-0.000 282 005 913 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 37(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
-0.000 282 005 913 37(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:

|-0.000 282 005 913 37| = 0.000 282 005 913 37


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: 0.000 282 005 913 37.

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) 0.000 282 005 913 37 × 2 = 0 + 0.000 564 011 826 74;
  • 2) 0.000 564 011 826 74 × 2 = 0 + 0.001 128 023 653 48;
  • 3) 0.001 128 023 653 48 × 2 = 0 + 0.002 256 047 306 96;
  • 4) 0.002 256 047 306 96 × 2 = 0 + 0.004 512 094 613 92;
  • 5) 0.004 512 094 613 92 × 2 = 0 + 0.009 024 189 227 84;
  • 6) 0.009 024 189 227 84 × 2 = 0 + 0.018 048 378 455 68;
  • 7) 0.018 048 378 455 68 × 2 = 0 + 0.036 096 756 911 36;
  • 8) 0.036 096 756 911 36 × 2 = 0 + 0.072 193 513 822 72;
  • 9) 0.072 193 513 822 72 × 2 = 0 + 0.144 387 027 645 44;
  • 10) 0.144 387 027 645 44 × 2 = 0 + 0.288 774 055 290 88;
  • 11) 0.288 774 055 290 88 × 2 = 0 + 0.577 548 110 581 76;
  • 12) 0.577 548 110 581 76 × 2 = 1 + 0.155 096 221 163 52;
  • 13) 0.155 096 221 163 52 × 2 = 0 + 0.310 192 442 327 04;
  • 14) 0.310 192 442 327 04 × 2 = 0 + 0.620 384 884 654 08;
  • 15) 0.620 384 884 654 08 × 2 = 1 + 0.240 769 769 308 16;
  • 16) 0.240 769 769 308 16 × 2 = 0 + 0.481 539 538 616 32;
  • 17) 0.481 539 538 616 32 × 2 = 0 + 0.963 079 077 232 64;
  • 18) 0.963 079 077 232 64 × 2 = 1 + 0.926 158 154 465 28;
  • 19) 0.926 158 154 465 28 × 2 = 1 + 0.852 316 308 930 56;
  • 20) 0.852 316 308 930 56 × 2 = 1 + 0.704 632 617 861 12;
  • 21) 0.704 632 617 861 12 × 2 = 1 + 0.409 265 235 722 24;
  • 22) 0.409 265 235 722 24 × 2 = 0 + 0.818 530 471 444 48;
  • 23) 0.818 530 471 444 48 × 2 = 1 + 0.637 060 942 888 96;
  • 24) 0.637 060 942 888 96 × 2 = 1 + 0.274 121 885 777 92;
  • 25) 0.274 121 885 777 92 × 2 = 0 + 0.548 243 771 555 84;
  • 26) 0.548 243 771 555 84 × 2 = 1 + 0.096 487 543 111 68;
  • 27) 0.096 487 543 111 68 × 2 = 0 + 0.192 975 086 223 36;
  • 28) 0.192 975 086 223 36 × 2 = 0 + 0.385 950 172 446 72;
  • 29) 0.385 950 172 446 72 × 2 = 0 + 0.771 900 344 893 44;
  • 30) 0.771 900 344 893 44 × 2 = 1 + 0.543 800 689 786 88;
  • 31) 0.543 800 689 786 88 × 2 = 1 + 0.087 601 379 573 76;
  • 32) 0.087 601 379 573 76 × 2 = 0 + 0.175 202 759 147 52;
  • 33) 0.175 202 759 147 52 × 2 = 0 + 0.350 405 518 295 04;
  • 34) 0.350 405 518 295 04 × 2 = 0 + 0.700 811 036 590 08;
  • 35) 0.700 811 036 590 08 × 2 = 1 + 0.401 622 073 180 16;
  • 36) 0.401 622 073 180 16 × 2 = 0 + 0.803 244 146 360 32;
  • 37) 0.803 244 146 360 32 × 2 = 1 + 0.606 488 292 720 64;
  • 38) 0.606 488 292 720 64 × 2 = 1 + 0.212 976 585 441 28;
  • 39) 0.212 976 585 441 28 × 2 = 0 + 0.425 953 170 882 56;
  • 40) 0.425 953 170 882 56 × 2 = 0 + 0.851 906 341 765 12;
  • 41) 0.851 906 341 765 12 × 2 = 1 + 0.703 812 683 530 24;
  • 42) 0.703 812 683 530 24 × 2 = 1 + 0.407 625 367 060 48;
  • 43) 0.407 625 367 060 48 × 2 = 0 + 0.815 250 734 120 96;
  • 44) 0.815 250 734 120 96 × 2 = 1 + 0.630 501 468 241 92;
  • 45) 0.630 501 468 241 92 × 2 = 1 + 0.261 002 936 483 84;
  • 46) 0.261 002 936 483 84 × 2 = 0 + 0.522 005 872 967 68;
  • 47) 0.522 005 872 967 68 × 2 = 1 + 0.044 011 745 935 36;
  • 48) 0.044 011 745 935 36 × 2 = 0 + 0.088 023 491 870 72;
  • 49) 0.088 023 491 870 72 × 2 = 0 + 0.176 046 983 741 44;
  • 50) 0.176 046 983 741 44 × 2 = 0 + 0.352 093 967 482 88;
  • 51) 0.352 093 967 482 88 × 2 = 0 + 0.704 187 934 965 76;
  • 52) 0.704 187 934 965 76 × 2 = 1 + 0.408 375 869 931 52;
  • 53) 0.408 375 869 931 52 × 2 = 0 + 0.816 751 739 863 04;
  • 54) 0.816 751 739 863 04 × 2 = 1 + 0.633 503 479 726 08;
  • 55) 0.633 503 479 726 08 × 2 = 1 + 0.267 006 959 452 16;
  • 56) 0.267 006 959 452 16 × 2 = 0 + 0.534 013 918 904 32;
  • 57) 0.534 013 918 904 32 × 2 = 1 + 0.068 027 837 808 64;
  • 58) 0.068 027 837 808 64 × 2 = 0 + 0.136 055 675 617 28;
  • 59) 0.136 055 675 617 28 × 2 = 0 + 0.272 111 351 234 56;
  • 60) 0.272 111 351 234 56 × 2 = 0 + 0.544 222 702 469 12;
  • 61) 0.544 222 702 469 12 × 2 = 1 + 0.088 445 404 938 24;
  • 62) 0.088 445 404 938 24 × 2 = 0 + 0.176 890 809 876 48;
  • 63) 0.176 890 809 876 48 × 2 = 0 + 0.353 781 619 752 96;
  • 64) 0.353 781 619 752 96 × 2 = 0 + 0.707 563 239 505 92;

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:


0.000 282 005 913 37(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1101 1010 0001 0110 1000 1000(2)

6. Positive number before normalization:

0.000 282 005 913 37(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1101 1010 0001 0110 1000 1000(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:


0.000 282 005 913 37(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1101 1010 0001 0110 1000 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1101 1010 0001 0110 1000 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1100 1101 1010 0001 0110 1000 1000(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 0100 0110 0010 1100 1101 1010 0001 0110 1000 1000


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 0100 0110 0010 1100 1101 1010 0001 0110 1000 1000 =

0010 0111 1011 0100 0110 0010 1100 1101 1010 0001 0110 1000 1000


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 0100 0110 0010 1100 1101 1010 0001 0110 1000 1000


Decimal number -0.000 282 005 913 37 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1100 1101 1010 0001 0110 1000 1000

Share

» Convert decimal -0.000 282 005 914 07 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: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 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