-0.180 000 000 032 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.180 000 000 032 9(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.180 000 000 032 9(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.180 000 000 032 9| = 0.180 000 000 032 9


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: 0.180 000 000 032 9.

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) 0.180 000 000 032 9 × 2 = 0 + 0.360 000 000 065 8;
  • 2) 0.360 000 000 065 8 × 2 = 0 + 0.720 000 000 131 6;
  • 3) 0.720 000 000 131 6 × 2 = 1 + 0.440 000 000 263 2;
  • 4) 0.440 000 000 263 2 × 2 = 0 + 0.880 000 000 526 4;
  • 5) 0.880 000 000 526 4 × 2 = 1 + 0.760 000 001 052 8;
  • 6) 0.760 000 001 052 8 × 2 = 1 + 0.520 000 002 105 6;
  • 7) 0.520 000 002 105 6 × 2 = 1 + 0.040 000 004 211 2;
  • 8) 0.040 000 004 211 2 × 2 = 0 + 0.080 000 008 422 4;
  • 9) 0.080 000 008 422 4 × 2 = 0 + 0.160 000 016 844 8;
  • 10) 0.160 000 016 844 8 × 2 = 0 + 0.320 000 033 689 6;
  • 11) 0.320 000 033 689 6 × 2 = 0 + 0.640 000 067 379 2;
  • 12) 0.640 000 067 379 2 × 2 = 1 + 0.280 000 134 758 4;
  • 13) 0.280 000 134 758 4 × 2 = 0 + 0.560 000 269 516 8;
  • 14) 0.560 000 269 516 8 × 2 = 1 + 0.120 000 539 033 6;
  • 15) 0.120 000 539 033 6 × 2 = 0 + 0.240 001 078 067 2;
  • 16) 0.240 001 078 067 2 × 2 = 0 + 0.480 002 156 134 4;
  • 17) 0.480 002 156 134 4 × 2 = 0 + 0.960 004 312 268 8;
  • 18) 0.960 004 312 268 8 × 2 = 1 + 0.920 008 624 537 6;
  • 19) 0.920 008 624 537 6 × 2 = 1 + 0.840 017 249 075 2;
  • 20) 0.840 017 249 075 2 × 2 = 1 + 0.680 034 498 150 4;
  • 21) 0.680 034 498 150 4 × 2 = 1 + 0.360 068 996 300 8;
  • 22) 0.360 068 996 300 8 × 2 = 0 + 0.720 137 992 601 6;
  • 23) 0.720 137 992 601 6 × 2 = 1 + 0.440 275 985 203 2;
  • 24) 0.440 275 985 203 2 × 2 = 0 + 0.880 551 970 406 4;
  • 25) 0.880 551 970 406 4 × 2 = 1 + 0.761 103 940 812 8;
  • 26) 0.761 103 940 812 8 × 2 = 1 + 0.522 207 881 625 6;
  • 27) 0.522 207 881 625 6 × 2 = 1 + 0.044 415 763 251 2;
  • 28) 0.044 415 763 251 2 × 2 = 0 + 0.088 831 526 502 4;
  • 29) 0.088 831 526 502 4 × 2 = 0 + 0.177 663 053 004 8;
  • 30) 0.177 663 053 004 8 × 2 = 0 + 0.355 326 106 009 6;
  • 31) 0.355 326 106 009 6 × 2 = 0 + 0.710 652 212 019 2;
  • 32) 0.710 652 212 019 2 × 2 = 1 + 0.421 304 424 038 4;
  • 33) 0.421 304 424 038 4 × 2 = 0 + 0.842 608 848 076 8;
  • 34) 0.842 608 848 076 8 × 2 = 1 + 0.685 217 696 153 6;
  • 35) 0.685 217 696 153 6 × 2 = 1 + 0.370 435 392 307 2;
  • 36) 0.370 435 392 307 2 × 2 = 0 + 0.740 870 784 614 4;
  • 37) 0.740 870 784 614 4 × 2 = 1 + 0.481 741 569 228 8;
  • 38) 0.481 741 569 228 8 × 2 = 0 + 0.963 483 138 457 6;
  • 39) 0.963 483 138 457 6 × 2 = 1 + 0.926 966 276 915 2;
  • 40) 0.926 966 276 915 2 × 2 = 1 + 0.853 932 553 830 4;
  • 41) 0.853 932 553 830 4 × 2 = 1 + 0.707 865 107 660 8;
  • 42) 0.707 865 107 660 8 × 2 = 1 + 0.415 730 215 321 6;
  • 43) 0.415 730 215 321 6 × 2 = 0 + 0.831 460 430 643 2;
  • 44) 0.831 460 430 643 2 × 2 = 1 + 0.662 920 861 286 4;
  • 45) 0.662 920 861 286 4 × 2 = 1 + 0.325 841 722 572 8;
  • 46) 0.325 841 722 572 8 × 2 = 0 + 0.651 683 445 145 6;
  • 47) 0.651 683 445 145 6 × 2 = 1 + 0.303 366 890 291 2;
  • 48) 0.303 366 890 291 2 × 2 = 0 + 0.606 733 780 582 4;
  • 49) 0.606 733 780 582 4 × 2 = 1 + 0.213 467 561 164 8;
  • 50) 0.213 467 561 164 8 × 2 = 0 + 0.426 935 122 329 6;
  • 51) 0.426 935 122 329 6 × 2 = 0 + 0.853 870 244 659 2;
  • 52) 0.853 870 244 659 2 × 2 = 1 + 0.707 740 489 318 4;
  • 53) 0.707 740 489 318 4 × 2 = 1 + 0.415 480 978 636 8;
  • 54) 0.415 480 978 636 8 × 2 = 0 + 0.830 961 957 273 6;
  • 55) 0.830 961 957 273 6 × 2 = 1 + 0.661 923 914 547 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:


0.180 000 000 032 9(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 1011 1101 1010 1001 101(2)

6. Positive number before normalization:

0.180 000 000 032 9(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 1011 1101 1010 1001 101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.180 000 000 032 9(10) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 1011 1101 1010 1001 101(2) =

0.0010 1110 0001 0100 0111 1010 1110 0001 0110 1011 1101 1010 1001 101(2) × 20 =

1.0111 0000 1010 0011 1101 0111 0000 1011 0101 1110 1101 0100 1101(2) × 2-3


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -3

Mantissa (not normalized):
1.0111 0000 1010 0011 1101 0111 0000 1011 0101 1110 1101 0100 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 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) =

1020(10) =

011 1111 1100(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 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0111 0000 1010 0011 1101 0111 0000 1011 0101 1110 1101 0100 1101 =

0111 0000 1010 0011 1101 0111 0000 1011 0101 1110 1101 0100 1101


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

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

Exponent (11 bits) =
011 1111 1100

Mantissa (52 bits) =
0111 0000 1010 0011 1101 0111 0000 1011 0101 1110 1101 0100 1101


Decimal number -0.180 000 000 032 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0111 0000 1010 0011 1101 0111 0000 1011 0101 1110 1101 0100 1101

» Convert decimal -0.180 000 000 035 2 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double 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 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 from the previous 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 multiplying operations, starting from the top of the list constructed 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, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' 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 -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

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

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 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:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. 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) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) 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:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

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

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

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

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100