312 499 998 504 96 × 2 = 0 + 0.000 244 140 624 999 997 009 92;
  • 12) 0.000 244 140 624 999 997 009 92 × 2 = 0 + 0.000 488 281 249 999 994 019 84;
  • 13) 0.000 488 281 249 999 994 019 84 × 2 = 0 + 0.000 976 562 499 999 988 039 68;
  • 14) 0.000 976 562 499 999 988 039 68 × 2 = 0 + 0.001 953 124 999 999 976 079 36;
  • 15) 0.001 953 124 999 999 976 079 36 × 2 = 0 + 0.003 906 249 999 999 952 158 72;
  • 16) 0.003 906 249 999 999 952 158 72 × 2 = 0 + 0.007 812 499 999 999 904 317 44;
  • 17) 0.007 812 499 999 999 904 317 44 × 2 = 0 + 0.015 624 999 999 999 808 634 88;
  • 18) 0.015 624 999 999 999 808 634 88 × 2 = 0 + 0.031 249 999 999 999 617 269 76;
  • 19) 0.031 249 999 999 999 617 269 76 × 2 = 0 + 0.062 499 999 999 999 234 539 52;
  • 20) 0.062 499 999 999 999 234 539 52 × 2 = 0 + 0.124 999 999 999 998 469 079 04;
  • 21) 0.124 999 999 999 998 469 079 04 × 2 = 0 + 0.249 999 999 999 996 938 158 08;
  • 22) 0.249 999 999 999 996 938 158 08 × 2 = 0 + 0.499 999 999 999 993 876 316 16;
  • 23) 0.499 999 999 999 993 876 316 16 × 2 = 0 + 0.999 999 999 999 987 752 632 32;
  • 24) 0.999 999 999 999 987 752 632 32 × 2 = 1 + 0.999 999 999 999 975 505 264 64;
  • 25) 0.999 999 999 999 975 505 264 64 × 2 = 1 + 0.999 999 999 999 951 010 529 28;
  • 26) 0.999 999 999 999 951 010 529 28 × 2 = 1 + 0.999 999 999 999 902 021 058 56;
  • 27) 0.999 999 999 999 902 021 058 56 × 2 = 1 + 0.999 999 999 999 804 042 117 12;
  • 28) 0.999 999 999 999 804 042 117 12 × 2 = 1 + 0.999 999 999 999 608 084 234 24;
  • 29) 0.999 999 999 999 608 084 234 24 × 2 = 1 + 0.999 999 999 999 216 168 468 48;
  • 30) 0.999 999 999 999 216 168 468 48 × 2 = 1 + 0.999 999 999 998 432 336 936 96;
  • 31) 0.999 999 999 998 432 336 936 96 × 2 = 1 + 0.999 999 999 996 864 673 873 92;
  • 32) 0.999 999 999 996 864 673 873 92 × 2 = 1 + 0.999 999 999 993 729 347 747 84;
  • 33) 0.999 999 999 993 729 347 747 84 × 2 = 1 + 0.999 999 999 987 458 695 495 68;
  • 34) 0.999 999 999 987 458 695 495 68 × 2 = 1 + 0.999 999 999 974 917 390 991 36;
  • 35) 0.999 999 999 974 917 390 991 36 × 2 = 1 + 0.999 999 999 949 834 781 982 72;
  • 36) 0.999 999 999 949 834 781 982 72 × 2 = 1 + 0.999 999 999 899 669 563 965 44;
  • 37) 0.999 999 999 899 669 563 965 44 × 2 = 1 + 0.999 999 999 799 339 127 930 88;
  • 38) 0.999 999 999 799 339 127 930 88 × 2 = 1 + 0.999 999 999 598 678 255 861 76;
  • 39) 0.999 999 999 598 678 255 861 76 × 2 = 1 + 0.999 999 999 197 356 511 723 52;
  • 40) 0.999 999 999 197 356 511 723 52 × 2 = 1 + 0.999 999 998 394 713 023 447 04;
  • 41) 0.999 999 998 394 713 023 447 04 × 2 = 1 + 0.999 999 996 789 426 046 894 08;
  • 42) 0.999 999 996 789 426 046 894 08 × 2 = 1 + 0.999 999 993 578 852 093 788 16;
  • 43) 0.999 999 993 578 852 093 788 16 × 2 = 1 + 0.999 999 987 157 704 187 576 32;
  • 44) 0.999 999 987 157 704 187 576 32 × 2 = 1 + 0.999 999 974 315 408 375 152 64;
  • 45) 0.999 999 974 315 408 375 152 64 × 2 = 1 + 0.999 999 948 630 816 750 305 28;
  • 46) 0.999 999 948 630 816 750 305 28 × 2 = 1 + 0.999 999 897 261 633 500 610 56;
  • 47) 0.999 999 897 261 633 500 610 56 × 2 = 1 + 0.999 999 794 523 267 001 221 12;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.000 000 119 209 289 550 779 79(10) =

0.0000 0000 0000 0000 0000 0001 1111 1111 1111 1111 1111 111(2)

5. Positive number before normalization:

0.000 000 119 209 289 550 779 79(10) =

0.0000 0000 0000 0000 0000 0001 1111 1111 1111 1111 1111 111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 119 209 289 550 779 79(10) =

0.0000 0000 0000 0000 0000 0001 1111 1111 1111 1111 1111 111(2) =

0.0000 0000 0000 0000 0000 0001 1111 1111 1111 1111 1111 111(2) × 20 =

1.1111 1111 1111 1111 1111 111(2) × 2-24


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): -24

Mantissa (not normalized):
1.1111 1111 1111 1111 1111 111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

103(10) =

0110 0111(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 111 1111 1111 1111 1111 1111 =

111 1111 1111 1111 1111 1111


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) =
0110 0111

Mantissa (23 bits) =
111 1111 1111 1111 1111 1111


The base ten decimal number 0.000 000 119 209 289 550 779 79 converted and written in 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 111 1111 1111 1111 1111 1111

» Decimal number in base ten 0.000 000 119 209 289 550 778 83 converted and written as 32 bit single precision IEEE 754 binary floating point representation standard

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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:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

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

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

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

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

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

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111