-0.000 282 005 914 412 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 412(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
-0.000 282 005 914 412(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:

|-0.000 282 005 914 412| = 0.000 282 005 914 412


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: 0.000 282 005 914 412.

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) 0.000 282 005 914 412 × 2 = 0 + 0.000 564 011 828 824;
  • 2) 0.000 564 011 828 824 × 2 = 0 + 0.001 128 023 657 648;
  • 3) 0.001 128 023 657 648 × 2 = 0 + 0.002 256 047 315 296;
  • 4) 0.002 256 047 315 296 × 2 = 0 + 0.004 512 094 630 592;
  • 5) 0.004 512 094 630 592 × 2 = 0 + 0.009 024 189 261 184;
  • 6) 0.009 024 189 261 184 × 2 = 0 + 0.018 048 378 522 368;
  • 7) 0.018 048 378 522 368 × 2 = 0 + 0.036 096 757 044 736;
  • 8) 0.036 096 757 044 736 × 2 = 0 + 0.072 193 514 089 472;
  • 9) 0.072 193 514 089 472 × 2 = 0 + 0.144 387 028 178 944;
  • 10) 0.144 387 028 178 944 × 2 = 0 + 0.288 774 056 357 888;
  • 11) 0.288 774 056 357 888 × 2 = 0 + 0.577 548 112 715 776;
  • 12) 0.577 548 112 715 776 × 2 = 1 + 0.155 096 225 431 552;
  • 13) 0.155 096 225 431 552 × 2 = 0 + 0.310 192 450 863 104;
  • 14) 0.310 192 450 863 104 × 2 = 0 + 0.620 384 901 726 208;
  • 15) 0.620 384 901 726 208 × 2 = 1 + 0.240 769 803 452 416;
  • 16) 0.240 769 803 452 416 × 2 = 0 + 0.481 539 606 904 832;
  • 17) 0.481 539 606 904 832 × 2 = 0 + 0.963 079 213 809 664;
  • 18) 0.963 079 213 809 664 × 2 = 1 + 0.926 158 427 619 328;
  • 19) 0.926 158 427 619 328 × 2 = 1 + 0.852 316 855 238 656;
  • 20) 0.852 316 855 238 656 × 2 = 1 + 0.704 633 710 477 312;
  • 21) 0.704 633 710 477 312 × 2 = 1 + 0.409 267 420 954 624;
  • 22) 0.409 267 420 954 624 × 2 = 0 + 0.818 534 841 909 248;
  • 23) 0.818 534 841 909 248 × 2 = 1 + 0.637 069 683 818 496;
  • 24) 0.637 069 683 818 496 × 2 = 1 + 0.274 139 367 636 992;
  • 25) 0.274 139 367 636 992 × 2 = 0 + 0.548 278 735 273 984;
  • 26) 0.548 278 735 273 984 × 2 = 1 + 0.096 557 470 547 968;
  • 27) 0.096 557 470 547 968 × 2 = 0 + 0.193 114 941 095 936;
  • 28) 0.193 114 941 095 936 × 2 = 0 + 0.386 229 882 191 872;
  • 29) 0.386 229 882 191 872 × 2 = 0 + 0.772 459 764 383 744;
  • 30) 0.772 459 764 383 744 × 2 = 1 + 0.544 919 528 767 488;
  • 31) 0.544 919 528 767 488 × 2 = 1 + 0.089 839 057 534 976;
  • 32) 0.089 839 057 534 976 × 2 = 0 + 0.179 678 115 069 952;
  • 33) 0.179 678 115 069 952 × 2 = 0 + 0.359 356 230 139 904;
  • 34) 0.359 356 230 139 904 × 2 = 0 + 0.718 712 460 279 808;
  • 35) 0.718 712 460 279 808 × 2 = 1 + 0.437 424 920 559 616;
  • 36) 0.437 424 920 559 616 × 2 = 0 + 0.874 849 841 119 232;
  • 37) 0.874 849 841 119 232 × 2 = 1 + 0.749 699 682 238 464;
  • 38) 0.749 699 682 238 464 × 2 = 1 + 0.499 399 364 476 928;
  • 39) 0.499 399 364 476 928 × 2 = 0 + 0.998 798 728 953 856;
  • 40) 0.998 798 728 953 856 × 2 = 1 + 0.997 597 457 907 712;
  • 41) 0.997 597 457 907 712 × 2 = 1 + 0.995 194 915 815 424;
  • 42) 0.995 194 915 815 424 × 2 = 1 + 0.990 389 831 630 848;
  • 43) 0.990 389 831 630 848 × 2 = 1 + 0.980 779 663 261 696;
  • 44) 0.980 779 663 261 696 × 2 = 1 + 0.961 559 326 523 392;
  • 45) 0.961 559 326 523 392 × 2 = 1 + 0.923 118 653 046 784;
  • 46) 0.923 118 653 046 784 × 2 = 1 + 0.846 237 306 093 568;
  • 47) 0.846 237 306 093 568 × 2 = 1 + 0.692 474 612 187 136;
  • 48) 0.692 474 612 187 136 × 2 = 1 + 0.384 949 224 374 272;
  • 49) 0.384 949 224 374 272 × 2 = 0 + 0.769 898 448 748 544;
  • 50) 0.769 898 448 748 544 × 2 = 1 + 0.539 796 897 497 088;
  • 51) 0.539 796 897 497 088 × 2 = 1 + 0.079 593 794 994 176;
  • 52) 0.079 593 794 994 176 × 2 = 0 + 0.159 187 589 988 352;
  • 53) 0.159 187 589 988 352 × 2 = 0 + 0.318 375 179 976 704;
  • 54) 0.318 375 179 976 704 × 2 = 0 + 0.636 750 359 953 408;
  • 55) 0.636 750 359 953 408 × 2 = 1 + 0.273 500 719 906 816;
  • 56) 0.273 500 719 906 816 × 2 = 0 + 0.547 001 439 813 632;
  • 57) 0.547 001 439 813 632 × 2 = 1 + 0.094 002 879 627 264;
  • 58) 0.094 002 879 627 264 × 2 = 0 + 0.188 005 759 254 528;
  • 59) 0.188 005 759 254 528 × 2 = 0 + 0.376 011 518 509 056;
  • 60) 0.376 011 518 509 056 × 2 = 0 + 0.752 023 037 018 112;
  • 61) 0.752 023 037 018 112 × 2 = 1 + 0.504 046 074 036 224;
  • 62) 0.504 046 074 036 224 × 2 = 1 + 0.008 092 148 072 448;
  • 63) 0.008 092 148 072 448 × 2 = 0 + 0.016 184 296 144 896;
  • 64) 0.016 184 296 144 896 × 2 = 0 + 0.032 368 592 289 792;

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:


0.000 282 005 914 412(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1111 0110 0010 1000 1100(2)

6. Positive number before normalization:

0.000 282 005 914 412(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1111 0110 0010 1000 1100(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:


0.000 282 005 914 412(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1111 0110 0010 1000 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1111 0110 0010 1000 1100(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1111 1111 0110 0010 1000 1100(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 1101 1111 1111 0110 0010 1000 1100


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 1101 1111 1111 0110 0010 1000 1100 =

0010 0111 1011 0100 0110 0010 1101 1111 1111 0110 0010 1000 1100


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 1101 1111 1111 0110 0010 1000 1100


Decimal number -0.000 282 005 914 412 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

Share

» Convert decimal -0.000 282 005 914 476 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