Decimal to 32 Bit IEEE 754 Binary: Convert Number -0.000 566 999 951 843 172 322 8 to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number -0.000 566 999 951 843 172 322 8(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. Start with the positive version of the number:

|-0.000 566 999 951 843 172 322 8| = 0.000 566 999 951 843 172 322 8


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 566 999 951 843 172 322 8.

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) 0.000 566 999 951 843 172 322 8 × 2 = 0 + 0.001 133 999 903 686 344 645 6;
  • 2) 0.001 133 999 903 686 344 645 6 × 2 = 0 + 0.002 267 999 807 372 689 291 2;
  • 3) 0.002 267 999 807 372 689 291 2 × 2 = 0 + 0.004 535 999 614 745 378 582 4;
  • 4) 0.004 535 999 614 745 378 582 4 × 2 = 0 + 0.009 071 999 229 490 757 164 8;
  • 5) 0.009 071 999 229 490 757 164 8 × 2 = 0 + 0.018 143 998 458 981 514 329 6;
  • 6) 0.018 143 998 458 981 514 329 6 × 2 = 0 + 0.036 287 996 917 963 028 659 2;
  • 7) 0.036 287 996 917 963 028 659 2 × 2 = 0 + 0.072 575 993 835 926 057 318 4;
  • 8) 0.072 575 993 835 926 057 318 4 × 2 = 0 + 0.145 151 987 671 852 114 636 8;
  • 9) 0.145 151 987 671 852 114 636 8 × 2 = 0 + 0.290 303 975 343 704 229 273 6;
  • 10) 0.290 303 975 343 704 229 273 6 × 2 = 0 + 0.580 607 950 687 408 458 547 2;
  • 11) 0.580 607 950 687 408 458 547 2 × 2 = 1 + 0.161 215 901 374 816 917 094 4;
  • 12) 0.161 215 901 374 816 917 094 4 × 2 = 0 + 0.322 431 802 749 633 834 188 8;
  • 13) 0.322 431 802 749 633 834 188 8 × 2 = 0 + 0.644 863 605 499 267 668 377 6;
  • 14) 0.644 863 605 499 267 668 377 6 × 2 = 1 + 0.289 727 210 998 535 336 755 2;
  • 15) 0.289 727 210 998 535 336 755 2 × 2 = 0 + 0.579 454 421 997 070 673 510 4;
  • 16) 0.579 454 421 997 070 673 510 4 × 2 = 1 + 0.158 908 843 994 141 347 020 8;
  • 17) 0.158 908 843 994 141 347 020 8 × 2 = 0 + 0.317 817 687 988 282 694 041 6;
  • 18) 0.317 817 687 988 282 694 041 6 × 2 = 0 + 0.635 635 375 976 565 388 083 2;
  • 19) 0.635 635 375 976 565 388 083 2 × 2 = 1 + 0.271 270 751 953 130 776 166 4;
  • 20) 0.271 270 751 953 130 776 166 4 × 2 = 0 + 0.542 541 503 906 261 552 332 8;
  • 21) 0.542 541 503 906 261 552 332 8 × 2 = 1 + 0.085 083 007 812 523 104 665 6;
  • 22) 0.085 083 007 812 523 104 665 6 × 2 = 0 + 0.170 166 015 625 046 209 331 2;
  • 23) 0.170 166 015 625 046 209 331 2 × 2 = 0 + 0.340 332 031 250 092 418 662 4;
  • 24) 0.340 332 031 250 092 418 662 4 × 2 = 0 + 0.680 664 062 500 184 837 324 8;
  • 25) 0.680 664 062 500 184 837 324 8 × 2 = 1 + 0.361 328 125 000 369 674 649 6;
  • 26) 0.361 328 125 000 369 674 649 6 × 2 = 0 + 0.722 656 250 000 739 349 299 2;
  • 27) 0.722 656 250 000 739 349 299 2 × 2 = 1 + 0.445 312 500 001 478 698 598 4;
  • 28) 0.445 312 500 001 478 698 598 4 × 2 = 0 + 0.890 625 000 002 957 397 196 8;
  • 29) 0.890 625 000 002 957 397 196 8 × 2 = 1 + 0.781 250 000 005 914 794 393 6;
  • 30) 0.781 250 000 005 914 794 393 6 × 2 = 1 + 0.562 500 000 011 829 588 787 2;
  • 31) 0.562 500 000 011 829 588 787 2 × 2 = 1 + 0.125 000 000 023 659 177 574 4;
  • 32) 0.125 000 000 023 659 177 574 4 × 2 = 0 + 0.250 000 000 047 318 355 148 8;
  • 33) 0.250 000 000 047 318 355 148 8 × 2 = 0 + 0.500 000 000 094 636 710 297 6;
  • 34) 0.500 000 000 094 636 710 297 6 × 2 = 1 + 0.000 000 000 189 273 420 595 2;

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).


5. 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:


0.000 566 999 951 843 172 322 8(10) =

0.0000 0000 0010 0101 0010 1000 1010 1110 01(2)

6. Positive number before normalization:

0.000 566 999 951 843 172 322 8(10) =

0.0000 0000 0010 0101 0010 1000 1010 1110 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 11 positions to the right, so that only one non zero digit remains to the left of it:


0.000 566 999 951 843 172 322 8(10) =

0.0000 0000 0010 0101 0010 1000 1010 1110 01(2) =

0.0000 0000 0010 0101 0010 1000 1010 1110 01(2) × 20 =

1.0010 1001 0100 0101 0111 001(2) × 2-11


8. 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 1 (a negative number)

Exponent (unadjusted): -11

Mantissa (not normalized):
1.0010 1001 0100 0101 0111 001


9. 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) =

116(10)


10. 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;
  • 116 ÷ 2 = 58 + 0;
  • 58 ÷ 2 = 29 + 0;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

116(10) =

0111 0100(2)


12. 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. 001 0100 1010 0010 1011 1001 =

001 0100 1010 0010 1011 1001


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

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

Exponent (8 bits) =
0111 0100

Mantissa (23 bits) =
001 0100 1010 0010 1011 1001


The base ten decimal number -0.000 566 999 951 843 172 322 8 converted and written in 32 bit single precision IEEE 754 binary floating point representation:

1 - 0111 0100 - 001 0100 1010 0010 1011 1001

» Decimal number in base ten -0.000 566 999 951 843 172 330 5 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