-0.000 566 999 951 843 016 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 566 999 951 843 016(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 566 999 951 843 016(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 566 999 951 843 016| = 0.000 566 999 951 843 016


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: 0.000 566 999 951 843 016.

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) 0.000 566 999 951 843 016 × 2 = 0 + 0.001 133 999 903 686 032;
  • 2) 0.001 133 999 903 686 032 × 2 = 0 + 0.002 267 999 807 372 064;
  • 3) 0.002 267 999 807 372 064 × 2 = 0 + 0.004 535 999 614 744 128;
  • 4) 0.004 535 999 614 744 128 × 2 = 0 + 0.009 071 999 229 488 256;
  • 5) 0.009 071 999 229 488 256 × 2 = 0 + 0.018 143 998 458 976 512;
  • 6) 0.018 143 998 458 976 512 × 2 = 0 + 0.036 287 996 917 953 024;
  • 7) 0.036 287 996 917 953 024 × 2 = 0 + 0.072 575 993 835 906 048;
  • 8) 0.072 575 993 835 906 048 × 2 = 0 + 0.145 151 987 671 812 096;
  • 9) 0.145 151 987 671 812 096 × 2 = 0 + 0.290 303 975 343 624 192;
  • 10) 0.290 303 975 343 624 192 × 2 = 0 + 0.580 607 950 687 248 384;
  • 11) 0.580 607 950 687 248 384 × 2 = 1 + 0.161 215 901 374 496 768;
  • 12) 0.161 215 901 374 496 768 × 2 = 0 + 0.322 431 802 748 993 536;
  • 13) 0.322 431 802 748 993 536 × 2 = 0 + 0.644 863 605 497 987 072;
  • 14) 0.644 863 605 497 987 072 × 2 = 1 + 0.289 727 210 995 974 144;
  • 15) 0.289 727 210 995 974 144 × 2 = 0 + 0.579 454 421 991 948 288;
  • 16) 0.579 454 421 991 948 288 × 2 = 1 + 0.158 908 843 983 896 576;
  • 17) 0.158 908 843 983 896 576 × 2 = 0 + 0.317 817 687 967 793 152;
  • 18) 0.317 817 687 967 793 152 × 2 = 0 + 0.635 635 375 935 586 304;
  • 19) 0.635 635 375 935 586 304 × 2 = 1 + 0.271 270 751 871 172 608;
  • 20) 0.271 270 751 871 172 608 × 2 = 0 + 0.542 541 503 742 345 216;
  • 21) 0.542 541 503 742 345 216 × 2 = 1 + 0.085 083 007 484 690 432;
  • 22) 0.085 083 007 484 690 432 × 2 = 0 + 0.170 166 014 969 380 864;
  • 23) 0.170 166 014 969 380 864 × 2 = 0 + 0.340 332 029 938 761 728;
  • 24) 0.340 332 029 938 761 728 × 2 = 0 + 0.680 664 059 877 523 456;
  • 25) 0.680 664 059 877 523 456 × 2 = 1 + 0.361 328 119 755 046 912;
  • 26) 0.361 328 119 755 046 912 × 2 = 0 + 0.722 656 239 510 093 824;
  • 27) 0.722 656 239 510 093 824 × 2 = 1 + 0.445 312 479 020 187 648;
  • 28) 0.445 312 479 020 187 648 × 2 = 0 + 0.890 624 958 040 375 296;
  • 29) 0.890 624 958 040 375 296 × 2 = 1 + 0.781 249 916 080 750 592;
  • 30) 0.781 249 916 080 750 592 × 2 = 1 + 0.562 499 832 161 501 184;
  • 31) 0.562 499 832 161 501 184 × 2 = 1 + 0.124 999 664 323 002 368;
  • 32) 0.124 999 664 323 002 368 × 2 = 0 + 0.249 999 328 646 004 736;
  • 33) 0.249 999 328 646 004 736 × 2 = 0 + 0.499 998 657 292 009 472;
  • 34) 0.499 998 657 292 009 472 × 2 = 0 + 0.999 997 314 584 018 944;

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 - 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:


0.000 566 999 951 843 016(10) =

0.0000 0000 0010 0101 0010 1000 1010 1110 00(2)

6. Positive number before normalization:

0.000 566 999 951 843 016(10) =

0.0000 0000 0010 0101 0010 1000 1010 1110 00(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:


0.000 566 999 951 843 016(10) =

0.0000 0000 0010 0101 0010 1000 1010 1110 00(2) =

0.0000 0000 0010 0101 0010 1000 1010 1110 00(2) × 20 =

1.0010 1001 0100 0101 0111 000(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 000


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 1000 =

001 0100 1010 0010 1011 1000


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 1000


Decimal number -0.000 566 999 951 843 016 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0111 0100 - 001 0100 1010 0010 1011 1000

Share

» Convert decimal -0.000 566 999 951 842 929 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: The Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 minutes

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