-0.000 282 005 914 49 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 49(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
-0.000 282 005 914 49(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:

|-0.000 282 005 914 49| = 0.000 282 005 914 49


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: 0.000 282 005 914 49.

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) 0.000 282 005 914 49 × 2 = 0 + 0.000 564 011 828 98;
  • 2) 0.000 564 011 828 98 × 2 = 0 + 0.001 128 023 657 96;
  • 3) 0.001 128 023 657 96 × 2 = 0 + 0.002 256 047 315 92;
  • 4) 0.002 256 047 315 92 × 2 = 0 + 0.004 512 094 631 84;
  • 5) 0.004 512 094 631 84 × 2 = 0 + 0.009 024 189 263 68;
  • 6) 0.009 024 189 263 68 × 2 = 0 + 0.018 048 378 527 36;
  • 7) 0.018 048 378 527 36 × 2 = 0 + 0.036 096 757 054 72;
  • 8) 0.036 096 757 054 72 × 2 = 0 + 0.072 193 514 109 44;
  • 9) 0.072 193 514 109 44 × 2 = 0 + 0.144 387 028 218 88;
  • 10) 0.144 387 028 218 88 × 2 = 0 + 0.288 774 056 437 76;
  • 11) 0.288 774 056 437 76 × 2 = 0 + 0.577 548 112 875 52;
  • 12) 0.577 548 112 875 52 × 2 = 1 + 0.155 096 225 751 04;
  • 13) 0.155 096 225 751 04 × 2 = 0 + 0.310 192 451 502 08;
  • 14) 0.310 192 451 502 08 × 2 = 0 + 0.620 384 903 004 16;
  • 15) 0.620 384 903 004 16 × 2 = 1 + 0.240 769 806 008 32;
  • 16) 0.240 769 806 008 32 × 2 = 0 + 0.481 539 612 016 64;
  • 17) 0.481 539 612 016 64 × 2 = 0 + 0.963 079 224 033 28;
  • 18) 0.963 079 224 033 28 × 2 = 1 + 0.926 158 448 066 56;
  • 19) 0.926 158 448 066 56 × 2 = 1 + 0.852 316 896 133 12;
  • 20) 0.852 316 896 133 12 × 2 = 1 + 0.704 633 792 266 24;
  • 21) 0.704 633 792 266 24 × 2 = 1 + 0.409 267 584 532 48;
  • 22) 0.409 267 584 532 48 × 2 = 0 + 0.818 535 169 064 96;
  • 23) 0.818 535 169 064 96 × 2 = 1 + 0.637 070 338 129 92;
  • 24) 0.637 070 338 129 92 × 2 = 1 + 0.274 140 676 259 84;
  • 25) 0.274 140 676 259 84 × 2 = 0 + 0.548 281 352 519 68;
  • 26) 0.548 281 352 519 68 × 2 = 1 + 0.096 562 705 039 36;
  • 27) 0.096 562 705 039 36 × 2 = 0 + 0.193 125 410 078 72;
  • 28) 0.193 125 410 078 72 × 2 = 0 + 0.386 250 820 157 44;
  • 29) 0.386 250 820 157 44 × 2 = 0 + 0.772 501 640 314 88;
  • 30) 0.772 501 640 314 88 × 2 = 1 + 0.545 003 280 629 76;
  • 31) 0.545 003 280 629 76 × 2 = 1 + 0.090 006 561 259 52;
  • 32) 0.090 006 561 259 52 × 2 = 0 + 0.180 013 122 519 04;
  • 33) 0.180 013 122 519 04 × 2 = 0 + 0.360 026 245 038 08;
  • 34) 0.360 026 245 038 08 × 2 = 0 + 0.720 052 490 076 16;
  • 35) 0.720 052 490 076 16 × 2 = 1 + 0.440 104 980 152 32;
  • 36) 0.440 104 980 152 32 × 2 = 0 + 0.880 209 960 304 64;
  • 37) 0.880 209 960 304 64 × 2 = 1 + 0.760 419 920 609 28;
  • 38) 0.760 419 920 609 28 × 2 = 1 + 0.520 839 841 218 56;
  • 39) 0.520 839 841 218 56 × 2 = 1 + 0.041 679 682 437 12;
  • 40) 0.041 679 682 437 12 × 2 = 0 + 0.083 359 364 874 24;
  • 41) 0.083 359 364 874 24 × 2 = 0 + 0.166 718 729 748 48;
  • 42) 0.166 718 729 748 48 × 2 = 0 + 0.333 437 459 496 96;
  • 43) 0.333 437 459 496 96 × 2 = 0 + 0.666 874 918 993 92;
  • 44) 0.666 874 918 993 92 × 2 = 1 + 0.333 749 837 987 84;
  • 45) 0.333 749 837 987 84 × 2 = 0 + 0.667 499 675 975 68;
  • 46) 0.667 499 675 975 68 × 2 = 1 + 0.334 999 351 951 36;
  • 47) 0.334 999 351 951 36 × 2 = 0 + 0.669 998 703 902 72;
  • 48) 0.669 998 703 902 72 × 2 = 1 + 0.339 997 407 805 44;
  • 49) 0.339 997 407 805 44 × 2 = 0 + 0.679 994 815 610 88;
  • 50) 0.679 994 815 610 88 × 2 = 1 + 0.359 989 631 221 76;
  • 51) 0.359 989 631 221 76 × 2 = 0 + 0.719 979 262 443 52;
  • 52) 0.719 979 262 443 52 × 2 = 1 + 0.439 958 524 887 04;
  • 53) 0.439 958 524 887 04 × 2 = 0 + 0.879 917 049 774 08;
  • 54) 0.879 917 049 774 08 × 2 = 1 + 0.759 834 099 548 16;
  • 55) 0.759 834 099 548 16 × 2 = 1 + 0.519 668 199 096 32;
  • 56) 0.519 668 199 096 32 × 2 = 1 + 0.039 336 398 192 64;
  • 57) 0.039 336 398 192 64 × 2 = 0 + 0.078 672 796 385 28;
  • 58) 0.078 672 796 385 28 × 2 = 0 + 0.157 345 592 770 56;
  • 59) 0.157 345 592 770 56 × 2 = 0 + 0.314 691 185 541 12;
  • 60) 0.314 691 185 541 12 × 2 = 0 + 0.629 382 371 082 24;
  • 61) 0.629 382 371 082 24 × 2 = 1 + 0.258 764 742 164 48;
  • 62) 0.258 764 742 164 48 × 2 = 0 + 0.517 529 484 328 96;
  • 63) 0.517 529 484 328 96 × 2 = 1 + 0.035 058 968 657 92;
  • 64) 0.035 058 968 657 92 × 2 = 0 + 0.070 117 937 315 84;

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:


0.000 282 005 914 49(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0101 0101 0111 0000 1010(2)

6. Positive number before normalization:

0.000 282 005 914 49(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0101 0101 0111 0000 1010(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:


0.000 282 005 914 49(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0101 0101 0111 0000 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 0101 0101 0111 0000 1010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0001 0101 0101 0111 0000 1010(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 1110 0001 0101 0101 0111 0000 1010


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 1110 0001 0101 0101 0111 0000 1010 =

0010 0111 1011 0100 0110 0010 1110 0001 0101 0101 0111 0000 1010


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 1110 0001 0101 0101 0111 0000 1010


Decimal number -0.000 282 005 914 49 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0001 0101 0101 0111 0000 1010

Share

» Convert decimal -0.000 282 005 913 8 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 Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 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