-0.000 282 035 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 035 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
-0.000 282 035 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:

|-0.000 282 035 9| = 0.000 282 035 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: 0.000 282 035 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) 0.000 282 035 9 × 2 = 0 + 0.000 564 071 8;
  • 2) 0.000 564 071 8 × 2 = 0 + 0.001 128 143 6;
  • 3) 0.001 128 143 6 × 2 = 0 + 0.002 256 287 2;
  • 4) 0.002 256 287 2 × 2 = 0 + 0.004 512 574 4;
  • 5) 0.004 512 574 4 × 2 = 0 + 0.009 025 148 8;
  • 6) 0.009 025 148 8 × 2 = 0 + 0.018 050 297 6;
  • 7) 0.018 050 297 6 × 2 = 0 + 0.036 100 595 2;
  • 8) 0.036 100 595 2 × 2 = 0 + 0.072 201 190 4;
  • 9) 0.072 201 190 4 × 2 = 0 + 0.144 402 380 8;
  • 10) 0.144 402 380 8 × 2 = 0 + 0.288 804 761 6;
  • 11) 0.288 804 761 6 × 2 = 0 + 0.577 609 523 2;
  • 12) 0.577 609 523 2 × 2 = 1 + 0.155 219 046 4;
  • 13) 0.155 219 046 4 × 2 = 0 + 0.310 438 092 8;
  • 14) 0.310 438 092 8 × 2 = 0 + 0.620 876 185 6;
  • 15) 0.620 876 185 6 × 2 = 1 + 0.241 752 371 2;
  • 16) 0.241 752 371 2 × 2 = 0 + 0.483 504 742 4;
  • 17) 0.483 504 742 4 × 2 = 0 + 0.967 009 484 8;
  • 18) 0.967 009 484 8 × 2 = 1 + 0.934 018 969 6;
  • 19) 0.934 018 969 6 × 2 = 1 + 0.868 037 939 2;
  • 20) 0.868 037 939 2 × 2 = 1 + 0.736 075 878 4;
  • 21) 0.736 075 878 4 × 2 = 1 + 0.472 151 756 8;
  • 22) 0.472 151 756 8 × 2 = 0 + 0.944 303 513 6;
  • 23) 0.944 303 513 6 × 2 = 1 + 0.888 607 027 2;
  • 24) 0.888 607 027 2 × 2 = 1 + 0.777 214 054 4;
  • 25) 0.777 214 054 4 × 2 = 1 + 0.554 428 108 8;
  • 26) 0.554 428 108 8 × 2 = 1 + 0.108 856 217 6;
  • 27) 0.108 856 217 6 × 2 = 0 + 0.217 712 435 2;
  • 28) 0.217 712 435 2 × 2 = 0 + 0.435 424 870 4;
  • 29) 0.435 424 870 4 × 2 = 0 + 0.870 849 740 8;
  • 30) 0.870 849 740 8 × 2 = 1 + 0.741 699 481 6;
  • 31) 0.741 699 481 6 × 2 = 1 + 0.483 398 963 2;
  • 32) 0.483 398 963 2 × 2 = 0 + 0.966 797 926 4;
  • 33) 0.966 797 926 4 × 2 = 1 + 0.933 595 852 8;
  • 34) 0.933 595 852 8 × 2 = 1 + 0.867 191 705 6;
  • 35) 0.867 191 705 6 × 2 = 1 + 0.734 383 411 2;
  • 36) 0.734 383 411 2 × 2 = 1 + 0.468 766 822 4;
  • 37) 0.468 766 822 4 × 2 = 0 + 0.937 533 644 8;
  • 38) 0.937 533 644 8 × 2 = 1 + 0.875 067 289 6;
  • 39) 0.875 067 289 6 × 2 = 1 + 0.750 134 579 2;
  • 40) 0.750 134 579 2 × 2 = 1 + 0.500 269 158 4;
  • 41) 0.500 269 158 4 × 2 = 1 + 0.000 538 316 8;
  • 42) 0.000 538 316 8 × 2 = 0 + 0.001 076 633 6;
  • 43) 0.001 076 633 6 × 2 = 0 + 0.002 153 267 2;
  • 44) 0.002 153 267 2 × 2 = 0 + 0.004 306 534 4;
  • 45) 0.004 306 534 4 × 2 = 0 + 0.008 613 068 8;
  • 46) 0.008 613 068 8 × 2 = 0 + 0.017 226 137 6;
  • 47) 0.017 226 137 6 × 2 = 0 + 0.034 452 275 2;
  • 48) 0.034 452 275 2 × 2 = 0 + 0.068 904 550 4;
  • 49) 0.068 904 550 4 × 2 = 0 + 0.137 809 100 8;
  • 50) 0.137 809 100 8 × 2 = 0 + 0.275 618 201 6;
  • 51) 0.275 618 201 6 × 2 = 0 + 0.551 236 403 2;
  • 52) 0.551 236 403 2 × 2 = 1 + 0.102 472 806 4;
  • 53) 0.102 472 806 4 × 2 = 0 + 0.204 945 612 8;
  • 54) 0.204 945 612 8 × 2 = 0 + 0.409 891 225 6;
  • 55) 0.409 891 225 6 × 2 = 0 + 0.819 782 451 2;
  • 56) 0.819 782 451 2 × 2 = 1 + 0.639 564 902 4;
  • 57) 0.639 564 902 4 × 2 = 1 + 0.279 129 804 8;
  • 58) 0.279 129 804 8 × 2 = 0 + 0.558 259 609 6;
  • 59) 0.558 259 609 6 × 2 = 1 + 0.116 519 219 2;
  • 60) 0.116 519 219 2 × 2 = 0 + 0.233 038 438 4;
  • 61) 0.233 038 438 4 × 2 = 0 + 0.466 076 876 8;
  • 62) 0.466 076 876 8 × 2 = 0 + 0.932 153 753 6;
  • 63) 0.932 153 753 6 × 2 = 1 + 0.864 307 507 2;
  • 64) 0.864 307 507 2 × 2 = 1 + 0.728 615 014 4;

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:


0.000 282 035 9(10) =

0.0000 0000 0001 0010 0111 1011 1100 0110 1111 0111 1000 0000 0001 0001 1010 0011(2)

6. Positive number before normalization:

0.000 282 035 9(10) =

0.0000 0000 0001 0010 0111 1011 1100 0110 1111 0111 1000 0000 0001 0001 1010 0011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 035 9(10) =

0.0000 0000 0001 0010 0111 1011 1100 0110 1111 0111 1000 0000 0001 0001 1010 0011(2) =

0.0000 0000 0001 0010 0111 1011 1100 0110 1111 0111 1000 0000 0001 0001 1010 0011(2) × 20 =

1.0010 0111 1011 1100 0110 1111 0111 1000 0000 0001 0001 1010 0011(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 0111 1011 1100 0110 1111 0111 1000 0000 0001 0001 1010 0011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 0111 1011 1100 0110 1111 0111 1000 0000 0001 0001 1010 0011 =

0010 0111 1011 1100 0110 1111 0111 1000 0000 0001 0001 1010 0011


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 0011

Mantissa (52 bits) =
0010 0111 1011 1100 0110 1111 0111 1000 0000 0001 0001 1010 0011


Decimal number -0.000 282 035 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 1100 0110 1111 0111 1000 0000 0001 0001 1010 0011

Share

» Convert decimal -0.000 282 039 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 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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